Appendix B Hints, Answers, and Solutions to Exercises
1 Solving Systems of Linear Equations
1.1 Systems of Linear Equations
Exercises
1.1.5.
1.1.6.
Writing Exercises
1.1.7.
Answer.
-
This system has no solution because the lines are parallel:\begin{align*} -2x + y \amp =0\\ -2x + y \amp =1\text{.} \end{align*}
-
This system has one solution because the lines only intersect at the origin:\begin{align*} -2x + y \amp =0\\ 2x+y \amp =0\text{.} \end{align*}
-
This system has infinitely many solutions because the equations describe the same line:\begin{align*} -2x+y \amp =0\\ 2x-y \amp =0\text{.} \end{align*}
1.1.8.
1.2 Matrices
Exercises
1.2.4.
1.2.5.
1.2.6.
1.2.7.
1.3 Solving Linear Systems With Matrices
Exercises
1.3.1.
Answer.
-
The only solution is \((-4,5,0)\text{.}\)
-
The solutions can be written parametrically as\begin{equation*} \begin{cases} x_1 = 2x_2 + 2 \\ x_2 \text{ is free} \\ x_3 = -3. \end{cases} \end{equation*}
-
There are no solutions.
-
The solutions can be written parametrically as\begin{equation*} \begin{cases} x_1 = -2x_3 + 4x_4 + 7 \\ x_2 = -9x_3 + x_4 - 4 \\ x_3 \text{ is free} \\ x_4 \text{ is free} \\ x_5 = 4. \end{cases} \end{equation*}
1.3.5.
1.3.8.
1.3.9.
Writing Exercises
1.3.10.
1.3.13.
1.4 Vectors
1.4.4 Exercises
1.4.4.1.
1.4.4.2.
1.4.4.3.
Answer.
There are an infinite number of correct answers! Three vectors that are in the span are \(\bfu = 1\bfu + 0\bfv\text{,}\) \(\bfv = 0\bfu + 1\bfv\text{,}\) and \(\bfo = 0\bfu + 0\bfv\text{.}\) Here are two additional vectors in the span:
\begin{equation*}
\begin{bmatrix} 2 \\ 3 \\ 2 \end{bmatrix} = \bfu + \bfv, \hspace{6pt} \text{and} \hspace{6pt}
\begin{bmatrix} -4 \\ 1 \\ 6 \end{bmatrix} = \bfu - \bfv\text{.}
\end{equation*}
1.4.4.4.
Answer.
If \(\bfx = \begin{bmatrix} -1 \\ 1 \end{bmatrix}\text{,}\) \(\bfu = \begin{bmatrix} 2 \\ 3 \end{bmatrix}\text{,}\) \(\bfv = \begin{bmatrix} -1 \\ 0 \end{bmatrix}\text{,}\) and \(\bfw = \begin{bmatrix} 4 \\ 1 \end{bmatrix}\text{,}\) then \(\bfx\) can be written as
\begin{equation*}
\bfx = a \bfu + b \bfv + c \bfw\text{,}
\end{equation*}
where \(a = \frac{1}{3} - \frac{1}{3}c\text{,}\) \(b = \frac{5}{3} + \frac{10}{3}c\text{,}\) and \(c\) can be any real number.
1.4.4.7.
Writing Exercises
1.4.4.9.
1.4.4.12.
Solution.
We must prove that for all real numbers \(c\) and all \(\bfu\) and \(\bfv\) in \(\rr^n\) we have
\begin{equation*}
c(\bfu + \bfv) = c\bfu + c\bfv\text{.}
\end{equation*}
Let \(\bfu\) and \(\bfv\) be denoted in the following way:
\begin{equation*}
\bfu = \begin{bmatrix} u_1 \\ \vdots \\ u_n \end{bmatrix} \hspace{6pt} \text{and} \hspace{6pt}
\bfv = \begin{bmatrix} v_1 \\ \vdots \\ v_n \end{bmatrix}\text{.}
\end{equation*}
Then we have
\begin{equation*}
c(\bfu + \bfv) = c \begin{bmatrix} u_1 + v_1 \\ \vdots \\ u_n + v_n \end{bmatrix} =
\begin{bmatrix} c(u_1 + v_1) \\ \vdots \\ c(u_n + v_n) \end{bmatrix}\text{.}
\end{equation*}
Now, since the distributive law holds for real numbers, we have \(c(u_i + v_i) = cu_i + cv_i\) for all \(i = 1, \ldots, n\text{.}\) This means that
\begin{align*}
c(\bfu + \bfv) \amp = \begin{bmatrix} cu_1 + cv_1 \\ \vdots \\ cu_n + cv_n \end{bmatrix} =
\begin{bmatrix} cu_1 \\ \vdots \\ cu_n \end{bmatrix} + \begin{bmatrix} cv_1 \\ \vdots \\ cv_n \end{bmatrix}\\
\amp = c\begin{bmatrix} u_1 \\ \vdots \\ u_n \end{bmatrix} + c \begin{bmatrix} v_1 \\ \vdots \\ v_n \end{bmatrix} = c\bfu + c\bfv \text{.}
\end{align*}
This completes the proof.
2 Fields and Vector Spaces
2.1 Fields
Exercises
Writing Exercises
2.1.6.
Answer.
-
No, \(\zz[x]\) is not a field. All of the integers are contained in \(\zz[x]\text{,}\) and since (for example) \(2 \in \zz\) has no multiplicative inverse in \(\zz\text{,}\) it will not have a multiplicative inverse in \(\zz[x]\text{.}\)
-
No, \(\rr[x]\) is not a field. The element \(x \in \rr[x]\) has no multiplicative inverse. We can argue this by contradiction. If \(b = a_0 + a_1x + a_2x^2 + \cdots\) was the multiplicative inverse of \(x\text{,}\) then\begin{equation*} 1 = xb = a_0x + a_1x^2 + a_2x^3 + \cdots\text{.} \end{equation*}But since the constant term on the right side of this equation is \(0\) and the constant term on the left side is \(1\text{,}\) and since those are not equal, we have a contradiction.
2.1.8.
Answer.
-
Yes, this is a field. The most difficult axiom to check is the one about multiplicative inverses. It turns out that the multiplicative inverse of \(a + b\alpha\) is \(c + d\alpha\) where\begin{equation*} c = \frac{1}{a^2-2b^2}, \hspace{6pt} \text{and} \hspace{6pt} d = -\frac{b}{a^2-2b^2}\text{.} \end{equation*}One needs to check that for all non-zero elements \(a + b\alpha\) of \(\ff_5[\alpha]\) we have \(a^2-2b^2 \neq 0\text{.}\) But this is relatively easy to verify.
2.1.10.
2.1.12.
Solution.
Let \(b \in \ff\) be a non-zero element with multiplicative inverse \(b'\text{.}\) Suppose that the element \(a \in \ff\) also has the properties of a multiplicative inverse of \(b\text{.}\) This means that \(ab=1\text{.}\) If we multiply both sides of this equation by \(b'\text{,}\) we have
\begin{align*}
(ab)b' \amp =b'\\
a(bb') \amp =b'\\
a(1) \amp =b'\\
a \amp =b'\text{.}
\end{align*}
This proves that \(a=b'\text{,}\) so multiplicative inverses are unique.
2.2 Solving Linear Systems Over Fields
Exercises
2.2.1.
2.2.2.
2.2.5.
2.2.6.
Writing Exercises
2.2.9.
2.2.10.
Answer.
-
If all of the coefficients and constants begin as integers, all of the elementary row operations will produce coefficients and constants within \(\qq\text{.}\) None of the elementary row operations can produce an irrational number when beginning with integers.
-
If there is a unique solution, there are no free variables, so whether one considers the solution in \(\qq^n\) or \(\rr^n\text{,}\) there cannot be another solution. Thus, the numbers in the solution set must be rational numbers.
2.3 Vector Spaces
Exercises
2.3.1.
Answer.
-
No, this is not a vector space. For example, it is not closed under scalar multiplication. We can see that \(-2(1,1) = (-2,-2) \not \in V\text{.}\)
-
No, this is not a vector space. For example, it is not closed under addition. We can see that \((1,4) + (-2,-1)= (-1,3) \not \in V\text{.}\)
-
Yes, this is a vector space. All of the vector space axioms hold.
2.3.3.
2.3.4.
2.3.6.
2.3.9.
Answer.
No, \(\bfw\) is not in \(\spn\{\bfu,\bfv\}\text{.}\) When we form the matrix with \(\bfu\text{,}\) \(\bfv\text{,}\) and \(\bfw\) as its columns, it reduces to the following:
\begin{equation*}
\begin{bmatrix} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 1 \end{bmatrix}\text{.}
\end{equation*}
Since there is a pivot in the final column, the corresponding vector equation \(x_1\bfu + x_2\bfv = \bfw\) has no solutions.
Writing Exercises
2.3.12.
2.3.15.
Solution.
Suppose that \(\bfo\) is a zero vector in \(V\) and that \(\mathbf{z}\) also has the properties of a zero vector. Since \(\bfo\) is a zero vector, we have \(\bfo + \mathbf{z} = \mathbf{z}\text{.}\) Since \(\mathbf{z}\) has the properties of a zero vector, we have \(\bfo + \mathbf{z} = \bfo\text{.}\) Therefore, we have
\begin{equation*}
\mathbf{z} = \bfo + \mathbf{z} = \bfo\text{,}
\end{equation*}
so \(\mathbf{z} = \bfo\text{.}\) This proves that the zero vector is unique.
2.4 Subspaces
Exercises
2.4.2.
2.4.4.
Answer.
-
Yes, \(W\) is a subspace as it can be written as the span of two vectors.
-
No, \(W\) is not a subspace. It is not closed under scalar multiplication, for example.
-
Yes, \(W\) is a subspace as it can be written as the span of four vectors.
-
No, \(W\) is not a subspace as it does not contain the zero vector.
Writing Exercises
2.4.6.
Answer.
Let \(V\) denote the set of functions described in the statement of the problem. We first observe that the zero function is in \(V\) since it integrates to zero over any interval \([a,b]\text{.}\) If \(f, g \in V\text{,}\) then
\begin{align*}
\int_a^b (f+g)(x)\;dx \amp = \int_a^b (f(x) + g(x))\;dx\\
\amp = \int_a^b f(x)\;dx + \int_a^b g(x)\;dx = 0 + 0 = 0\text{,}
\end{align*}
which proves that \(f + g \in V\text{.}\) Finally, if \(c \in \rr\) and \(f \in V\text{,}\) we have
\begin{equation*}
\int_a^b (cf)(x)\;dx = \int_a^b cf(x)\;dx = c\int_a^b f(x)\;dx = c \cdot 0 = 0\text{,}
\end{equation*}
which shows that \(cf \in V\text{.}\) (These last two calculations rely on the linear properties of the definite integral.) This completes the proof that \(V\) is a subspace.
2.4.9.
Answer.
No. Let \(W_1\) and \(W_2\) be the following two subspaces of \(\rr^2\text{:}\)
\begin{equation*}
W_1 = \{(x,2x) \mid x \in \rr\}, \hspace{12pt} W_2 = \{(x,3x) \mid x \in \rr\}\text{.}
\end{equation*}
Then \((1,2) \in W_1\) and \((1,3) \in W_2\text{,}\) so both of these elements are in \(W_1 \cup W_2\text{.}\) However, \((1,2) + (1,3) = (2,5)\text{,}\) and \((2,5) \not \in W_1 \cup W_2\text{.}\) This shows that \(W_1 \cup W_2\) is not closed under addition, so it is not a subspace.
3 Linear Transformations
3.1 Linear Transformations
3.1.7 Exercises
3.1.7.2.
3.1.7.4.
Answer.
No, \(T\) is not a linear transformation. Consider the following example:
\begin{align*}
T(2(1,1,0)) \amp = T(2,2,0) = (6+2-0,-4) \equiv (3,1)\\
2T(1,1,0) \amp = 2(3+1-0,-1) \equiv 2(4,4) \equiv (3,3)\text{.}
\end{align*}
This shows that the scalar multiplication property does not hold, as \(2T(1,1,0) \neq T(2(1,1,0))\text{.}\)
3.1.7.6.
3.1.7.8.
Answer.
-
No, since\begin{equation*} \left[\begin{array}{@{}cc|c@{}} 3 \amp -2 \amp 1 \\ 1 \amp 4 \amp 1 \\ -1 \amp 0 \amp 1 \end{array}\right] \sim \begin{bmatrix} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 1 \end{bmatrix}\text{.} \end{equation*}The pivot in the final column shows that there is no solution to the matrix-vector equation \(A\bfx = \bfv\text{,}\) where \(\bfv = (1,1,1)\text{.}\)
-
No. Since in part a we identified a vector which is not in the image of \(T\text{,}\) this means that \(T\) cannot be surjective.
Writing Exercises
3.1.7.11.
Answer.
Let \(V = C[0,\infty)\) and let \(f, g \in V\text{.}\) Then we have
\begin{align*}
T(f+g)(x) \amp = \int_0^x (f+g)(y)\;dy = \int_0^x (f(y) + g(y))\;dy\\
\amp = \int_0^x f(y)\;dy + \int_0^x g(y)\;dy = T(f)(x) + T(g)(x)\text{.}
\end{align*}
We now let \(c \in \rr\) and \(f \in V\text{.}\) Then
\begin{align*}
T(cf)(x) \amp = \int_0^x (cf)(y)\;dy = \int_0^x cf(y)\;dy\\
\amp = c\int_0^x f(y)\;dy = cT(f)(x)\text{.}
\end{align*}
Both of these calculations rely on the linear properties of the definite integral.
3.1.7.12.
Solution.
Let \(\bfu\) and \(\bfv\) be vectors in \(U\text{.}\) Then, since \(S\) and \(T\) are both linear transformations, we have
\begin{align*}
(S\circ T)(\bfu + \bfv) \amp = S(T(\bfu + \bfv)) = S(T(\bfu) + T(\bfv))\\
\amp = S(T(\bfu)) + S(T(\bfv)) = (S\circ T)(\bfu) + (S\circ T)(\bfv)\text{.}
\end{align*}
This proves that \(S\circ T\) has the first property of a linear transformation.
We now let \(c \in \ff\) and \(\bfu \in U\text{.}\) Then, since \(S\) and \(T\) are both linear transformations, we have
\begin{align*}
(S\circ T)(c\bfu) \amp = S(T(c\bfu)) = S(cT(\bfu))\\
\amp = cS(T(\bfu)) = c(S\circ T)(\bfu)\text{.}
\end{align*}
This shows that \(S\circ T\) is a linear transformation.
3.1.7.13.
Solution.
-
We assume that \(S\circ T\) is injective, and we let \(\bfu_1\) and \(\bfu_2\) be vectors in \(U\) with \(T(\bfu_1) = T(\bfu_2)\text{.}\) Then, since \(S \circ T\) is injective, we have\begin{align*} S(T(\bfu_1)) \amp = S(T(\bfu_2))\\ \bfu_1 \amp = \bfu_2\text{.} \end{align*}This proves that \(T\) is injective.
-
We assume that \(S\circ T\) is surjective, and we let \(\bfw \in W\text{.}\) Since \(S\circ T\) is surjective, there is a vector \(\bfu \in U\) such that \((S\circ T)(\bfu) = \bfw\text{.}\) But this means that \(S(T(\bfu)) = \bfw\text{;}\) in other words, \(S\) sends \(T(\bfu)\) to \(\bfw\text{.}\) This proves that \(S\) is surjective, since we have found a vector \(\bfv=T(\bfu) \in V\) such that \(S(\bfv) = \bfw\text{.}\)
3.2 The Matrix of a Linear Transformation
3.2.5 Exercises
3.2.5.3.
3.2.5.5.
3.3 Inverting a Matrix
3.3.5 Exercises
3.3.5.2.
3.3.5.4.
Answer.
-
The matrix \(A\) is invertible and the inverse of \(A\) is\begin{equation*} \begin{bmatrix} 0 \amp 2 \amp 1 \\ 2 \amp 4 \amp 2 \\ 4 \amp 0 \amp 2 \end{bmatrix}\text{.} \end{equation*}
-
The matrix \(A\) is not invertible since its RREF is\begin{equation*} \begin{bmatrix} 1 \amp 0 \amp 1 \\ 0 \amp 1 \amp 3 \\ 0 \amp 0 \amp 0 \end{bmatrix}\text{.} \end{equation*}
3.3.5.6.
Answer.
If \(A = \begin{bmatrix} 2 \amp 4 \\ 1 \amp 3 \end{bmatrix}\text{,}\) then the inverse over \(\ff_5\) is \(A^{-1} = \begin{bmatrix} 4 \amp 3 \\ 2 \amp 1 \end{bmatrix}\text{.}\) Then we have
\begin{equation*}
\begin{bmatrix} 4 \amp 3 \\ 2 \amp 1 \end{bmatrix} \begin{bmatrix} 1 \\ 4 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}\text{.}
\end{equation*}
Writing Exercises
3.3.5.12.
Solution.
Let \(A \in M_n(\ff)\) be invertible. This means that \(A^{-1}\) exists and that \(AA^{-1} = A^{-1}A = I_n\text{.}\) These equalities involving \(A\) and \(A^{-1}\) show that \(A^{-1}\) is invertible and that \(A\) is a matrix which has the properties of the inverse of \(A^{-1}\text{.}\) Since the inverse of a matrix is unique, it must be that \((A^{-1})^{-1} = A\text{.}\)
3.3.5.14.
Answer.
-
We suppose that \((B-C)D = 0\) where \(D\) is invertible. Then we have\begin{align*} (B-C)D \amp = 0 \\ (B-C)D D^{-1} \amp = 0 D^{-1}\\ B-C \amp = 0\\ B \amp = C\text{.} \end{align*}
-
Suppose that \(B = \begin{bmatrix} 1 \amp 3 \\ 1 \amp 2 \end{bmatrix}\text{,}\) \(C = \begin{bmatrix} 1 \amp 1 \\ 1 \amp 0 \end{bmatrix}\text{,}\) and \(D = \begin{bmatrix} 1 \amp 1 \\ 0 \amp 0 \end{bmatrix}\text{.}\) Then, it is true that \(D\) is not invertible. We obviously have \(B \neq C\text{.}\) But we do have \(B - C = \begin{bmatrix} 0 \amp 2 \\ 0 \amp 2 \end{bmatrix}\) and then\begin{equation*} (B-C)D = \begin{bmatrix} 0 \amp 2 \\ 0 \amp 2 \end{bmatrix} \begin{bmatrix} 1 \amp 1 \\ 0 \amp 0 \end{bmatrix} = \begin{bmatrix} 0 \amp 0 \\ 0 \amp 0 \end{bmatrix}\text{.} \end{equation*}
3.3.5.16.
Answer.
Suppose that \(A\) is invertible and upper triangular. Then we can use elementary row operations to reduce \(\left[\begin{array}{@{}c|c@{}} A \amp I \end{array}\right]\) to \(\left[\begin{array}{@{}c|c@{}} I \amp A^{-1} \end{array}\right]\text{.}\) But the elementary row operations that are needed for this reduction will be only the scaling operation and the replacement operation, where we add a multiple of one row to another higher up in the matrix. (Specifically, we will only need to add a multiple of a row \(i\) to a row \(j\) where \(i > j\text{.}\)) The elementary matrices that correspond to these specific elementary row operations are all upper triangular. And since the product of upper triangular matrices is upper triangular (this should probably be proved), this shows that \(A^{-1}\) will be upper triangular.
3.4 Subspaces and Linear Transformations
3.4.4 Exercises
3.4.4.1.
3.4.4.4.
Answer.
-
It is easy to calculate \(A\bfx\) and see that \(A\bfx = \bfo\text{.}\) To show that \(\bfx\) is not in \(\col(A)\text{,}\) we form the matrix \(\left[\begin{array}{@{}c|c@{}} A \amp \bfx \end{array}\right]\) and row reduce:\begin{equation*} \left[\begin{array}{@{}cc|c@{}} 3 \amp -1 \amp 2 \\ -9 \amp 3 \amp 6 \end{array}\right] \sim \left[\begin{array}{@{}cc|c@{}} 1 \amp -1/3 \amp 0 \\ 0 \amp 0 \amp 1 \end{array}\right]\text{.} \end{equation*}The pivot in the last column shows that \(\bfx \not \in \col(A)\text{.}\)
-
Since\begin{equation*} A \sim \begin{bmatrix} 1 \amp -1/3 \\ 0 \amp 0 \end{bmatrix}\text{,} \end{equation*}we can see that any vector \(\bfv\) in the null space of \(A\) must be a multiple of \(\bfx\text{.}\) And since \(\bfx\) is not in \(\col(A)\text{,}\) no non-zero multiple of \(\bfx\) can be in \(\col(A)\text{.}\) This shows that the only vector in \(\nll(A) \cap \col(A)\) is the zero vector.
-
Yes, this is possible. Consider the matrix \(A = \begin{bmatrix} 1 \amp -1 \\ 1 \amp -1 \end{bmatrix}\text{.}\) The vector \(\bfv = \begin{bmatrix} 1 \\ 1 \end{bmatrix}\) is in the null space of \(A\text{.}\) (This can easily be checked.) Also, since \(\bfv\) is a column of \(A\text{,}\) it is in \(\col(A)\text{.}\)
3.4.4.7.
Solution.
-
This linear transformation is injective. Observe that if \(T\) sends the vector \((a,b,c)\) to the zero polynomial, then we have the following linear system in which each equation must hold:\begin{align*} a + b \amp =0\\ b + c \amp = 0\\ a + b + c \amp =0\text{.} \end{align*}We can solve this using methods from earlier in this book. We find that\begin{equation*} \begin{bmatrix} 1 \amp 1 \amp 0 \\ 0 \amp 1 \amp 1 \\ 1 \amp 1 \amp 1 \end{bmatrix} \sim \begin{bmatrix} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 1 \end{bmatrix}\text{.} \end{equation*}This shows that the only solution to this linear system is the trivial one—that is, where \(a = b = c = 0\text{.}\) But this means that the only vector that \(T\) sends to the zero polynomial is the zero vector.
-
We claim that \(T\) is surjective. Let \(p = d + et + ft^2\) be an arbitrary element of \(P_2\text{.}\) In order to demonstrate that there is an element \(\bfx \in \rr^3\) such that \(T(\bfx) = p\text{,}\) we need to solve the following linear system:\begin{align*} a + b \amp =d\\ b + c \amp = e\\ a + b + c \amp =f\text{.} \end{align*}However, the RREF of the coefficient matrix for this system looks like this:\begin{equation*} \begin{bmatrix} 1 \amp 1 \amp 0 \\ 0 \amp 1 \amp 1 \\ 1 \amp 1 \amp 1 \end{bmatrix} \sim \begin{bmatrix} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 1 \end{bmatrix}\text{.} \end{equation*}The fact that there is a pivot in each row of this RREF means that a combination of \(a\text{,}\) \(b\text{,}\) and \(c\) can be found such that \(T(\bfx) = p\text{.}\) This proves that \(T\) is surjective.
3.4.4.9.
Solution.
-
This linear transformation is not injective. We can calculate that \(T(-\frac{2}{3} - \frac{1}{3}t + t^2) = \bfo\text{.}\) Since this shows that \(T\) has a nonzero vector in the kernel, \(T\) is not injective.
-
This linear transformation is surjective. Let \(\bfx = \begin{bmatrix} d \\ e \end{bmatrix}\) be an element of \(\rr^2\text{.}\) We claim that we can find an element \(p \in P_2\) such that \(T(p) = \bfx\text{.}\) If \(p = a + bt + ct^2\text{,}\) we will need to solve the following system of equations to find the coefficients of \(p\text{:}\)\begin{align*} a - 2b \amp = d\\ 3b + c \amp = e\text{.} \end{align*}But the coefficient matrix of this system has the following RREF:\begin{equation*} \begin{bmatrix} 1 \amp -2 \amp 0 \\ 0 \amp 3 \amp 1 \end{bmatrix} \sim \begin{bmatrix} 1 \amp 0 \amp 2/3 \\ 0 \amp 1 \amp 1/3 \end{bmatrix}\text{.} \end{equation*}The important thing to notice here is that there is a pivot in each row of this RREF. This means that the linear system always has a solution, so we can always find values of \(a\text{,}\) \(b\text{,}\) and \(c\) such that \(T(p) = \bfx\text{.}\) This proves that \(T\) is surjective.
Writing Exercises
3.4.4.11.
Solution.
Since \(U'\) is a subspace of \(U\text{,}\) we must have \(\bfo \in U'\text{.}\) Further, we know that \(T(\bfo) = \bfo\) for all linear transformations. This proves that \(\bfo \in T(U')\text{.}\)
We now let \(\bfv_1, \bfv_2 \in T(U')\text{.}\) This means that there exist \(\bfu_1, \bfu_2 \in U'\) with \(T(\bfu_1) = \bfv_1\) and \(T(\bfu_2) = \bfv_2\text{.}\) Since \(U'\) is a subspace of \(U\text{,}\) we know that \(\bfu_1 + \bfu_2 \in U'\text{,}\) so \(T(\bfu_1 + \bfu_2) \in T(U')\text{.}\) However, because of the properties of a linear transformation, we have
\begin{equation*}
T(\bfu_1 + \bfu_2) = T(\bfu_1) + T(\bfu_2) = \bfv_1 + \bfv_2\text{.}
\end{equation*}
This proves that \(\bfv_1 + \bfv_2 \in T(U')\text{,}\) which shows that \(T(U')\) is closed under addition.
Finally, we let \(\bfv \in T(U')\) and \(c \in \ff\text{.}\) This means that there exists \(\bfu \in U'\) such that \(T(\bfu) = \bfv\text{.}\) Since \(U'\) is a subspace of \(U\text{,}\) we know that \(c\bfu \in U'\text{,}\) so \(T(c\bfu) \in T(U')\text{.}\) Using the properties of a linear transformation, we see that
\begin{equation*}
T(c\bfu) = cT(\bfu) = c\bfv\text{.}
\end{equation*}
This proves that \(c\bfv \in T(U')\text{,}\) showing that \(T(U')\) is closed under scalar multiplication. This completes the proof that \(T(U')\) is a subspace of \(V\text{.}\)
4 Determinants
4.1 Defining the Determinant
4.1.4 Exercises
4.1.4.2.
4.1.4.4.
Answer.
-
The elementary matrix is\begin{equation*} E = \begin{bmatrix} 1 \amp 0 \amp 0 \\ 0 \amp -3 \amp 0 \\ 0 \amp 0 \amp 1 \end{bmatrix}\text{,} \end{equation*}and \(\det(E) = -3\text{.}\)
-
The elementary matrix is\begin{equation*} E = \begin{bmatrix} 1 \amp 0 \amp 0 \\ 0 \amp 0 \amp 1 \\ 0 \amp 1 \amp 0 \end{bmatrix}\text{,} \end{equation*}and \(\det(E) = -1\text{.}\)
-
The elementary matrix is\begin{equation*} E = \begin{bmatrix} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ 7 \amp 0 \amp 1 \end{bmatrix}\text{,} \end{equation*}and \(\det(E) = 1\text{.}\)
4.1.4.6.
4.1.4.8.
4.1.4.9.
Writing Exercises
4.1.4.12.
Answer.
We let \(A = \begin{bmatrix} a \amp b \\ c \amp d \end{bmatrix}\text{,}\) and we consider the case of adding \(k\) times row 1 to row 2. We will call the resulting matrix \(B\text{,}\) and it has this appearance:
\begin{equation*}
B = \begin{bmatrix} a \amp b \\ ka + c \amp kb + d \end{bmatrix}\text{.}
\end{equation*}
We know that \(\det(A) = ad-bc\text{,}\) and we will now calculate \(\det(B)\text{:}\)
\begin{align*}
\det(B) \amp = a(kb+d) - b(ka+c)\\
\amp = kab + ad - kab - bc\\
\amp = ad - bc = \det(A)\text{.}
\end{align*}
Technically we have only proved one of the two necessary cases to complete this proof.
4.2 Properties of the Determinant
4.2.5 Exercises
4.2.5.2.
4.2.5.4.
4.2.5.6.
Writing Exercises
4.2.5.8.
Solution.
If \(A\) has identical columns, then \(A^T\) has identical rows. We can use an elementary row operation to add \(-1\) times one of these rows to the other, producing a row of zeros in this matrix we will call \(B\text{.}\) Since we used the replace row operation to go from \(A^T\) to \(B\text{,}\) we have \(\det(A^T) = \det(B)\) by Theorem 4.2.5. Since \(B\) has a row of zeros, we know that \(\det(B) = 0\) by Proposition 4.2.1. This means that \(\det(A^T)=0\text{,}\) and since we have \(\det(A)=\det(A^T)\) by Theorem 4.1.9, this means that \(\det(A)=0\text{,}\) as desired.
4.2.5.11.
Solution.
If \(A, B \in M_n(\ff)\text{,}\) we have
\begin{equation*}
\det(AB) = \det(A)\det(B) = \det(B)\det(A) = \det(BA)\text{.}
\end{equation*}
This string of equations uses Theorem 4.2.13 twice as well as the fact that the determinant of a matrix is an element of \(\ff\text{,}\) and elements of \(\ff\) commute via multiplication.
5 The Dimension of a Vector Space
5.1 Linear Independence
Exercises
5.1.3.
5.1.4.
5.1.5.
Writing Exercises
5.1.9.
Solution.
Suppose that
\begin{equation*}
V_1 = \{\bfv_1,\ldots, \bfv_k\}
\end{equation*}
and
\begin{equation*}
V_2 = \{\bfv_1,\ldots,\bfv_k,\bfv_{k+1},\ldots,\bfv_n\}\text{.}
\end{equation*}
Since \(V_1\) is linearly dependent, we know that there exist scalars \(c_1, \ldots, c_k\text{,}\) not all of which are zero, such that
\begin{equation*}
c_1\bfv_1 + \cdots + c_k\bfv_k = \bfo\text{.}
\end{equation*}
Then it is easy to produce a linear dependence relation for the set \(V_2\text{:}\)
\begin{equation*}
c_1\bfv_1 + \cdots + c_k\bfv_k + 0\bfv_{k+1} + \cdots + 0\bfv_n = \bfo\text{.}
\end{equation*}
This proves that \(V_2\) is linearly dependent.
5.1.10.
Solution.
-
If \(A \in M_n(\ff)\text{,}\) then \(\nll(A) = \{\bfo\}\) if and only if the RREF of \(A\) is \(I_n\text{.}\) It is always true that the null space is \(\{\bfo \}\) if and only if and we have a pivot in each column of the RREF; when \(A\) is square, this means that the RREF must be \(I_n\text{.}\)We also know that \(\col(A) = \ff^n\) if and only if the RREF of \(A\) is \(I_n\text{.}\) It is always true that the column space is \(\ff^n\) if and only if and we have a pivot in each row of the RREF; when \(A\) is square, this means that the RREF must be \(I_n\text{.}\)Putting these two paragraphs together, we conclude that \(\nll(A) = \{\bfo\}\) if and only if \(\col(A) = \ff^n\text{.}\)
-
If \(T\) is a linear transformation \(\ff^n \to \ff^n\text{,}\) then by Theorem 3.2.2 there is a matrix \(B \in M_n(\ff)\) such that \(T\) is multiplication by \(B\text{.}\) Then the kernel of \(T\) is the null space of \(B\) and the range of \(T\) is the column space of \(B\text{.}\) So, \(T\) is injective if and only if \(\nll(B) = \{\bfo\}\text{,}\) and \(T\) is surjective if and only if \(\col(B) = \ff^n\text{.}\) Then the result from part a completes the argument.
5.2 Basis of a Vector Space
5.2.4 Exercises
5.2.4.2.
5.2.4.4.
Answer.
Any matrix that has two pivot columns and two non-pivot columns will work as an answer here. This is one of many such matrices:
\begin{equation*}
\begin{bmatrix} 1 \amp 2 \amp 0 \amp 3 \\ 0 \amp 0 \amp 1 \amp 3 \\ 0 \amp 0 \amp 0 \amp 0 \end{bmatrix}\text{.}
\end{equation*}
Of course, any matrix row equivalent to a matrix like this one would also work as an answer.
5.2.4.6.
5.2.4.7.
Answer.
We will argue using the (contrapositive of the) Linear Dependence Lemma. Since \(\sin(t)\) is not the zero function, then \(\{\sin(t)\}\) is a linearly independent set. Also, since \(\sin(2t)\) is not a scalar multiple of \(\sin(t)\) (this can be verified by comparing graphs of the two functions), the set \(\{\sin(t), \sin(2t)\}\) is also linearly independent. However, there is a trig identity which says that \(\sin(2t) = 2\sin(t)\cos(t)\text{,}\) meaning that the set \(\{\sin(t), \sin(2t), \sin(t)\cos(t) \}\) is linearly dependent (as one vector is a multiple of another within this set). Therefore, one basis for \(H\) is \(\{\sin(t), \sin(2t)\}\text{.}\)
Writing Exercises
5.2.4.12.
Solution.
We let \(B = \{ \bfv_1 + \bfv_2, \bfv_2 + \bfv_3, \ldots, \bfv_{n-1} + \bfv_n, \bfv_n \}\text{.}\) We will first show that \(B\) is linearly independent.
We first suppose that \(c_1, \ldots, c_n\) are scalars such that
\begin{equation*}
c_1(\bfv_1 + \bfv_2) + \cdots + c_{n-1}(\bfv_{n-1} + \bfv_n) + c_n\bfv_n = \bfo\text{.}
\end{equation*}
Rearranging this equation, we see that it is equivalent to
\begin{equation*}
c_1\bfv_1 + (c_1+c_2)\bfv_2 + \cdots + (c_{n-1} + c_n)\bfv_n = \bfo\text{.}
\end{equation*}
However, since we were given that \(B' = \{\bfv_1, \ldots, \bfv_n \}\) is a basis for \(V\text{,}\) this means that the coefficients in this last equation must all be zero, since \(B'\) is linearly independent. This means that \(c_1 = 0\text{,}\) and then since we must also have \(c_1+c_2 = 0\text{,}\) we have \(c_2 = 0\text{,}\) and so on. The result is that \(c_i = 0\) for all \(i\text{,}\) \(i=1,\ldots,n\text{.}\) This proves that \(B\) is linearly independent.
We will now show that \(B\) spans \(V\text{.}\) Let \(\bfv \in V\text{.}\) We want to argue that \(\bfv\) can be written as a linear combination of the vectors in \(B\text{.}\) Since \(B'\) is a basis for \(V\text{,}\) there exist scalars \(d_1,\ldots, d_n\) such that
\begin{equation}
\bfv = d_1\bfv_1 + \cdots + d_n\bfv_n\text{.}\tag{5.3}
\end{equation}
We want to argue that we can always find scalars \(c_1, \ldots, c_n\) such that
\begin{equation*}
\bfv = c_1(\bfv_1 + \bfv_2) + \cdots + c_n\bfv_n\text{.}
\end{equation*}
This equation can be rewritten as
\begin{equation}
\bfv = c_1\bfv_1 + (c_1+c_2)\bfv_2 + \cdots + (c_{n-1}+c_n)\bfv_n\text{,}\tag{5.4}
\end{equation}
and by The Unique Representation Theorem (Theorem 5.2.11), we know that the coefficients on the right sides of (5.3) and (5.4) must be equal. Immediately we see that \(c_1 = d_1\) and then since we must have \(c_1 + c_2 = d_2\text{,}\) we conclude \(c_2 = d_2 - d_1\text{.}\) We can continue on in this way, eventually producing an expression for each \(c_i\) in terms of the \(d_i\) coefficients.
This proves that \(B\) spans \(V\) which concludes the proof that \(B\) is a basis of \(V\text{.}\)
5.3 Dimension
5.3.4 Exercises
5.3.4.1.
5.3.4.4.
5.3.4.6.
Answer.
Since \(\dim(P_3)=4\) and this is a set of four polynomials, we only need to argue that this set is linearly independent (see Theorem 5.3.17). If we label the polynomials as \(p_1 = 1\text{,}\) \(p_2 = 2t\text{,}\) \(p_3=-2+4t^2\text{,}\) and \(p_4 = -12t+8t^3\text{,}\) then we can argue that the set \(\{p_1,p_2,p_3,p_4\}\) is linearly independent by the contrapositive of the Linear Dependence Lemma. The set containing only \(p_1\) is linearly independent since \(p_1 \neq 0\text{.}\) Then the set \(\{p_1,p_2\}\) is linearly independent since neither polynomial is a scalar multiple of the other. Then since \(p_3\) cannot be a linear combination of \(p_1\) and \(p_2\) for degree reasons, \(\{p_1,p_2,p_3\}\) is linearly independent. Similarly, since \(p_4\) cannot be a linear combination of \(p_1\text{,}\) \(p_2\text{,}\) and \(p_3\) for degree reasons, \(\{p_1,p_2,p_3,p_4\}\) is linearly independent.
Writing Exercises
5.3.4.8.
Answer.
-
The null space of a matrix has a basis vector for each column in the RREF which does not contain a pivot. Therefore, the dimension of the null space is the number of non-pivot columns.
-
By Algorithm 5.2.14, we see that there is a vector in the basis for \(\col(A)\) for each pivot in the RREF of \(A\text{.}\) Therefore, \(\dim(\col(A))\) is the number of pivot columns of \(A\text{.}\)
5.3.4.9.
Answer.
Since \(V\) contains the vector space of all polynomials, and since the vector space of all polynomials is infinite-dimensional (see Example 5.3.3), then \(V\) must be infinite-dimensional.
5.4 Rank and Nullity
5.4.4 Exercises
5.4.4.4.
Answer.
This is not possible. A \(9\times 10\) linear system has a \(9\times 10\) coefficient matrix which we will call \(A\text{.}\) This situation can be reinterpreted as a linear transformation from \(\ff^{10} \to \ff^9\) which is multiplication by \(A\text{.}\) The information given in the problem tells us that this transformation is surjective, meaning that the rank of \(A\) is as large as possible, which is 9 in this case. Therefore, the Rank-Nullity Theorem says that \(\dim(\nll(A)) = 1\text{.}\) This means that every nonzero vector in the null space of \(A\) is a scalar multiple of a single basis vector. Therefore, the associated homogeneous linear system only has nonzero solutions which are multiples of each other.
Writing Exercises
5.4.4.7.
Answer.
-
If \(T\) is surjective, then \(\img(T) = W\text{,}\) which means that \(\dim(\img(T)) = \dim(W)\text{.}\) This implies that \(\rank(T) = \dim(W)\text{.}\) On the other hand, if \(\rank(T) = \dim(W)\text{,}\) then \(\dim(\img(T)) = \dim(W)\text{.}\) Since \(\img(T)\) is a subspace of \(W\text{,}\) Theorem 5.3.18 implies that \(\img(T) = W\text{,}\) which means that \(T\) is surjective.
-
We will be able to argue both implications at once. A linear transformation \(T\) is injective if and only if \(\ker(T) = \{\bfo\}\text{.}\) This occurs if and only if \(\dim(\ker(T)) = 0\text{.}\) The Rank-Nullity Theorem then says that \(\dim(\ker(T)) = 0\) if and only if \(\rank(T) = \dim(V)\text{.}\) This proves the result.
5.4.4.8.
Answer.
The given information means there is no pivot in the last column of the augmented matrix but that there is at least one non-pivot column among the first six columns. The coefficient matrix \(A\) then has at least one non-pivot column, meaning that \(\rank(A) \le 5\text{.}\) But this means that the linear transformation \(T:\ff^6 \to \ff^6\) which is multiplication by \(A\) cannot be surjective, since the rank of \(A\) is less than six. Therefore, we can find a vector \(\mathbf{c} \in \ff^6\) such that \(A\bfx = \mathbf{c}\) is inconsistent.
5.4.4.10.
Answer.
We know that \(\rank(A)\) is the number of pivots in the RREF of \(A\text{,}\) but the number of pivots must be less than both the number of columns (\(n\)) and the number of rows (\(m\)) of \(A\text{.}\) Then if \(\rank(A) \le m\) and \(\rank(A) \le n\text{,}\) we have \(\rank(A) \le \min \{m,n \}\text{.}\)
5.5 Coordinates
5.5.4 Exercises
5.5.4.1.
5.5.4.3.
5.5.4.6.
5.5.4.7.
Writing Exercises
5.5.4.13.
Solution.
Let \(V\) be an \(n\)-dimensional vector space over \(\ff\text{,}\) and let
\begin{equation*}
\mcb = \{\bfv_1,\ldots,\bfv_n\}
\end{equation*}
be a basis for \(V\text{.}\) Let \(C_{\mcb}:V \to \ff^n\) be the coordinate mapping. We will prove that \(C_{\mcb}\) is injective by showing that it has a trivial kernel.
Suppose that \(\bfv \in \ker(C_{\mcb})\text{.}\) This means that \(C_{\mcb}(\bfv) = \bfo \in \ff^n\text{,}\) so \([\bfv]_{\mcb} = \bfo\text{.}\) Since this is the coordinate vector of \(\bfv\text{,}\) this tells us that
\begin{equation*}
\bfv = 0 \bfv_1 + \cdots + 0\bfv_n\text{.}
\end{equation*}
This proves that \(\bfv = \bfo \in V\text{,}\) and therefore \(\ker(C_{\mcb}) = \{\bfo\}\text{.}\) This proves that the coordinate mapping is injective.
5.6 Change of Basis
5.6.3 Exercises
5.6.3.1.
5.6.3.3.
Answer.
-
Here are the change-of-basis matrices:\begin{equation*} P_{\mcb,\mce} = \begin{bmatrix} 1 \amp 0 \\ 2 \amp 2 \end{bmatrix}, \hspace{6pt} \text{and} \hspace{6pt} P_{\mce,\mcb} = \begin{bmatrix} 1 \amp 0 \\ 2 \amp 2 \end{bmatrix}\text{.} \end{equation*}(Yes, they are the same!)
-
We find that\begin{equation*} [\bfv]_{\mcb} = P_{\mce,\mcb}[\bfv]_{\mce} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}\text{.} \end{equation*}
5.6.3.8.
Answer.
-
We will use the basis \(\mcb = \{\bfv_1,\bfv_2\}\text{,}\) where\begin{equation*} \bfv_1 = \begin{bmatrix} 4 \\ -1 \end{bmatrix}, \hspace{6pt} \text{and} \hspace{6pt} \bfv_2 = \begin{bmatrix} 1 \\ 4 \end{bmatrix}\text{.} \end{equation*}
-
\(\displaystyle [T]_{\mcb} = \begin{bmatrix} 1 \amp 0 \\ 0 \amp -1 \end{bmatrix}\)
-
We find that\begin{equation*} P_{\mcb,\mce} = \begin{bmatrix} 4 \amp 1 \\ -1 \amp 4 \end{bmatrix}, \hspace{6pt} \text{and} \hspace{6pt} P_{\mce,\mcb} = \begin{bmatrix} 4/17 \amp -1/17 \\ 1/17 \amp 4/17 \end{bmatrix}\text{.} \end{equation*}
-
\(\displaystyle [T]_{\mce} = \begin{bmatrix} 15/17 \amp -8/17 \\ -8/17 \amp -15/17 \end{bmatrix}\)
Writing Exercises
5.6.3.10.
Solution.
The \(i\)th column of \(P_{\mcb,\mcc}\) is \([\bfv_i]_{\mcc}\text{.}\) Or, stated differently, if \(C_{\mcc}:V \to \ff^n\) is the coordinate mapping, then the \(i\)th column of \(P_{\mcb,\mcc}\) is \(C_{\mcc}(\bfv_i)\text{.}\)
We know (by Theorem 5.5.3) that the coordinate mapping is an isomorphism. Since \(\mcb\) is a basis for \(V\text{,}\) \(\mcb\) is a spanning set for \(V\text{.}\) But then the set containing the columns of \(P_{\mcb,\mcc}\) is \(C_{\mcc}(\mcb)\text{,}\) and since \(\mcb\) spans \(V\) we know that \(C_{\mcc}(\mcb)\) will span \(\ff^n\text{.}\)
6 Eigenvalues and Eigenvectors
6.1 Eigenvalues and Eigenvectors
6.1.4 Exercises
6.1.4.3.
Answer.
-
The RREF of \(A + 4I\) has one non-pivot column, so \(\lambda = -4\) is an eigenvalue for \(A\text{.}\) A basis for \(\mathrm{eig}_{-4}(A)\) is\begin{equation*} \left\{ \begin{bmatrix} 5 \\ -2 \\ 6 \end{bmatrix} \right\}\text{.} \end{equation*}
-
The RREF of \(A - 2I\) has two non-pivot columns, so \(\lambda = 2\) is an eigenvalue for \(A\text{.}\) A basis for \(\mathrm{eig}_{2}(A)\) is\begin{equation*} \left\{ \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} \right\}\text{.} \end{equation*}
6.1.4.5.
Answer.
-
The RREF of \(A - 4I\) has two non-pivot columns, so \(\lambda = 4\) is an eigenvalue for \(A\text{.}\) A basis for \(\mathrm{eig}_{4}(A)\) is\begin{equation*} \left\{ \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 3 \\ 0 \\ 1 \end{bmatrix} \right\}\text{.} \end{equation*}
-
The RREF of \(A - 2I\) has one non-pivot column, so \(\lambda = 2\) is an eigenvalue for \(A\text{.}\) A basis for \(\mathrm{eig}_{2}(A)\) is\begin{equation*} \left\{ \begin{bmatrix} 0 \\ 4 \\ 1 \end{bmatrix} \right\}\text{.} \end{equation*}
6.1.4.9.
Answer.
Since \(A\) is clearly not invertible—its columns are linearly dependent—it has 0 as an eigenvalue by Theorem 6.1.12. A basis for \(\mathrm{eig}_0(A)\) is
\begin{equation*}
\left\{ \begin{bmatrix} -1 \\ 1 \\ 0 \\ \vdots \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} -1 \\ 0 \\ 1 \\ 0 \\ \vdots \\ 0 \end{bmatrix}, \ldots, \begin{bmatrix} -1 \\ 0 \\ \vdots \\ 0 \\ 0 \\ 1 \end{bmatrix} \right\}\text{.}
\end{equation*}
The dimension of \(\mathrm{eig}_0(A)\) is \(n-1\text{.}\)
The other eigenvalue is \(\lambda = n\text{,}\) and a basis for \(\mathrm{eig}_n(A)\) is
\begin{equation*}
\left\{ \begin{bmatrix} 1 \\ \vdots \\ 1 \end{bmatrix} \right\}\text{.}
\end{equation*}
Writing Exercises
6.1.4.10.
Solution.
Suppose that \(\lambda\) is an eigenvalue of \(A\text{.}\) Then there exists a nonzero vector \(\bfv\) such that \(A\bfv = \lambda\bfv\text{.}\) But then
\begin{equation*}
A^2\bfv = A(\lambda\bfv) = \lambda \cdot A\bfv = \lambda^2\bfv\text{.}
\end{equation*}
If \(A^2 = 0\text{,}\) then \(A^2\bfv = \bfo\text{,}\) so we have \(\lambda^2\bfv = \bfo\text{.}\) But since \(\bfv\) is nonzero, and since \(\ff\) is a field, we must have \(\lambda = 0\text{.}\) Therefore, the only eigenvalue of \(A\) is 0.
6.1.4.11.
Solution.
We will argue by contradiction. Suppose that an \(n\times n\) matrix \(A\) has \(n+1\) distinct eigenvalues. This means that there are corresponding eigenvectors \(\bfv_1,\ldots, \bfv_{n+1}\text{.}\) By Theorem 6.1.13, we know that the set \(S = \{\bfv_1,\ldots, \bfv_{n+1}\}\) is linearly independent. However, this gives us a linearly independent set of \(n+1\) vectors in \(\ff^n\text{,}\) which is a contradiction by Corollary 5.1.13. This proves that \(A\) can have at most \(n\) distinct eigenvalues.
6.2 The Characteristic Equation
6.2.4 Exercises
6.2.4.2.
6.2.4.4.
6.2.4.6.
Answer.
The characteristic polynomial is \(p_A(\lambda) = -\lambda^3 - 6\lambda^2 - 9\lambda - 4\text{,}\) but it is easier in this problem NOT to expand the calculation to this form. The characteristic polynomial as calculated is \(p_A(\lambda) = (-1-\lambda)(\lambda^2 + 5\lambda + 4)\text{.}\) The eigenvalues are \(\lambda=-4\) and \(\lambda = -1\text{.}\)
6.2.4.8.
Writing Exercises
6.2.4.10.
Solution.
For the purposes of notation, we will say that \(A \sim B\) if \(A\) is similar to \(B\text{.}\) We first note that a matrix \(A \in M_n(\ff)\) is similar to itself since \(A = IPI^{-1}\text{,}\) where \(I\) is the \(n\times n\) identity matrix. This proves that the relation is reflexive.
We now suppose that \(A, B \in M_n(\ff)\) with \(A \sim B\text{.}\) This means that there is an invertible matrix \(P\) such that \(A = PBP^{-1}\text{.}\) We can manipulate this equation to give \(B = P^{-1}AP\text{,}\) and since \(P^{-1}\) is invertible (and since \((P^{-1})^{-1}=P\)), this proves that \(B \sim A\text{.}\) This proves that the relation is symmetric.
We now suppose that \(A, B, C \in M_n(\ff)\) with \(A \sim B\) and \(B \sim C\text{.}\) This means that we have invertible matrices \(P\) and \(Q\) such that \(A = PBP^{-1}\) and \(B = QCQ^{-1}\text{.}\) Putting these equations together, we have
\begin{equation*}
A = P(QCQ^{-1})P^{-1} = (PQ)C(Q^{-1}P^{-1}) = (PQ)C(PQ)^{-1}\text{.}
\end{equation*}
In this derivation we have used Proposition 3.3.3. Since \(PQ\) is invertible (again by Proposition 3.3.3), this proves that \(A \sim C\text{,}\) which shows that the relation is transitive.
We have shown that the relation of similarity on the set \(M_n(\ff)\) is reflexive, symmetric, and transitive, so it is an equivalence relation.
6.3 Diagonalization
6.3.5 Exercises
6.3.5.3.
6.3.5.5.
Answer.
This matrix is diagonalizable using the following matrices \(P\) and \(D\text{:}\)
\begin{equation*}
P = \begin{bmatrix} -1 \amp -2 \amp -2 \\
1 \amp 0 \amp -1 \\
0 \amp 1 \amp 1 \end{bmatrix}, \hspace{6pt}
D = \begin{bmatrix} -1 \amp 0 \amp 0 \\
0 \amp -1 \amp 0 \\
0 \amp 0 \amp -5 \end{bmatrix}\text{.}
\end{equation*}
6.3.5.6.
Answer.
No, that is not possible. The matrix \(A\) must be diagonalizable. The dimension of the eigenspace that is not yet specified must be at least one. The sum of all dimensions of the eigenspaces must be at most four (since \(A\) is \(4\times 4\)), and the given information tells us that this sum will be exactly four. This means that \(A\) is diagonalizable.
6.3.5.7.
Answer.
Here is one pair of matrices that diagonalizes \(A\text{:}\)
\begin{equation*}
P = \begin{bmatrix} 1 \amp -1 \\ 2 \amp 2 \end{bmatrix}, \hspace{6pt}
D = \begin{bmatrix} 2 \amp 0 \\ 0 \amp 6 \end{bmatrix}\text{.}
\end{equation*}
We can make subtle manipulations to these matrices to find another pair which diagonalizes \(A\text{:}\)
\begin{equation*}
P = \begin{bmatrix} -1 \amp 1 \\ 2 \amp 2 \end{bmatrix}, \hspace{6pt}
D = \begin{bmatrix} 6 \amp 0 \\ 0 \amp 2 \end{bmatrix}\text{.}
\end{equation*}
There are many, many other pairs of matrices that diagonalize \(A\text{,}\) mainly because the matrix \(P\) can have infinitely many different columns.
Writing Exercises
6.3.5.9.
Solution.
If \(A\) has \(n\) linearly independent eigenvectors, then it is diagonalizable by Theorem 6.3.2. This means there exist matrices \(P\) and \(D\) such that \(A = PDP^{-1}\text{.}\) If we take the transpose of both sides of this equation, we get
\begin{equation*}
A^T = (PDP^{-1})^T = (P^{-1})^TD^TP^T\text{.}
\end{equation*}
Since \(D\) is diagonal, \(D^T=D\text{.}\) Also, Exercise 3.3.5.8 tells us that \((P^{-1})^T=(P^T)^{-1}\text{.}\) So, we have
\begin{equation*}
A^T = (P^T)^{-1}DP^T\text{.}
\end{equation*}
This proves that \(A^T\) is diagonalizable, and Theorem 6.3.2 allows us to conclude that \(A^T\) has \(n\) linearly independent eigenvectors.
6.4 Invariants
6.4.4 Exercises
6.4.4.2.
Answer.
6.4.4.3.
Answer.
-
We find that \(\tr(A)=\tr(B)=3\text{,}\) \(\rank(A)=\rank(B)=2\text{,}\) and \(\det(A)=\det(B)=0\text{,}\) so the trace, rank, and determinant do not give us enough information to determine whether or not \(A\) and \(B\) are similar.
-
Some calculation shows that the eigenvalues for \(A\) are \(\lambda=0, -1, 4\text{,}\) while the eigenvalues for \(B\) are \(\lambda=0, 1, 2\text{.}\) Since these are different (and therefore the characteristic polynomials are different), we see that \(A\) and \(B\) are not similar.
6.4.4.6.
Answer.
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We find that \(\rank(A)=\rank(B)=3\text{,}\) \(\tr(A)=\tr(B)=5\text{,}\) and \(\det(A)=\det(B)=4\text{.}\) Further, the eigenvalues for both \(A\) and \(B\) are \(\lambda = 1, 2, 2\text{.}\) These four invariants do not provide enough information to determine whether or not \(A\) and \(B\) are similar.
-
We can show that \(A\) is not diagonalizable but \(B\) is. Since diagonalizability is an invariant for similarity, this proves that \(A\) and \(B\) are not similar.
Writing Exercises
6.4.4.14.
Answer.
No, trace is not an invariant for row equivalence. Consider the following matrix \(A \in M_2(\rr)\text{:}\)
\begin{equation*}
A = \begin{bmatrix} 2 \amp 1 \\ 1 \amp 1 \end{bmatrix}\text{.}
\end{equation*}
The matrix \(A\) is invertible, so it is row equivalent to \(I_2\text{.}\) However, \(\tr(A)=3\) and \(\tr(I_2)=2\text{.}\) This proves that trace is not an invariant for row equivalence.
6.4.4.17.
Answer.
Yes, invertibility is an invariant for similarity. This argument relies on the fact that the determinant is an invariant for similarity.
We will denote the similarity relation by \(A \sim B\text{.}\) Suppose that \(A \sim B\) and that \(A\) is invertible. This means that \(\det(A)\neq 0\text{,}\) and since \(A \sim B\text{,}\) this means that \(\det(B) \neq 0\) also. But the fact that \(\det(B) \neq 0\) means that \(B\) is invertible.
We now suppose that \(A \sim B\) and that \(A\) is not invertible. This means that \(\det(A)=0\text{,}\) and since \(A\sim B\text{,}\) this means that \(\det(B)=0\text{.}\) But \(\det(B)=0\) means that \(B\) is not invertible.
This proves that invertibility is an invariant for similarity.
7 Inner Products
7.1 Inner Products
7.1.6 Exercises
7.1.6.2.
7.1.6.3.
Answer.
-
We let \(\bfu\) and \(\bfv\) be the following vectors:\begin{equation*} \bfu = \begin{bmatrix} 1 \\ 2 \end{bmatrix}, \hspace{6pt} \text{and} \hspace{6pt} \bfv = \begin{bmatrix} -2 \\ 1 \end{bmatrix}\text{.} \end{equation*}Then it is easy to see that \(\bfu \cdot \bfv = 0\text{,}\) but \(\ip{\bfu, \bfv} = 4\text{.}\)
-
We let \(\bfu\) and \(\bfv\) be the following vectors:\begin{equation*} \bfu = \begin{bmatrix} 1 \\ 2 \end{bmatrix}, \hspace{6pt} \text{and} \hspace{6pt} \bfv = \begin{bmatrix} -6 \\ 1 \end{bmatrix}\text{.} \end{equation*}Then it is easy to see that \(\ip{\bfu, \bfv} = 0\text{,}\) but \(\bfu \cdot \bfv = -4\text{.}\)
7.1.6.6.
Answer.
Let \(A = \begin{bmatrix} 1 \amp 1 \\ 2 \amp 2 \end{bmatrix}\text{.}\) It is fairly easy to see that \(\det(A) = 0\text{,}\) so we have
\begin{equation*}
\ip{A, A} = 0 \cdot 0 = 0\text{.}
\end{equation*}
However, since \(A\) is not the zero matrix (i.e., the zero vector for the vector space \(M_2(\rr)\)), the fourth axiom of the inner product does not hold for this function.
Writing Exercises
7.1.6.8.
Solution.
This function is not an inner product. Consider the vector \(\bfu = \begin{bmatrix} 1 \\ 0 \end{bmatrix}\text{.}\) We have
\begin{equation*}
\ip{\bfu, \bfu} = (1)(0) + (0)(1) = 0\text{.}
\end{equation*}
However, since \(\bfu\) is not the zero vector in \(\rr^2\text{,}\) the fourth axiom of the inner product does not hold.
7.1.6.12.
Solution.
We will show that all four of the inner product axioms hold. For the first axiom, we note that everything here is real-valued, so we do not need to worry about any complex conjugates. Since the order of the functions within a definite integral can be switched, we have
\begin{align*}
\ip{g, f} \amp = \int_0^1 g(x)f(x)\;dx\\
\amp = \int_0^1 f(x)g(x)\;dx = \ip{f,g}\text{.}
\end{align*}
This shows that the first axiom holds.
For the second axiom, we let \(f, g, h \in C([0,1])\text{.}\) The definite integral is linear with respect to the sum of functions, so we have
\begin{align*}
\ip{f+g,h} \amp = \int_0^1 (f(x)+g(x))h(x)\;dx\\
\amp = \int_0^1 (f(x)h(x) + g(x)h(x))\;dx\\
\amp = \int_0^1 f(x)h(x)\;dx + \int_0^1 g(x)h(x)\;dx\\
\amp = \ip{f,h} + \ip{g,h}\text{.}
\end{align*}
This proves that the second axiom holds.
Let \(f,g \in C([0,1])\) and let \(c \in \rr\text{.}\) The definite integral is linear with respect to scalar multiplication by a real number, so we have
\begin{align*}
\ip{cf,g} \amp = \int_0^1 (cf(x))g(x)\;dx\\
\amp = c \int_0^1 f(x)g(x)\;dx\\
\amp = c \ip{f,g}\text{.}
\end{align*}
This proves that the third axiom holds.
Let \(f\in C([0,1])\text{.}\) We observe that \(f(x)^2\) is a function with values that are always non-negative (since each value of this function is a real number squared). Since the definite integral can be interpreted as calculating signed area between the graph of a function and the \(x\)-axis, we know that
\begin{equation*}
\ip{f,f} = \int_0^1 f(x)^2\;dx \ge 0\text{.}
\end{equation*}
Finally, using this same signed area interpretation of the definite integral, the only way a non-negative function could produce a zero value for the definite integral is if the function was uniformly the zero function. This means that if \(\ip{f,f}=0\text{,}\) we must have \(f(x)^2 = 0\) and therefore \(f(x) = 0\text{.}\) This proves that the fourth axiom holds.
7.2 Orthonormal Bases
7.2.5 Exercises
7.2.5.2.
7.2.5.3.
Answer.
The orthonormal basis we obtain is
\begin{equation*}
\left\{ \begin{bmatrix} 1 / \sqrt{2} \\ i/\sqrt{2} \\ 0 \end{bmatrix},
\begin{bmatrix} i/\sqrt{6} \\ 1/\sqrt{6} \\ 2i / \sqrt{6} \end{bmatrix},
\frac{\sqrt{3}}{2\sqrt{2}}\begin{bmatrix} -\tfrac{1}{3} + \tfrac{4}{3}i \\ -\tfrac{2}{3} + \tfrac{1}{3}i \\ \tfrac{1}{3} - \tfrac{1}{3}i \end{bmatrix} \right\}\text{.}
\end{equation*}
7.3 Orthogonal Projections
7.3.5 Exercises
7.3.5.1.
Answer.
If we let \(\bfw\) and \(\bfw'\) be the following vectors,
\begin{equation*}
\bfw = \begin{bmatrix} 35/34\\ 21/34 \end{bmatrix}, \hspace{6pt} \text{and} \hspace{6pt}
\bfw' = \begin{bmatrix} -69/34 \\ 115/34 \end{bmatrix}\text{,}
\end{equation*}
then \(\bfv = \bfw + \bfw'\) where \(\bfw \in L\) and \(\bfw' \in L^{\perp}\text{.}\)
7.3.5.2.
Answer.
-
Our matrix is\begin{equation*} [\proj_U]_{\mce} = \begin{bmatrix} \frac{1}{11} \amp -\frac{3}{11} \amp \frac{1}{11} \\[2pt] -\frac{3}{11} \amp \frac{101}{110} \amp \frac{3}{110} \\[2pt] \frac{1}{11} \amp \frac{3}{110} \amp \frac{109}{110} \end{bmatrix}\text{.} \end{equation*}
-
The vector in \(U\) which is closest to \(\bfv\) is\begin{equation*} \begin{bmatrix} \frac{3}{11} \\[2pt] \frac{15}{22} \\[2pt] \frac{105}{22} \end{bmatrix}\text{.} \end{equation*}
7.3.5.3.
Answer.
Let \(q\) and \(q'\) be the following polynomials:
\begin{align*}
q \amp = \tfrac{11}{10} - \tfrac{17}{20}t + \tfrac{61}{20}t^2, \hspace{3pt} \text{and}\\
q' \amp = \tfrac{9}{10} - \tfrac{3}{20}t - \tfrac{21}{20}t^2\text{.}
\end{align*}
Then \(p = q + q'\text{,}\) where \(q \in U\) and \(q' \in U^{\perp}\text{.}\)
Writing Exercises
7.3.5.5.
Solution.
First, we know that \(\ip{\bfo, \bfu} = 0\) for all \(\bfu \in U\) by Proposition 7.1.17. This proves that \(\bfo \in U^{\perp}\text{.}\)
Next, we let \(\bfw_1, \bfw_2 \in U^{\perp}\) and \(\bfu \in U\text{.}\) Then, by the properties of the inner product, we have
\begin{equation*}
\ip{\bfw_1 + \bfw_2, \bfu} = \ip{\bfw_1, \bfu} + \ip{\bfw_2, \bfu} = 0 + 0 = 0\text{.}
\end{equation*}
This shows that \(\bfw_1 + \bfw_2 \in U^{\perp}\text{,}\) meaning that \(U^{\perp}\) is closed under addition.
Finally, we let \(\bfw \in U^{\perp}\) and \(c \in \ff\text{.}\) (Here, \(\ff\) is either \(\rr\) or \(\cc\text{.}\)) We also let \(\bfu \in U\text{.}\) Then, by the properties of the inner product, we have
\begin{equation*}
\ip{c\bfw, \bfu} = c\ip{\bfw, \bfu} = c \cdot 0 = 0\text{.}
\end{equation*}
This proves that \(c\bfw \in U^{\perp}\) so that \(U^{\perp}\) is closed under scalar multiplication.
Since \(U^{\perp}\) contains the zero vector and is closed under addition and scalar multiplication, \(U^{\perp}\) is a subspace of \(V\text{.}\)
7.3.5.6.
Solution.
We will prove that each of these sets is a subset of the other. First, we let \(\bfx \in \nll(A)\text{,}\) so that \(A\bfx = \bfo\text{.}\) The fact that \(A\bfx = \bfo\) means that \(\mathbf{r} \cdot \bfx = 0\) for all rows \(\mathbf{r}\) of \(A\text{.}\) (Entry \(i\) in \(A\bfx\) is the dot product of the \(i\)th row of \(A\) with \(\bfx\text{.}\)) Since the rows of \(A\) span \(\row(A)\text{,}\) the fact that \(\mathbf{r} \cdot \bfx = 0\) for each row \(\mathbf{r}\) of \(A\) means that \(\bfx \cdot \bfv = 0\) for all \(\bfv \in \row(A)\text{.}\) This proves that \(\bfx \in (\row(A))^{\perp}\text{,}\) so \(\nll(A) \subseteq (\row(A))^{\perp}\text{.}\)
We now let \(\bfx \in (\row(A))^{\perp}\text{.}\) We want to show that \(\bfx \in \nll(A)\text{.}\) Since \(\bfx \in (\row(A))^{\perp}\text{,}\) we know that \(\bfx \cdot \mathbf{r}=0\) for each row \(\mathbf{r}\) of \(A\text{.}\) This shows that \(A\bfx = \bfo\text{,}\) which proves that \(\bfx \in \nll(A)\text{.}\) Therefore, \((\row(A))^{\perp} \subseteq \nll(A)\text{.}\)
Since we have shown that \(\nll(A) \subseteq (\row(A))^{\perp}\) and \((\row(A))^{\perp} \subseteq \nll(A)\text{,}\) we can conclude that \((\row(A))^{\perp} = \nll(A)\text{,}\) as desired.
