# 5.4: Dimension

- Page ID
- 1625

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

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.

### Contributors

- Isaiah Lankham, Mathematics Department at UC Davis
- Bruno Nachtergaele, Mathematics Department at UC Davis
- Anne Schilling, Mathematics Department at UC Davis

Both hardbound and softbound versions of this textbook are available online at WorldScientific.com.