Vector Sapces

4.4 The Dimension of a Vector Space

Theorem 9:

If a vector space $V$ has a basis $B=\left\{ \textbf{b}_1, \cdots, \textbf{b}_n \right\}$, then any set in $V$ containing more than $n$ vectors must be linearly dependent.

Proof:

• Let $\left\{ \textbf{u}_1, \cdots, \textbf{u}_p \right\}$ be a set in $V$ with more than $n$ vectors.

• The coordinate vectors $[\textbf{u}_1]_B, \cdots,[\textbf{u}_p]_B$ form a linearly dependent set in $\mathbb{R}^n$, because there are more vectors ($p$) than entries ($n$) in each vector.

• So there exist scalars $c_1, \cdots, c_p$, not all zero, such that

$$c_1[\textbf{u}_1]_B + \cdots + c_p [\textbf{u}_p]_B = \begin{bmatrix} 0 \\ \vdots \\ 0 \end{bmatrix} \text{The zero vector in }\mathbb{R}^n$$

• Since the coordinate mapping is a linear transformation,

$$\left[ c_1 \textbf{u}_1 + \cdots + c_p \textbf{u}_p \right]_B = \begin{bmatrix} 0 \\ \vdots \\ 0 \end{bmatrix}$$

• The zero vector on the right displays the $n$ weights needed to build the vector $c_1 \textbf{u}_1 + \cdots + c_p \textbf{u}_p$ from the basis vectors in $B$.
• That is, $c_1 \textbf{u}_1 + \cdots + c_p \textbf{u}_p = 0\textbf{b}_1 +\cdots 0\textbf{b}_n = \textbf{0}$.
• Since the $c_i$ are not all zero, $\left\{ \textbf{u}_1, \cdots, \textbf{u}_p \right\}$ is linearly dependent.

Theorem 9 implies that if a vector space $V$ has a basis $B=\left\{ \textbf{b}_1, \cdots, \textbf{b}_n \right\}$ then each linearly independent set in $V$ has no more than $n$ vectors.

Theorem 10:

If a vector space $V$ has a basis of $n$ vectors, then every basis of $V$ must consist of exactly $n$ vectors.

Proof:

• Let $B_1$ be a basis of $n$ vectors and $B_2$ be any other basis (of $V$).
• Since $B_1$ is a basis and $B_2$ is linearly independent, $B_2$ has no more than $n$ vectors, by Theorem 9.
• Also, since $B_2$ is a basis and $B_1$ is linearly independent, $B_2$ has at least $n$ vectors. ref.
• Thus $B_2$ consists of exactly $n$ vectors.

Definition

• If $V$ is spanned by a finite set, then $V$ is said to be finite-dimensional, and
• the dimension of $V$, written as dim $V$, is the number of vectors in a basis for $V$.
• The dimension of the zero vector space $\left\{\textbf{0}\right\}$ is defined to be zero.
• If $V$ is not spanned by a finite set, then $V$ is said to be infinite-dimensional.

Example 3

Find the dimension of the subspace

$$H = \left\{ \begin{bmatrix} a-3b+6c \\ 5a+4d \\ b-2c-d \\ 5d \end{bmatrix} :a,b,c,d \in \mathbb{R} \right\}$$

Solution:

• $H$ is the set of all linear combinations of the vectors $$\textbf{v}_1 = \begin{bmatrix} 1 \\ 5 \\ 0 \\ 0 \end{bmatrix}, \textbf{v}_2 = \begin{bmatrix} -3 \\ 0 \\ 1 \\ 0 \end{bmatrix}, \textbf{v}_3 = \begin{bmatrix} 6 \\ 0 \\ -2 \\ 0 \end{bmatrix}, \textbf{v}_4 = \begin{bmatrix} 0 \\ 4 \\ -1 \\ 5 \end{bmatrix}$$

• Clearly $\textbf{v}_1 \ne \textbf{0}, \textbf{v}_2$ is not a multiple of $\textbf{v}_1$, but $\textbf{v}_3$ is a multiple of $\textbf{v}_2$.

• By the Spanning Set Theorem, we may discard $\textbf{v}_3$ and still have a set that spans $H$.
• Finally $\textbf{v}_4$ is not a linear combination of $\textbf{v}_1$ and $\textbf{v}_2$. So $\left\{ \textbf{v}_1, \textbf{v}_2, \textbf{v}_4 \right\}$ is linearly independent and hence is a basis for H.
• Thus dim $H$=3.

Theorem 11:

Let $H$ be a subspace of a finite-dimensional vector space $V$. Any linearly independent set in $H$ can be expanded, if necessary, to a basis for $H$. Also, $H$ is finite-dimensional and $\text{dim }H \le \text{dim }V$.

Proof:

• If $H=\left\{ \textbf{0} \right\}$, then certainly $\text{dim }H=0 \le \text{dim }V$.
• Otherwise, let $S=\left\{ \textbf{u}_1, \cdots, \textbf{u}_k \right\}$ be any linearly independent set in $H$.
• If $S$ spans $H$, then $S$ is a basis for $H$.
• Otherwise, there is some $\textbf{u}_{k+1}$ in $H$ that is not in Span $S$.
• But then $\left\{ \textbf{u}_1, \cdots ,\textbf{u}_k, \textbf{u}_{k+1} \right\}$ will be linearly independent, because no vector in the set can be a linear combination of vectors that precede it (by Theorem 4.
• So long as the new set does not span $H$, we can continue this process of expanding $S$ to a larger linearly independent set in $H$.
• But the number of vectors in a linearly independent expansion of $S$ can never exceed the dimension of $V$, by Theorem 9.
• So eventually the expansion of $S$ will span $H$ and hence will be a basis for $H$, and $\text{dim }H \le \text{dim }V$.

Theorem 12: Basis Theorem

Let $V$ be a $p$-dimensional vector space, $p\ge 1$. Any linearly independent set of exactly $p$ elements in $V$ is automatically a basis for $V$. Any set of exactly $p$ elements that spans $V$ is automatically a basis for $V$.

Proof:

• By Theorem 11, a linearly independent set $S$ of $p$ elements can be extended to a basis for $V$.
• But that basis must contain exactly $p$ elements, since $\text{dim }V=p$.
• So $S$ must already be a basis for $V$.
• Now suppose that $S$ has $p$ elements and spans $V$.
• Since $V$ is nonzero, the Spanning Set Theorem implies that a subset $S'$ of $S$ is a basis of $V$.
• Since $\text{dim }V=p, S'$ must contain $p$ vectors.
• Hence $S=S'$.

The Dimensions of Nul A and Col A

• Let $A$ be an $m \times n$ matrix, and suppose the equation $A\textbf{x}=\textbf{0}$ has $k$ free variables.
• A spanning set for Nul $A$ will produce exactly $k$ linearly independent vectors—say, $\textbf{u}_1, \cdots, \textbf{u}_k$ —one for each free variable.
• So $\left\{ \textbf{u}_1, \cdots, \textbf{u}_k \right\}$ is a basis for Nul $A$, and the number of free variables determines the size of the basis.
• Thus, the dimension of Nul $A$ is the number of free variables in the equation $A\textbf{x}=\textbf{0}$ and the dimension of Col $A$ is the number of pivot columns in $A$.

Example 5:

Find the dimensions of the null space and the column space of

$$A=\begin{bmatrix} -3 & 6 & -1 & 1 & -7 \\ 1 & -2 & 2 & 3 & -1 \\ 2 & -4 & 5 & 8 & -4 \end{bmatrix}$$

Solution:

• Row reduce the augmented matrix $\begin{bmatrix} A & \textbf{0} \end{bmatrix}$ to echelon form:

$$\begin{bmatrix} 1 & -2 & 2 & 3 & -1 & 0 \\ 0 & 0 & 1 & 2 & -2 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix}$$

• There are three free variables— $x_2, x_4$, and $x_5$.
• Hence the dimension of Nul $A$ is 3.
• Also dim Col $A$ =2 because $A$ has two pivot columns.