Definition 3.4.1.
If \(T:V\to W\) is a linear transformation between vector spaces, then the kernel of \(T\) is the set
\begin{equation*}
\kerr(T) = \{ \bfv \in V \mid T(\bfv) = \bfo \}\text{.}
\end{equation*}
Section 3.4 Subspaces and Linear Transformations
Every linear transformation between vector spaces brings with it some descriptions of related subspaces of the domain and codomain. We will explore some of these subspaces in this section.
Subsection 3.4.1 The Kernel of a Linear Transformation
The kernel of a linear transformation \(T\) is the set of all vectors that \(T\) sends to the zero vector.
While it may seem strange to single out the vectors that are sent to \(\bfo\text{,}\) this set reveals a lot about \(T\text{.}\)
Theorem 3.4.2.
Proof.
We will prove this theorem using the criteria for a subspace spelled out in Theorem 2.4.3. Since \(T(\bfo_V) = \bfo_W\text{,}\) we have \(\bfo_V \in \kerr(T)\text{.}\) (The fact that \(T(\bfo_V)=\bfo_W\) is found in Exercise 3.1.7.14.)
Let \(\bfv_1,\bfv_2 \in \kerr(T)\text{.}\) Using the additive property of \(T\) and the fact that these vectors are in the kernel, we have
\begin{equation*}
T(\bfv_1 + \bfv_2) = T(\bfv_1) + T(\bfv_2) = \bfo_W + \bfo_W = \bfo_W\text{.}
\end{equation*}
Finally, we let \(\bfv \in \kerr(T)\) and \(k \in \ff\text{.}\) Then we have
\begin{equation*}
T(k\bfv) = kT(\bfv) = k\bfo_W = \bfo_W\text{.}
\end{equation*}
This calculation used the fact that \(\bfv\) was assumed to be in the kernel and the scalar multiplication property of \(T\text{.}\)
Example 3.4.3.
Let \(V=D[a,b]\) be the set of all differentiable functions from \([a,b]\to \rr\text{.}\) Let \(T:V \to V\) be the linear transformation which takes the derivative. (We proved a very similar function was a linear transformation in Example 3.1.3.) What is the kernel of \(T\text{?}\)
We recall from calculus that a function \(f\) has \(f'(x)=0\) for all \(x\) in an interval if and only if \(f\) is a constant function. This proves that \(\kerr(T)\) is the set of all constant functions, and it further establishes that set as a subspace of \(D[a,b]\text{.}\)
While not all linear transformations are linked to matrices, some are. The kernel has an alternate name in those situations.
Definition 3.4.4.
If \(A \in M_{m,n}(\ff)\text{,}\) then the null space of \(A\) is
\begin{equation*}
\nll(A) = \{ \bfv \in \ff^n \mid A\bfv = \bfo \}\text{.}
\end{equation*}
Since we have shown that the kernel is a subspace, the word “space” in the term null space is justified.
We also recall the link between matrix-vector equations like \(A\bfx = \bfb\) and linear systems. The definition of the null space shows that the set of solutions to a homogeneous linear system can be described as the null space of a matrix.
Example 3.4.5.
Let \(A\) be the matrix
\begin{equation*}
A = \begin{bmatrix}
3 \amp 1 \amp -3 \\
-1 \amp 0 \amp 2 \\
2 \amp -2 \amp -10
\end{bmatrix}\text{.}
\end{equation*}
We can find \(\nll(A)\) by row reducing the matrix \([A \mid \bfo]\text{.}\) Here is the RREF:
\begin{equation*}
\left[ \begin{array}{@{}ccc|c@{}}
1 \amp 0 \amp -2 \amp 0 \\
0 \amp 1 \amp 3 \amp 0 \\
0 \amp 0 \amp 0 \amp 0
\end{array}\right]\text{.}
\end{equation*}
From this we see that \(x_1 = 2x_3\text{,}\) \(x_2=-3x_3\text{,}\) and \(x_3\) is free. In other words, any vector \(\bfx\) in \(\nll(A)\) looks like
\begin{equation*}
\bfx = \begin{bmatrix}
2x_3 \\ -3x_3 \\ x_3
\end{bmatrix} = x_3\begin{bmatrix}
2 \\ -3 \\ 1
\end{bmatrix}\text{.}
\end{equation*}
So we have
\begin{equation*}
\nll(A) = \spn\{\bfv\}
\end{equation*}
where
\begin{equation*}
\bfv = \begin{bmatrix}
2 \\ -3 \\ 1
\end{bmatrix}\text{.}
\end{equation*}
Note 3.4.6.
There is an important fact contained in this last example. When we have a homogeneous system, we can always pay attention to just the coefficient matrix instead of the augmented matrix. No elementary row operation can produce a non-zero entry in a column of zeros.
The following theorem is one of the reasons that the kernel is so useful.
Theorem 3.4.7.
Let \(T:V\to W\) be a linear transformation. Then \(T\) is injective if and only if \(\kerr(T) = \{\bfo\}\text{.}\)
Proof.
We first suppose that \(T\) is injective. (We recall the definition of injectivity from Definition 3.1.12.) Since \(T(\bfo_V) = \bfo_W\text{,}\) the injectivity of \(T\) implies that \(\bfv = \bfo_V\) if \(T(\bfv)=\bfo_W\) for any \(\bfv \in V\text{.}\) Therefore, \(\kerr(T) = \{\bfo\}\text{.}\)
Next, we suppose that \(\kerr(T) = \{\bfo\}\text{.}\) We want to prove that \(T\) is injective, so we let \(\bfv_1, \bfv_2 \in V\) with \(T(\bfv_1)=T(\bfv_2)\text{.}\) We want to show that \(\bfv_1=\bfv_2\text{.}\) By the linearity of \(T\) we have
\begin{align*}
T(\bfv_1)-T(\bfv_2) \amp = \bfo\\
T(\bfv_1-\bfv_2) \amp = \bfo\text{.}
\end{align*}
Since \(\kerr(T) = \{\bfo \}\text{,}\) we must have \(\bfv_1-\bfv_2 = \bfo\text{,}\) meaning that \(\bfv_1=\bfv_2\text{.}\) This proves that \(T\) is injective.
Example 3.4.8.
Consider the linear transformation \(T:P_1 \to \rr^2\) given by
\begin{equation*}
T(p) = \begin{bmatrix}
p(0) \\ p'(0)
\end{bmatrix}\text{.}
\end{equation*}
To examine \(\kerr(T)\text{,}\) we need to look at polynomials \(p \in P_1\) such that \(p(0)=0\) and \(p'(0)=0\text{.}\) If \(p(t) = a + bt\text{,}\) then \(a = p(0)\) and \(b = p'(0)\text{,}\) so if \(T(p)= \bfo\text{,}\) we must have \(a=0\) and \(b=0\text{.}\) This means that the only polynomial in \(\kerr(T)\) is the zero polynomial. Therefore, \(T\) is injective.
We now present one final fact related to the kernel.
Corollary 3.4.9.
Suppose that the linear system represented by the equation \(A\bfx = \bfb\) is consistent. Then this system has a unique solution if and only if \(\nll(A) = \{\bfo\} \text{.}\)
Proof.
We first assume that the system has a unique solution. Since the linear system is consistent, then there exists a vector \(\bfv\) such that \(A\bfv = \bfb\text{.}\) If \(\bfw \in \nll(A)\text{,}\) then \(A(\bfv + \bfw) = A\bfv + A\bfw = \bfb + \bfo = \bfb\text{,}\) so \(\bfv + \bfw\) is also a solution. But since there is a unique solution, we must have \(\bfv = \bfv+\bfw\text{,}\) so \(\bfw = \bfo\text{.}\) This shows that \(\nll(A) = \{\bfo\}\text{.}\)
We now assume that \(\nll(A) = \{\bfo\}\text{.}\) From Theorem 3.4.7 we know that the associated linear transformation is injective. Since the system is consistent, there must be only one vector that the transformation sends to \(\bfb\) so the system has a unique solution.
Subsection 3.4.2 The Image as a Subspace
We have examined the kernel as a subspace of the domain of a linear transformation. We now turn our attention to a well-known subset of the codomain. The reader will be familiar with the image (or range) of a linear transformation. We can now prove that this is a subspace.
Note 3.4.10.
Theorem 3.4.11.
Let \(T:V\to W\) be a linear transformation between vector spaces over \(\ff\text{.}\) Then \(\img(T)\) is a subspace of \(W\text{.}\)
Proof.
Since we know that \(T(\bfo_V) = \bfo_W\text{,}\) it follows that \(\bfo_W \in \img(T)\text{.}\) We now need to show the other properties demanded by Theorem 2.4.3.
If \(\bfw_1, \bfw_2 \in \img(T)\text{,}\) then there exist vectors \(\bfv_1, \bfv_2 \in V\) with \(T(\bfv_1)=\bfw_1\) and \(T(\bfv_2)=\bfw_2\text{.}\) Then, using the linearity of \(T\text{,}\) we have
\begin{equation*}
\bfw_1 + \bfw_2 = T(\bfv_1) + T(\bfv_2) = T(\bfv_1+\bfv_2)\text{.}
\end{equation*}
This proves that \(\bfw_1+\bfw_2 \in \img(T)\text{,}\) so \(\img(T)\) is closed under addition.
Finally, we let \(k \in \ff\) and \(\bfw \in \img(T)\text{,}\) so there exists a vector \(\bfv \in V\) such that \(T(\bfv) = \bfw\text{.}\) Then, using the fact that \(T\) is a linear transformation, we have
\begin{equation*}
k\bfw = kT(\bfv) = T(k\bfv)\text{,}
\end{equation*}
which proves that \(k\bfw \in \img(T)\text{.}\) Thus, \(\img(T)\) is closed under scalar multiplication. We conclude that \(\img(T)\) is a subspace of \(W\text{.}\)
As usual, when our linear transformation is linked to a matrix, we have more to say.
Definition 3.4.12.
If \(A \in M_{m,n}(\ff)\text{,}\) then the column space of \(A\text{,}\) written \(\col(A)\text{,}\) is the set of all linear combinations of the columns of \(A\text{.}\) If \({A = [\mathbf{a}_1 \cdots \mathbf{a}_n]}\text{,}\) then
\begin{equation*}
\col(A) = \spn\{\mathbf{a}_1, \ldots, \mathbf{a}_n \}\text{.}
\end{equation*}
When we introduced the matrix-vector product in Example 3.1.5, we noted that \(A\bfv\) is a linear combination of the columns of \(A\) with weights coming from the entries in \(\bfv\text{.}\) So, a vector in \(\col(A)\) can be written as \(A\bfx\) for some \(\bfx\text{.}\) Therefore, another way to write the column space is
\begin{equation*}
\col(A) = \{ \bfw \in \ff^m \mid \bfw = A\bfx \text{ for some } \bfx \in \ff^n \}\text{.}
\end{equation*}
From this description, we can see that the column space of a matrix and the image of a linear transformation are the same.
Fact 3.4.13.
If \(T:\ff^n \to \ff^m\) is a linear transformation given by multiplication by a matrix \(A\text{,}\) then
\begin{equation*}
\img(T) = \col(A)\text{.}
\end{equation*}
We can also restate the consistency of linear systems using the language of the column space.
Fact 3.4.14.
If \(A \in M_{m,n}(\ff)\text{,}\) then the linear system \(A\bfx = \bfb\) is consistent if and only if \(\bfb \in \col(A)\text{.}\)
Since the column space of a matrix is a subspace of the codomain of the associated linear transformation, there will be some occasions when that subspace is as large as it could be. The next theorem gives conditions for just that situation.
Theorem 3.4.15.
A set of vectors \(\{\bfv_1, \ldots, \bfv_n \}\) spans \(\ff^m\) if and only if the RREF of the matrix \(A = [\bfv_1 \cdots \bfv_n]\) has a pivot in every row.
Proof.
The set of vectors \(\{\bfv_1, \ldots, \bfv_n \}\) spans \(\ff^m\) if and only if \(\bfb \in \col(A)\) for every \(\bfb \in \ff^m\text{.}\) This happens when the linear system with augmented matrix \([A \mid \bfb]\) is consistent for each \(\bfb\text{.}\)
We know from Theorem 2.2.2 that a linear system over \(\ff\) is consistent if and only if there is no pivot in the final column of the augmented matrix. If the RREF of \(A\) has a pivot in every row, then there cannot be a pivot in the final column of the RREF of \([A \mid \bfb]\) since each row already contains one pivot.
We will prove the contrapositive of the other implication. Suppose that the RREF of \(A\) does not have a pivot in every row. We will create a vector in \(\ff^m\) which is not in the span of this set of vectors. Since the RREF of \(A\) does not have a pivot in every row, let the smallest row number with no pivot be \(k\text{.}\) Form the augmented matrix with the RREF of \(A\) and the vector \(\bfe_k\text{.}\) Now reverse the elementary row operations that were taken to reduce \(A\) to its RREF. The result will be an augmented matrix \([A \mid \bfb]\) which is related to an inconsistent system. (There will be a pivot in the final column of the RREF of this matrix. We constructed it this way!) This proves that the set \(\{\bfv_1, \ldots, \bfv_n \}\) does not span \(\ff^m\text{.}\)
Example 3.4.16.
Define the vectors \(\bfu, \bfv, \bfw\) in \(\ff_5^3\) as
\begin{equation*}
\bfu = \begin{bmatrix}
3 \\ 2 \\ 3
\end{bmatrix}, \hspace{6pt}
\bfv = \begin{bmatrix}
0 \\ 4 \\ 4
\end{bmatrix}, \hspace{6pt}
\bfw = \begin{bmatrix}
0 \\ 4 \\ 3
\end{bmatrix}\text{.}
\end{equation*}
How large is \(\spn\{\bfu,\bfv,\bfw\}\text{?}\)
We form the matrix \([\bfu\ \bfv\ \bfw]\) and find the RREF:
\begin{equation*}
\begin{bmatrix}
3 \amp 0 \amp 0 \\
2 \amp 4 \amp 4 \\
3 \amp 4 \amp 3
\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*}
Since there is a pivot in each row, the set \(\{\bfu, \bfv, \bfw\}\) spans \(\ff_5^3\text{.}\)
Example 3.4.17.
Consider the following vectors in \(\rr^3\text{:}\)
\begin{equation*}
\bfu = \begin{bmatrix}
-2 \\ 3 \\ -1
\end{bmatrix}, \hspace{6pt}
\bfv = \begin{bmatrix}
1 \\ -2 \\ 2
\end{bmatrix}, \hspace{6pt}
\bfw = \begin{bmatrix}
-5 \\ 6 \\ 2
\end{bmatrix}\text{.}
\end{equation*}
When we row reduce the matrix \([\bfu\ \bfv\ \bfw]\text{,}\) we get
\begin{equation*}
\begin{bmatrix}
-2 \amp 1 \amp -5 \\
3 \amp -2 \amp 6 \\
-1 \amp 2 \amp 2
\end{bmatrix} \sim
\begin{bmatrix}
1 \amp 0 \amp 4 \\
0 \amp 1 \amp 3 \\
0 \amp 0 \amp 0
\end{bmatrix}\text{.}
\end{equation*}
Since there is not a pivot in every row, the set \(\{\bfu, \bfv, \bfw\}\) does not span \(\rr^3\text{.}\)
Reading Questions 3.4.3 Reading Questions
1.
Let \(A\) be the following matrix:
\begin{equation*}
A = \begin{bmatrix}
2 \amp 2 \\ 3 \amp -2 \\ 3 \amp -5
\end{bmatrix}\text{.}
\end{equation*}
Let \(T:\rr^2 \to \rr^3\) be the linear transformation which is multiplication by \(A\text{.}\)
-
Calculate \(\nll(A)\text{.}\)
-
Is \(T\) injective? Explain.
2.
Consider the following three vectors in \(\ff_3^2\text{:}\)
\begin{equation*}
\bfu = \begin{bmatrix}
2 \\ 1
\end{bmatrix}, \hspace{6pt}
\bfv = \begin{bmatrix}
0 \\ 1
\end{bmatrix}, \hspace{6pt}
\bfw = \begin{bmatrix}
0 \\ 2
\end{bmatrix}\text{.}
\end{equation*}
Does the set \(\{\bfu, \bfv, \bfw \}\) span \(\ff_3^2\text{?}\) Explain.
Exercises 3.4.4 Exercises
1.
Consider the following matrix \(A\) in \(M_{3,4}(\ff_7)\text{:}\)
\begin{equation*}
A = \begin{bmatrix}
6 \amp 3 \amp 3 \amp 6 \\ 0 \amp 3 \amp 0 \amp 2 \\ 3 \amp 5 \amp 2 \amp 2
\end{bmatrix}\text{.}
\end{equation*}
For each of the following vectors \(\bfx \in \ff_7^4\text{,}\) determine whether or not \({\bfx \in \nll(A)}\text{.}\)
-
\(\displaystyle \bfx = \begin{bmatrix} 6 \\ 1 \\ 4 \\ 2 \end{bmatrix}\)
-
\(\displaystyle \bfx = \begin{bmatrix} 4 \\ 3 \\ 1 \\ 2 \end{bmatrix}\)
2.
For each of the following matrices \(A\) over \(\rr\text{,}\) find \(\nll(A)\) by producing a set of vectors that spans \(\nll(A)\text{.}\)
-
\(\displaystyle A = \begin{bmatrix} 2 \amp 3 \amp 4 \amp 2 \\ -5 \amp -1 \amp 5 \amp -3 \end{bmatrix}\)
-
\(\displaystyle A = \begin{bmatrix} 2 \amp -4 \amp -5 \amp -3 \amp 1 \\ 1 \amp 4 \amp -5 \amp 4 \amp 5 \\ 0 \amp -2 \amp 5 \amp 0 \amp 5 \end{bmatrix} \)
3.
Find a matrix \(A\) so that the following set is \(\col(A)\text{:}\)
\begin{equation*}
\left\{ \begin{bmatrix}
2r + 4s \\ 3s-7t \\ -r-s+5t
\end{bmatrix} \large\mid r,s,t \in \rr \right\}\text{.}
\end{equation*}
4.
Let \(A\) and \(\bfx\) be the following, with entries from \(\rr\text{:}\)
\begin{equation*}
A = \begin{bmatrix}
3 \amp -1 \\ -9 \amp 3
\end{bmatrix}, \hspace{12pt}
\bfx = \begin{bmatrix}
2 \\ 6
\end{bmatrix} \text{.}
\end{equation*}
-
Explain why it is not possible, for this particular matrix \(A\text{,}\) to find a non-zero vector in \(\nll(A) \cap \col(A)\text{.}\)
-
Is it possible to find a matrix \(A \in M_2(\rr)\) such that there exists a non-zero vector in \(\nll(A) \cap \col(A)\text{?}\) Justify your answer thoroughly.
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 \(\col(A)\text{.}\)
5.
Consider the following linear transformation \(T:\ff_7^4 \to \ff_7^3\text{:}\)
\begin{equation*}
T \left( \begin{bmatrix}
x_1 \\ x_2 \\ x_3 \\ x_4
\end{bmatrix} \right) = \begin{bmatrix}
2x_1 + 4x_2 + 6x_3 + x_4 \\
4x_1 + 6x_2 + 5x_3 \\
4x_1 + 2x_2 + 3x_3
\end{bmatrix} \text{.}
\end{equation*}
This \(T\) is not injective. Find distinct vectors \(\bfx_1\) and \(\bfx_2\) in \(\ff_7^4\) such that \(T(\bfx_1) = T(\bfx_2)\text{.}\)
6.
Consider the following linear transformation \(T:\ff_7^3 \to \ff_7^4\text{:}\)
\begin{equation*}
T \left( \begin{bmatrix}
x_1 \\ x_2 \\ x_3
\end{bmatrix} \right) = \begin{bmatrix}
x_1 + 3x_2 + 3x_3 \\
2x_1 + 6x_2 + 3x_3 \\
x_2 + 3x_3 \\
3x_1 + 5x_2 + 6x_3
\end{bmatrix} \text{.}
\end{equation*}
This \(T\) is not surjective. Find a vector \(\mathbf{y} \in \ff_7^4\) such that \(\mathbf{y}\) is not in the image of \(T\text{.}\)
7.
Consider the following linear transformation \(T: \rr^3 \to P_2\text{:}\)
\begin{equation*}
T \left( \begin{bmatrix} a \\ b \\ c \end{bmatrix} \right) =
(a+b) + (b+ c)t + (a+b+c)t^2\text{.}
\end{equation*}
-
Either prove or disprove that \(T\) is injective.
-
Either prove or disprove that \(T\) is surjective.
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.
8.
Consider the following linear transformation \(T: \rr^2 \to P_2\text{:}\)
\begin{equation*}
T \left( \begin{bmatrix} a \\ b \end{bmatrix} \right) =
a + bt + bt^2\text{.}
\end{equation*}
-
Either prove or disprove that \(T\) is injective.
-
Either prove or disprove that \(T\) is surjective.
9.
Consider the following linear transformation \(T: P_2 \to \rr^2\text{:}\)
\begin{equation*}
T(a+bt+ct^2) = \begin{bmatrix} a-2b \\ 3b + c \end{bmatrix} \text{.}
\end{equation*}
-
Either prove or disprove that \(T\) is injective.
-
Either prove or disprove that \(T\) is surjective.
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.
10.
For each of the following, consider a linear transformation \({T: \rr^n \to \rr^m}\) which is multiplication by the given \(m\times n\) matrix \(A\text{.}\) In each case, determine whether or not \(T\) is injective and whether or not \(T\) is surjective. Explain your answers.
-
\(\displaystyle A = \begin{bmatrix} 2 \amp -4 \amp -14 \\ -4 \amp -1 \amp 10 \\ -2 \amp 2 \amp 10 \end{bmatrix}\)
-
\(\displaystyle A = \begin{bmatrix} -3 \amp -5 \amp -2 \amp 3 \\ 1 \amp 1 \amp -5 \amp -5 \\ 1 \amp -3 \amp 1 \amp 2 \end{bmatrix} \)
-
\(\displaystyle A = \begin{bmatrix} 2 \amp -2 \amp 2 \\ 4 \amp -1 \amp 7 \\ -3 \amp -1 \amp -7 \\ 0 \amp 5 \amp 5 \end{bmatrix} \)
-
\(\displaystyle A = \begin{bmatrix} -2 \amp -3 \amp -1 \\ 5 \amp -5 \amp -2 \\ 0 \amp 4 \amp -4 \end{bmatrix} \)
Writing Exercises
11.
Suppose that \(T:U\to V\) is a linear transformation between vector spaces over \(\ff\text{.}\) If \(U'\) is a subspace of \(U\text{,}\) prove that \(T(U')\) is a subspace of \(V\text{,}\) where
\begin{equation*}
T(U') = \{ T(\bfu) \in V \mid \bfu \in U' \} \text{.}
\end{equation*}
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 \(V\text{.}\)
12.
Suppose that \(T:U\to V\) is a linear transformation between vector spaces over \(\ff\text{.}\) If \(V'\) is a subspace of \(V\text{,}\) define the set \(U_1\) by
\begin{equation*}
U_1 = \{ \bfu \in U \mid T(\bfu) \in V' \}\text{.}
\end{equation*}
Prove that \(U_1\) is a subspace of \(U\text{.}\)
13.
Let \(A \in M_{m,n}(\ff)\) and let \(B \in M_{n,p}(\ff)\text{.}\) Prove that \(\col(AB) \subseteq \col(A)\text{.}\)
