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5.4: Dimension

Last updated

Mar 5, 2021
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- 5.3: Bases
- 5.E: Exercises for Chapter 5

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We now come to the important definition of the dimension of a finite-dimensional vector space. Intuitively, we know that $\mathbb{R}^2$ has dimension 2, that $\mathbb{R}^3$ has dimension 3, and, more generally, that $\mathbb{R}^n$ has dimension $n$ . This is precisely the length of every basis for each of these vector spaces, which prompts the following definition.

Definition 5.4.1

We call the length of any basis for $V$ (which is well-defined by Theorem 5.4.2 below) the dimension of $V$ , and we denote this by $\dim(V)$ .

Note that Definition 5.4.1 only makes sense if, in fact, every basis for a given finite-dimensional vector space has the same length. This is true by the following theorem.

Theorem 5.4.2.

Let $V$ be a finite-dimensional vector space. Then any two bases of $V$ have the same length

Proof

Let $(v_1,\ldots,v_m)$ and $(w_1,\ldots,w_n)$ be two bases of $V$ . Both span $V$ .

By Theorem 5.2.9, we have $m\le n$ since $(v_1,\ldots,v_m)$ is linearly independent. By the same theorem, we also have $n\le m$ since $(w_1,\ldots,w_n)$ is linearly independent. Hence $n=m$ , as asserted.

Example 5.4.3. $\dim(\mathbb{F}^n)=n$ and $\dim(\mathbb{F}_m[z]) = m + 1$ . Note that $\dim(\mathbb{C}^n)=n$ as a complex vector space, whereas $\dim(\mathbb{C}^n)=2n$ as an real vector space. This comes from the fact that we can view $\mathbb{C}$ itself as an real vector space of dimension 2 with basis $(1,i)$ .

Theorem 5.4.4. Let $V$ be a finite-dimensional vector space with $\dim(V)=n$ . Then:

If $U\subset V$ is a subspace of $V$ , then $\dim(U) \le \dim(V)$ .
If $V=\Span(v_1,\ldots,v_n)$ , then $(v_1,\ldots,v_n)$ is a basis of $V$ .
If $(v_1,\ldots,v_n)$ is linearly independent in $V$ , then $(v_1,\ldots,v_n)$ is a basis of $V$ .

Point 1 implies, in particular, that every subspace of a finite-dimensional vector space is finite-dimensional. Points 2 and 3 show that if the dimension of a vector space is known to be $n$ , then, to check that a list of $n$ vectors is a basis, it is enough to check whether it spans $V$ (resp. is linearly independent).

Proof.

To prove Point~1, first note that $U$ is necessarily finite-dimensional (otherwise we could find a list of linearly independent vectors longer than $\dim(V)$ ). Therefore, by Corollary 5.3.6, $U$ has a basis $(u_1,\ldots,u_m)$ (say). This list is linearly independent in both $U$ and $V$ . By the Basis Extension Theorem 5.3.7, we can extend $(u_1,\ldots,u_m)$ to a basis for $V$ , which is of length $n$ since $\dim(V)=n$ . This implies that $m\le n$ , as desired.

To prove Point~2, suppose that $(v_1,\ldots,v_n)$ spans $V$ . Then, by the Basis Reduction Theorem 5.3.4, this list can be reduced to a basis. However, every basis of $V$ has length $n$ ; hence, no vector needs to be removed from $(v_1,\ldots,v_n)$ . It follows that $(v_1,\ldots,v_n)$ is already a basis of $V$ .

Point~3 is proven in a very similar fashion. Suppose $(v_1,\ldots,v_n)$ is linearly independent. By the Basis Extension Theorem 5.3.7, this list can be extended to a basis. However, every basis has length $n$ ; hence, no vector needs to be added to $(v_1,\ldots,v_n)$ . It follows that $(v_1,\ldots,v_n)$ is already a basis of $V$ .

We conclude this chapter with some additional interesting results on bases and dimensions. The first one combines the concepts of basis and direct sum.

Theorem 5.4.5. Let $U\subset V$ be a subspace of a finite-dimensional vector space $V$ . Then there exists a subspace $W\subset V$ such that $V=U\oplus W$ .

Proof.

Let $(u_1,\ldots,u_m)$ be a basis of $U$ . By Theorem 5.4.4(1), we know that $m\le \dim(V)$ . Hence, by the Basis Extension Theorem 5.3.7, $(u_1,\ldots,u_m)$ can be extended to a basis $(u_1,\ldots,u_m,w_1,\ldots,w_n)$ of $V$ . Let $W=\Span(w_1,\ldots,w_n)$ .

To show that $V=U\oplus W$ , we need to show that $V=U+W$ and $U\cap W=\{0\}$ . Since $V=\Span(u_1,\ldots,u_m,w_1,\ldots,w_n)$ where $(u_1,\ldots,u_m)$ spans $U$ and $(w_1,\ldots,w_n)$ spans $W$ , it is clear that $V=U+W$ .

To show that $U\cap W=\{0\}$ , let $v\in U\cap W$ . Then there exist scalars $a_1,\ldots,a_m, b_1,\ldots,b_n\in\mathbb{F}$ such that

$v=a_1 u_1+\cdots+ a_m u_m = b_1 w_1 + \cdots + b_n w_n,$

or equivalently that

$a_1 u_1+\cdots+ a_m u_m -b_1 w_1 - \cdots - b_n w_n =0.$

Since $(u_1,\ldots,u_m,w_1,\ldots,w_n)$ forms a basis of $V$ and hence is linearly independent, the only solution to this equation is $a_1=\cdots=a_m=b_1=\cdots=b_n=0$ . Hence $v=0$ , proving that indeed $U\cap W=\{0\}$ .

Theorem 5.4.6. If $U,W\subset V$ are subspaces of a finite-dimensional vector space, then

$\dim(U+W) = \dim(U) + \dim(W) - \dim(U\cap W).$

Proof.

Let $(v_1,\ldots,v_n)$ be a basis of $U\cap W$ . By the Basis Extension

Theorem 5.3.7, there exist $(u_1,\ldots,u_k)$ and $(w_1,\ldots,w_\ell)$ such that $(v_1,\ldots,v_n,u_1,\ldots,u_k)$ is a basis of $U$ and $(v_1,\ldots,v_n,w_1,\ldots,w_\ell)$ is a basis of $W$ . It suffices to show that

$\mathcal{B} = (v_1,\ldots,v_n,u_1,\ldots,u_k,w_1,\ldots,w_\ell)$

is a basis of $U+W$ since then

$\dim(U+W) = n+k+\ell= (n+k) + (n+\ell)-n=\dim(U) + \dim(W) -\dim(U\cap W).$

Clearly $\Span(v_1,\ldots,v_n,u_1,\ldots,u_k,w_1,\ldots,w_\ell)$ contains $U$ and $W$ , and hence $U+W$ . To show that $\mathcal{B}$ is a basis, it remains to show that $\mathcal{B}$ is linearly independent. Suppose

$a_1v_1+\cdots+a_n v_n + b_1u_1+\cdots +b_k u_k + c_1w_1+\cdots+c_\ell w_\ell =0, \tag{5.4.1}$

and let $u=a_1v_1+\cdots+a_n v_n + b_1u_1+\cdots +b_k u_k\in U$ . Then, by Equation (5.4.1), we also have that $u=-c_1 w_1-\cdots - c_\ell w_\ell\in W$ , which implies that $u\in U\cap W$ . Hence, there exist scalars $a_1',\ldots,a_n'\in\mathbb{F}$ such that $u=a_1'v_1+\cdots+a_n'v_n$ .

Since there is a unique linear combination of the linearly independent vectors $(v_1,\ldots,v_n,u_1,\ldots,u_k)$ that describes $u$ , we must have $b_1=\cdots=b_k=0$ and $a_1=a_1',\ldots,a_n=a_n'$ . Since $(v_1,\ldots,v_n,w_1,\ldots,w_\ell)$ is also linearly independent, it further follows that $a_1=\cdots=a_n=c_1=\cdots=c_\ell=0$ . Hence, Equation (5.4.1) only has the trivial solution, which implies that $\mathcal{B}$ is a basis.

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