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Linear Algebra Gathering and Sharpening Tools

Ryan Higginbottom

Section 5.3 Dimension

In this section we will define the dimension of a vector space, finally delivering on the promise made in the introduction to this chapter to describe an intrinsic quality of vector spaces that allows a comparison between spaces.
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Subsection 5.3.1 The Dimension of a Vector Space

We are on the threshold of the definition of dimension. We will first present a result that connects (for a finite-dimensional space) the number of vectors needed for a spanning set to the concept of linear independence. We will omit the proof.
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Proof.

Since \(\bfw_j \in \spn\{\bfv_1, \ldots, \bfv_m\}\) for each \(j=1, \ldots, n\text{,}\) there exist scalars \(c_{ij} \in \ff\) such that
\begin{equation*} \bfw_j = c_{1j}\bfv_1 + \cdots + c_{mj}\bfv_m\text{.} \end{equation*}
We now consider the homogeneous linear system represented by the equation \(A\bfx = \bfo\text{,}\) where \(A\) is the \(m\times n\) matrix \(A = [c_{ij}]\text{.}\) If \(\bfx\) is a solution to this system, then we have
\begin{equation*} 0 = c_{i1}x_1 + \cdots + c_{in}x_n \end{equation*}
for all \(i = 1,\ldots, m\text{.}\) This means that we have
\begin{align*} \bfo \amp = \sum_{i=1}^m \left(\sum_{j=1}^n c_{ij}x_j \right) \bfv_i\\ \amp = \sum_{j=1}^n x_j \left(\sum_{i=1}^m c_{ij} \bfv_i \right)\\ \amp = \sum_{j=1}^n x_j \bfw_j\text{.} \end{align*}
Since \(\{\bfw_1, \ldots, \bfw_n \}\) is a linearly independent set, we must have \(x_j=0\) for all \(j\text{,}\) meaning that \(\bfx = \bfo\) is the only solution to the equation \(A\bfx = \bfo\text{.}\) By Corollary 1.3.7 (or, rather, the corresponding corollary to Theorem 2.2.3), this means that \(m \ge n\text{,}\) as desired.
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We will now use this lemma to prove a result related to dimension.
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Proof.

We will prove the contrapositive. If \(V\) is finite-dimensional, then there exists a set \(V' = \{\bfv_1, \ldots, \bfv_m\}\) such that \(V = \spn(V')\text{.}\) By Lemma 5.3.1, for any \(n > m\text{,}\) \(V\) cannot contain a linearly independent subset of \(n\) vectors. Thus, there exists a natural number \(n\) such that \(V\) does not contain a linearly independent subset of that size. This completes the proof of the contrapositive.
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This theorem gives us an introduction to our first infinite-dimensional example.
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Example 5.3.3.

Let \(P\) be the vector space of all polynomials with real coefficients. (We do not restrict the degree of polynomials in \(P\text{.}\)) For any \(n \ge 1\text{,}\) \(P\) contains the linearly independent set
\begin{equation*} \{ 1, t, \ldots, t^n \}\text{.} \end{equation*}
Therefore, by Theorem 5.3.2, \(P\) is infinite-dimensional.
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We now come to the bedrock result of this section, the result that makes the definition of dimension possible.
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Proof.

Since \(V \subseteq \spn\{\bfv_1, \ldots, \bfv_m \}\) and \(\{ \bfw_1, \ldots, \bfw_n \}\) is a linearly independent set, Lemma 5.3.1 implies that \(m \ge n\text{.}\) However, since \({V \subseteq \spn\{\bfw_1, \ldots, \bfw_n \}}\) and \(\{ \bfv_1, \ldots, \bfv_m \}\) is a linearly independent set, Lemma 5.3.1 also implies that \(n \ge m\text{.}\) Therefore, \(m=n\text{.}\)
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Even though a vector space may have a huge number of bases, all of those bases have the same size. This is a number intrinsic to the vector space, not to any specific basis of that vector space. This is what we mean by the dimension of a vector space.
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Definition 5.3.5.

Let \(V\) be a finite-dimensional vector space. If \(V \neq \{\bfo\}\text{,}\) then the dimension of \(V\text{,}\) written \(\dim(V)\text{,}\) is the size of any basis of \(V\text{.}\) If \(\dim(V)=n\text{,}\) we say that \(V\) is \(n\)-dimensional.
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If \(V = \{\bfo\}\text{,}\) then we define the dimension of \(V\) to be 0.
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Two of the families of vector spaces we frequently discuss have easy-to-determine dimensions, as the next two examples illustrate.
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Example 5.3.6.

Since \(\{\bfe_1, \ldots, \bfe_n \}\) is a basis for \(\ff^n\text{,}\) then \(\dim(\ff^n)=n\text{.}\)
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Example 5.3.7.

Since \(\{1, t, \ldots, t^n\}\) is a basis for \(P_n\text{,}\) then \(\dim(P_n)=n+1\text{.}\)
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The proofs of the next two results are a consequence of Lemma 5.3.1 and will appear in the exercises.
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We will now begin to discuss dimension as a tool to compare vector spaces. Linear transformations are the main way we relate vector spaces to each other, so these next results rely on that machinery.
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Proof.

Given \(\bfv \in V\text{,}\) there exists a unique linear combination
\begin{equation*} \bfv = \sum_{i=1}^n c_i \bfv_i \end{equation*}
by Theorem 5.2.11. We define the function \(T\) by
\begin{equation*} T(\bfv) = \sum_{i=1}^n c_i \bfw_i\text{.} \end{equation*}
In words, we send a vector \(\bfv\) to the linear combination of the \(\bfw_i\) vectors using the same weights as those needed to form \(\bfv\) from the basis \(V'\text{.}\) This gives \(T(\bfv_i)=\bfw_i\) for each \(i\text{,}\) so we only need to show that \(T\) is a linear transformation.
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Suppose that \(\bfu, \bfv \in V\) with
\begin{equation*} \bfv = \sum_{i=1}^n c_i \bfv_i \hspace{6pt} \text{and} \hspace{6pt} \bfu = \sum_{i=1}^n d_i \bfv_i\text{.} \end{equation*}
Then we have
\begin{align*} T(\bfu + \bfv) \amp = T \left( \sum_{i=1}^n (c_i+d_i)\bfv_i \right)\\ \amp = \sum_{i=1}^n (c_i+d_i)\bfw_i\\ \amp = \sum_{i=1}^n c_i\bfw_i + \sum_{i=1}^n d_i\bfw_i\\ \amp = T(\bfu) + T(\bfv)\text{.} \end{align*}
Now we let \(\bfv \in V\) and \(d \in \ff\text{.}\) Then, if
\begin{equation*} \bfv = \sum_{i=1}^n c_i \bfv_i\text{,} \end{equation*}
we have
\begin{align*} T(d\bfv) \amp = T\left( \sum_{i=1}^n (dc_i) \bfv_i \right)\\ \amp = \sum_{i=1}^n (dc_i) \bfw_i\\ \amp = d\sum_{i=1}^n c_i \bfw_i\\ \amp = dT(\bfv)\text{.} \end{align*}
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We will complete the proof by justifying the claim that \(T\) is unique. Suppose that \(T' \in L(V,W)\) with \(T'(\bfv_i)=\bfw_i\) for each \(i\text{.}\) Then, if \(\bfv \in V\) with
\begin{equation*} \bfv = \sum_{i=1}^n c_i \bfv_i\text{,} \end{equation*}
we have
\begin{equation*} T'(\bfv) = T'\left( \sum_{i=1}^n c_i \bfv_i \right) = \sum_{i=1}^n c_i T'(\bfv_i) = \sum_{i=1}^n c_i \bfw_i\text{.} \end{equation*}
This shows that \(T'(\bfv)=T(\bfv)\) for every \(\bfv \in V\text{,}\) so \(T'=T\) and \(T\) is unique.
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The notion of alikeness that we use in linear algebra is when two vector spaces are isomorphic. The reader may wish to consult Definition 3.1.15 for a refresher.
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Proof.

We first suppose that \(T\) is an isomorphism. We want to show that \(T(B)\) is a basis for \(W\text{,}\) so we begin with linear independence. Suppose that \(c_1, \ldots, c_n \in \ff\) such that
\begin{equation*} \bfo = \sum_{i=1}^n c_i T(\bfv_i)\text{.} \end{equation*}
Then we have
\begin{equation*} \bfo = \sum_{i=1}^n T(c_i\bfv_i) = T\left( \sum_{i=1}^n c_i\bfv_i \right)\text{.} \end{equation*}
Since \(T\) is injective, by Theorem 3.4.7 we must have
\begin{equation*} \bfo = \sum_{i=1}^n c_i\bfv_i\text{.} \end{equation*}
But since \(B\) is a linearly independent set, we have \(c_i=0\) for all \(i\text{.}\) This proves that \(T(B)\) is linearly independent.
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We now prove that \(T(B)\) spans \(W\text{.}\) Let \(\bfw \in W\text{.}\) Since \(T\) is surjective, there exists \(\bfv \in V\) such that \(T(\bfv)=\bfw\text{.}\) Since \(B\) is a basis for \(V\text{,}\) we have
\begin{equation*} \bfv = \sum_{i=1}^n c_i\bfv_i\text{.} \end{equation*}
Then
\begin{equation*} \bfw = T(\bfv) = T\left( \sum_{i=1}^n c_i\bfv_i \right) = \sum_{i=1}^n c_iT(\bfv_i)\text{.} \end{equation*}
This proves that \(W = \spn(T(B))\text{,}\) so \(T(B)\) is a basis for \(W\text{.}\)
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We now need to prove the other implication, and we assume that \(T(B)\) is a basis for \(W\text{.}\) We need to show that \(T\) is an isomorphism. To show that \(T\) is injective, suppose that \(\bfv \in V\) such that \(T(\bfv)=\bfo\text{.}\) We have
\begin{equation*} \bfv = \sum_{i=1}^n c_i\bfv_i\text{,} \end{equation*}
so
\begin{equation*} \bfo = T(\bfv) = T\left( \sum_{i=1}^n c_i\bfv_i \right) = \sum_{i=1}^n c_iT(\bfv_i)\text{.} \end{equation*}
But since \(T(B)\) is a linearly independent set by assumption, this implies that \(c_i=0\) for all \(i\text{.}\) This means that \(\bfv = \bfo\text{,}\) so \(T\) is injective.
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To prove that \(T\) is surjective, we assume that \(\bfw \in W\text{.}\) Since \(T(B)\) spans \(W\text{,}\) we have
\begin{equation*} \bfw = \sum_{i=1}^n d_iT(\bfv_i) \end{equation*}
for some \(d_i \in \ff\text{.}\) We claim that if
\begin{equation*} \bfv = \sum_{i=1}^n d_i \bfv_i\text{,} \end{equation*}
then \(T(\bfv) = \bfw\text{.}\) Here is the justification:
\begin{equation*} T(\bfv) = T\left( \sum_{i=1}^n d_i \bfv_i \right) = \sum_{i=1}^n d_i T(\bfv_i) = \bfw\text{.} \end{equation*}
This proves that \(T\) is surjective and is thus an isomorphism.
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When we view dimension as an intrinsic quality of a vector space that allows comparison between spaces, we find something surprising about vector spaces with the same dimension. They are essentially the same!
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Proof.

Suppose that \(\dim(V) = \dim(W) = n\text{.}\) Let \(\{\bfv_1, \ldots, \bfv_n\}\) be a basis for \(V\) and let \(\{\bfw_1, \ldots, \bfw_n \}\) be a basis for \(w\text{.}\) By Theorem 5.3.10, we can find \(T \in L(V,W)\) such that \(T(\bfv_i) = \bfw_i\) for each \(i\text{,}\) \(1 \le i \le n\text{.}\) Then by Theorem 5.3.11, \(T\) is an isomorphism.
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To prove the claim in the other direction, suppose that \(T \in L(V,W)\) is an isomorphism. If \(\{\bfv_1, \ldots, \bfv_n \}\) is a basis for \(V\text{,}\) then \(\{T(\bfv_1), \ldots, T(\bfv_n) \}\) is a basis for \(W\) by Theorem 5.3.11. Thus \(\dim(V) = \dim(W)\text{.}\)
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Here is an immediate consequence of this result.
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Example 5.3.14.

Since \(P_2\) is a three-dimensional vector space over \(\rr\text{,}\) \(\rr^3\) and \(P_2\) are isomorphic.
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Subsection 5.3.2 Dimension and Subspaces

If we know the dimension of a vector space, then we sometimes have a quicker path to finding a basis for that space. This next result says that if we have a spanning set of the same size as a basis, then it must be a basis.
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Proof.

By The Spanning Set Theorem (Theorem 5.2.12), we know that a subset \(B'\) of \(B\) will be a basis for \(V\text{.}\) But since \(\dim(V)=n\text{,}\) the size of \(B'\) must be \(n\text{.}\) Therefore, \(B'=B\) and \(B\) is a basis for \(V\text{.}\)
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What is true in Theorem 5.3.15 for a spanning set is also true for a linearly independent set. To prove that, however, we first need the analog to The Spanning Set Theorem for linearly independent sets.
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Proof.

Let \(V' = \{\bfv_1, \ldots \bfv_n \}\) be a linearly independent set of vectors in \(V\text{.}\) If \(V = \spn(V')\text{,}\) then \(V'\) is a basis and we are done. If \(V \neq \spn(V')\text{,}\) then there exists some vector \(\bfv_{n+1} \in V - \spn(V')\text{.}\) By the Linear Dependence Lemma (Theorem 5.1.19), the set \(V_1 = V' \cup \{\bfv_{n+1} \}\) is linearly independent.
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We can repeat this process. If \(V = \spn(V_1)\text{,}\) we are done; otherwise, we create \(V_2 = V_1 \cup \{\bfv_{n+2} \}\) in the same fashion that we created \(V_1\text{.}\) We can continue doing this, adding one vector at a time to this set and maintaining linear independence. Eventually we must reach the point where \(V = \spn\{ \bfv_1, \ldots, \bfv_n, \bfv_{n+1}, \ldots, \bfv_{n+k} \}\text{,}\) since otherwise Lemma 5.3.1 would imply that \(V\) is infinite-dimensional.
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We now have the machinery necessary to state the following theorem. The proof will appear in the exercises.
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The final result of this section collects some facts about dimension and subspaces which we will use in some of the sections that follow.
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Proof.

We will prove these facts in order. If the subspace \(U\) is \(\{\bfo \}\text{,}\) then we have nothing to prove. If not, then there is some non-zero vector \(\bfu_1 \in U\text{.}\) If \(U = \spn\{\bfu_1\}\text{,}\) we are done; if not, then there exists \(\bfu_2 \in U - \spn\{\bfu_1\}\text{.}\) By Theorem 5.1.19, the set \(\{\bfu_1, \bfu_2\}\) is linearly independent. We can continue to repeat this process. At each stage we have a linearly independent set \({U_k = \{\bfu_1, \ldots, \bfu_k\}}\text{,}\) and this cannot continue indefinitely since \(U\) is a subspace of \(V\text{,}\) which is finite-dimensional. Thus this process must eventually stop when \(U = \spn(U_j)\) for some \(j\text{,}\) and that proves that \(U\) is finite dimensional.
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The space \(U\) is finite dimensional, so it has a basis \(B\text{.}\) This is a linearly independent set of vectors in \(V\text{,}\) so Theorem 5.3.16 says that \(B\) can be extended to a basis \(B'\) of \(V\text{.}\) This means that \(B'\) will have at least as many vectors in it as \(B\text{,}\) so \(\dim(V) \ge \dim(U)\text{.}\)
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If \(U=V\) it is obvious that \(\dim(U)=\dim(V)\text{,}\) so we only need to prove the claim in the other direction. We will prove the contrapositive, so we assume \(U \neq V\text{.}\) Let \(B = \{\bfu_1, \ldots, \bfu_n \}\) be a basis for \(U\text{.}\) Since \(U \neq V\text{,}\) there exists a vector \(\bfv_1 \in V - \spn(B)\text{.}\) By Theorem 5.1.19 the set \(\{\bfu_1, \ldots, \bfu_n, \bfv_1 \}\) is linearly independent in \(V\text{,}\) implying that \(\dim(V) \ge n+1\text{.}\) Therefore \({\dim(U) \neq \dim(V)}\text{.}\)
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Example 5.3.19.

We can apply this latest result to the vector space \(\rr^3\text{.}\) The familiar subspaces of \(\rr^3\) are all of the subspaces of \(\rr^3\text{.}\)
  1. The only subspace of dimension 0 in \(\rr^3\) is the zero subspace \(\{\bfo \}\text{.}\)
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  2. One-dimensional subspaces of \(\rr^3\) are lines through the origin. These can all be written as the span of a single (non-zero) vector.
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  3. Two-dimensional subspaces of \(\rr^3\) are planes through the origin. These are all spanned by sets of two linearly independent vectors.
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  4. The only three-dimensional subspace of \(\rr^3\) is \(\rr^3\) itself.
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Reading Questions 5.3.3 Reading Questions

1.

Consider the following vectors in \(\rr^2\text{:}\)
\begin{equation*} \bfv_1 = \begin{bmatrix} 3 \\ 2 \end{bmatrix} \hspace{6pt} \text{and} \hspace{6pt} \bfv_2 = \begin{bmatrix} -2 \\ 1 \end{bmatrix}\text{.} \end{equation*}
By inspection, why is the set \(\{\bfv_1, \bfv_2 \}\) a basis for \(\rr^2\text{?}\) Explain your answer.
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2.

Let \(\bfv_1\text{,}\) \(\bfv_2\text{,}\) \(\bfv_3\text{,}\) and \(\bfv_4\) be vectors in \(\rr^3\text{.}\)
  1. The set \(S = \{\bfv_1, \bfv_2, \bfv_3, \bfv_4 \}\) is not a basis for \(\rr^3\text{,}\) and there’s a very short argument why. What is that argument?
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  2. Must there be a subset of \(S\) which is a basis of \(\rr^3\text{?}\) Why or why not?
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Exercises 5.3.4 Exercises

1.

Find the dimension of the subspace of \(\rr^3\) consisting of all vectors whose first and third coordinates are equal.
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Answer.
The dimension is 2.
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2.

For each of the following sets of vectors in the given vector space, find the dimension of the subspace spanned by that set of vectors.
  1. \(\{\bfv_1, \bfv_2, \bfv_3, \bfv_4\}\) in \(\rr^3\) if
    \begin{equation*} \bfv_1 = \begin{bmatrix} 2 \\ -3 \\ 3 \end{bmatrix}, \hspace{6pt} \bfv_2 = \begin{bmatrix} 3 \\ 5 \\ -3 \end{bmatrix}, \hspace{6pt} \bfv_3 = \begin{bmatrix} 9 \\ -4 \\ 6 \end{bmatrix}, \hspace{6pt} \bfv_4 = \begin{bmatrix} 6.5 \\ -5 \\ 6 \end{bmatrix} \end{equation*}
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  2. \(\{\bfv_1, \bfv_2, \bfv_3, \bfv_4\}\) in \(\ff_5^3\) if
    \begin{equation*} \bfv_1 = \begin{bmatrix} 2 \\ 0 \\ 0 \end{bmatrix}, \hspace{6pt} \bfv_2 = \begin{bmatrix} 0 \\ 3 \\ 2 \end{bmatrix}, \hspace{6pt} \bfv_3 = \begin{bmatrix} 4 \\ 2 \\ 3 \end{bmatrix}, \hspace{6pt} \bfv_4 = \begin{bmatrix} 3 \\ 1 \\ 1 \end{bmatrix} \end{equation*}
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3.

For each of the following matrices \(A\text{,}\) determine the dimensions of \(\nll(A)\) and \(\col(A)\text{.}\)
  1. \(A \in M_{4,5}(\rr)\text{,}\)
    \begin{equation*} A = \begin{bmatrix} 1 \amp -2.5 \amp 4 \amp 1 \amp 2.5 \\ -2.5 \amp 6 \amp -9.5 \amp -4.5 \amp -8 \\ 5 \amp 3 \amp -11 \amp 3 \amp -5 \\ 2 \amp -1 \amp 0 \amp -4 \amp -5 \end{bmatrix} \end{equation*}
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  2. \(A \in M_{2,4}(\ff_5)\text{,}\)
    \begin{equation*} A = \begin{bmatrix} 3 \amp 1 \amp 3 \amp 0 \\ 1 \amp 2 \amp 0 \amp 1 \end{bmatrix} \end{equation*}
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4.

Determine whether the following statements are true or false. Justify your answer either way.
  1. If a set \(\{\bfv_1, \ldots, \bfv_m \}\) spans a finite-dimensional space \(V\text{,}\) and if \(V'\) is a set of more than \(m\) vectors in \(V\text{,}\) then \(V'\) is linearly dependent.
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  2. The vector space \(\rr^2\) is a subspace of \(\rr^3\text{.}\)
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  3. A vector space is infinite-dimensional if it is spanned by an infinite set.
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Answer.
  1. This is true by Lemma 5.3.1.
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  2. This is false, as \(\rr^2\) is not even a subset of \(\rr^3\text{.}\)
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  3. This is false. The vector space \(\rr^2\) is a counter-example, as we know that \(\dim(\rr^2)=2\) but it is spanned by an infinite set (the entire vector space).
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5.

Determine whether the following statements are true or false. Justify your answer either way.
  1. If \(\dim(V) = m\text{,}\) then there exists a spanning set of \(m+1\) vectors in \(V\text{.}\)
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  2. If every set of \(m\) vectors in \(V\) fails to span \(V\text{,}\) then \(\dim(V) > m\text{.}\)
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  3. If \(m \ge 2\) and \(\dim(V) = m\text{,}\) then every set of \(m-1\) non-zero vectors in \(V\) is linearly independent.
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6.

The first four Hermite polynomials are \(1\text{,}\) \(2t\text{,}\) \(-2 + 4t^2\text{,}\) and \(-12t+8t^3\text{.}\) Show that the set of these polynomials is a basis for \(P_3\text{.}\)
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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.
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7.

The first four Laguerre polynomials are \(1\text{,}\) \(1-t\text{,}\) \(2-4t+t^2\text{,}\) and \({6-18t+9t^2-t^3}\text{.}\) Show that the set of these polynomials is a basis for \(P_3\text{.}\)
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Writing Exercises

8.
Let \(A\) be a matrix.
  1. Prove that \(\dim(\nll(A))\) is the number of non-pivot columns in \(A\text{.}\)
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  2. Prove that \(\dim(\col(A))\) is the number of pivot columns of \(A\text{.}\)
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Answer.
  1. 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.
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  2. 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 \(A\text{.}\)
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9.
Let \(V\) be the set of all functions \(\rr\to\rr\text{.}\) Prove that \(V\) is infinite-dimensional.
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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.
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10.
Suppose that \(T:V\to W\) is a linear transformation between vector spaces and that \(V\) is finite-dimensional. Prove that \(\dim(\range(T)) \le \dim(V)\text{.}\)
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11.
Prove that \(\cc^2\) is two-dimensional as a vector space over \(\cc\) but four-dimensional as a vector space over \(\rr\text{.}\)
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