# 5.4: Dimension

- Page ID
- 1625

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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.

### Contributors

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

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