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4.6 Dimension: Class Notes

Theorem 4.6.2

Let $V$ be a finite-dimensional vector space, and let $\{{𝐯}_1,{𝐯}_2,\dots ,{𝐯}_n\}$ be any basis for $V$.
a) If a set in $V$ has more than $n$ vectors, then it is linearly dependent.
b) If a set in $V$ has fewer than $n$ vectors, then it does not span $V$.

The proof of this essentially involves counting variables and equations in a linear system.

Theorem 4.6.1

All bases for a finite-dimensional vector space have the same number of vectors.

Definition

The dimension of a finite-dimensional vector space $V$ is denoted by $\dim (V)$ and is defined to be the number of vectors in a basis for $V$ (in some physical contexts, “dimension” is referred to as degrees of freedom). In addition, the zero vector space is defined to have dimension zero.

Recall that the basis for the zero vector space is $\varnothing$.

Solution Spaces of Homogeneous Systems

Find a basis for the solution space of each homogeneous linear system, and find the dimension of that space.

#1

Solution

Row reduce the coefficient matrix:

$$\left[\begin{array}{ccc}1& 1& -1\\ -2& -1& 2\\ -1& 0& 1\end{array}\right]⟶\left[\begin{array}{ccc}1& 1& -1\\ 0& 1& 0\\ 0& 0& 0\end{array}\right]⟶\left[\begin{array}{ccc}1& 0& -1\\ 0& 1& 0\\ 0& 0& 0\end{array}\right]$$

So $x_1=x_3$ and $x_2=0$, with $x_3=t$ free:

$$\overrightarrow{x}=\left[\begin{array}{c}t\\ 0\\ t\end{array}\right]=\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]t.$$

#5

Solution

$$\left[\begin{array}{ccc}2& -6& 2\\ 3& -9& 3\end{array}\right]⟶\left[\begin{array}{ccc}1& -3& 1\\ 0& 0& 0\end{array}\right]$$

$x_1=3x_2-x_3$, so with $x_2=s$ and $x_3=t$,

$$\overrightarrow{x}=\left[\begin{array}{c}3s-t\\ s\\ t\end{array}\right]=\left[\begin{array}{c}3\\ 1\\ 0\end{array}\right]s+\left[\begin{array}{c}-1\\ 0\\ 1\end{array}\right]t.$$

#8

In each part, find a basis for the given subspace of $R^4$, and state its dimension.
a. All vectors of the form $(a,b,c,0)$.
b. All vectors of the form $(a,b,c,d)$, where $d=a+b$ and $c=a-b$.
c. All vectors of the form $(a,b,c,d)$, where $a=b=c=d$.

Solution

a)

$$(a,b,c,0)=a(1,0,0,0)+b(0,1,0,0)+c(0,0,1,0)$$

b)

$$\left[\begin{array}{c}a\\ b\\ c\\ d\end{array}\right]=\left[\begin{array}{c}a\\ b\\ a-b\\ a+b\end{array}\right]=\left[\begin{array}{c}1\\ 0\\ 1\\ 1\end{array}\right]a+\left[\begin{array}{c}0\\ 1\\ -1\\ 1\end{array}\right]b$$

c)

$$\left[\begin{array}{c}a\\ b\\ c\\ d\end{array}\right]=\left[\begin{array}{c}a\\ a\\ a\\ a\end{array}\right]=\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right]a$$

Theorem 4.6.4

Let $V$ be an $n$-dimensional vector space, and let $S$ be a set in $V$ with exactly $n$ vectors. Then $S$ is a basis for $V$ if and only if $S$ spans $V$ or $S$ is linearly independent.

Tip: This proof is optional.

Proof: ($\Rightarrow$) If $S$ is a basis for $V$, then $S$ spans $V$ and is linearly independent by definition.

($\Leftarrow$) Suppose $S$ has exactly $n$ vectors and spans $V$. If $S$ is not linearly independent, then $\exists$ (there exists) ${\overrightarrow{v}}_n\in S$ $\ni$ (such that) ${\overrightarrow{v}}_n=\sum _{i=1}^{n-1}{k}_i{\overrightarrow{v}}_i$. But then ${\overrightarrow{v}}_n$ can be removed without affecting $\text{span}(S)$, contradicting that $V$ is $n$-dimensional. Thus, $S$ is independent.

Suppose $S$ has exactly $n$ vectors and is linearly independent. If $S$ does not span $V$, then $\exists$ (there exists) $\overrightarrow{v}\in V$ $\ni$ (such that) $\overrightarrow{v}\notin \text{span}(S)$. Then $S\cup \{\overrightarrow{v}\}$ is independent, again contradicting that $V$ is $n$-dimensional. Thus $S$ spans $V$.

Theorem 4.6.5

Let $S$ be a finite set of vectors in a finite-dimensional vector space $V$.
a) If $S$ spans $V$ but is not a basis for $V$, then $S$ can be reduced to a basis for $V$ by removing appropriate vectors from $S$.
b) If $S$ is a linearly independent set that is not already a basis for $V$, then $S$ can be enlarged to a basis for $V$ by inserting appropriate vectors into $S$.

#12

Find standard basis vectors for $R^3$ that can be added to the set $\{{𝐯}_1,{𝐯}_2\}$ to produce a basis for $R^3$.
a. ${𝐯}_1=(-1,2,3)$, ${𝐯}_2=(1,-2,-2)$
b. ${𝐯}_1=(1,-1,0)$, ${𝐯}_2=(3,1,-2)$

Solution

a) Put ${\overrightarrow{v}}_1$, ${\overrightarrow{v}}_2$, and the standard basis vectors in the columns of a matrix and row reduce:

$$\left[\begin{array}{ccccc}\hfill {\displaystyle -1}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle -2}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 3}\hfill & \hfill {\displaystyle -2}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill \end{array}\right]$$

$R_2\to R_2+2R_1$ and $R_3\to R_3+3R_1$, then $R_2\leftrightarrow R_3$:

$$\left[\begin{array}{ccccc}\hfill {\displaystyle -1}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 3}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 2}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill \end{array}\right]$$

Pivot columns are ${\overrightarrow{c}}_1,{\overrightarrow{c}}_2,{\overrightarrow{c}}_3$, so the desired basis is

Note: this answer is not unique. $(0,1,0)$ would also work.

#18

Find a basis for the subspace of $R^4$ that is spanned by the vectors ${𝐯}_1=(1,1,1,1)$, ${𝐯}_2=(2,2,2,0)$, ${𝐯}_3=(0,0,0,3)$, ${𝐯}_4=(3,3,3,4)$.

Solution

Put the vectors in the columns of a matrix and row reduce:

$$\left[\begin{array}{cccc}1& 2& 0& 3\\ 1& 2& 0& 3\\ 1& 2& 0& 3\\ 1& 0& 3& 4\end{array}\right]⟶\left[\begin{array}{cccc}1& 2& 0& 3\\ 1& 0& 3& 4\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right]⟶\left[\begin{array}{cccc}1& 2& 0& 3\\ 0& 2& -3& -1\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right]$$

Since the pivot columns are ${\overrightarrow{c}}_1$ and ${\overrightarrow{c}}_2$, the desired basis is

Note: ${\displaystyle {\overrightarrow{v}}_3=3{\overrightarrow{v}}_1-\frac{3}{2}{\overrightarrow{v}}_2}$ and ${\displaystyle {\overrightarrow{v}}_4=4{\overrightarrow{v}}_1-\frac{1}{2}{\overrightarrow{v}}_2}$. Again, the answer is not unique.

#20

In each part, let $T_A$ be multiplication by $A$ and find the dimension of the subspace of $R^4$ consisting of all vectors $𝐱$ for which $T_A(𝐱)=\mathrm{𝟎}$.

a. $\left[\begin{array}{cccc}1& 0& 2& -1\\ -1& 4& 0& 0\end{array}\right]$    b. $\left[\begin{array}{cccc}0& 0& 1& 1\\ -1& 1& 0& 0\\ 1& 0& 0& 1\end{array}\right]$

Solution

Want: $\overrightarrow{x}$ such that $A\overrightarrow{x}=\overrightarrow{0}$.

a)

$$\left[\begin{array}{cccc}1& 0& 2& -1\\ 0& 4& 2& -1\end{array}\right]⟶\left[\begin{array}{cccc}1& 0& 2& -1\\ 0& 1& \frac{1}{2}& -\frac{1}{4}\end{array}\right]$$

$${\displaystyle \begin{array}{cc}\hfill {\displaystyle x_1}& {\displaystyle =-2x_3+x_4}\hfill \\ \hfill {\displaystyle x_2}& {\displaystyle =-\frac{1}{2}{x}_3+\frac{1}{4}{x}_4}\hfill \end{array}}$$

$${\displaystyle \overrightarrow{x}=\left[\begin{array}{c}-2s+t\\ -\frac{1}{2}s+\frac{1}{4}t\\ s\\ t\end{array}\right]=\left[\begin{array}{c}-2\\ -\frac{1}{2}\\ 1\\ 0\end{array}\right]s+\left[\begin{array}{c}1\\ \frac{1}{4}\\ 0\\ 1\end{array}\right]t}$$

b)

$$\left[\begin{array}{ccccc}\hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill \\ \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 0}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 1}\hfill & \hfill {\displaystyle 0}\hfill \end{array}\right]\Rightarrow \begin{array}{cc}\hfill {\displaystyle x_1}& {\displaystyle =-x_4}\hfill \\ \hfill {\displaystyle x_2}& {\displaystyle =-x_4}\hfill \\ \hfill {\displaystyle x_3}& {\displaystyle =-x_4}\hfill \end{array}$$

$$\overrightarrow{x}=\left[\begin{array}{c}-1\\ -1\\ -1\\ 1\end{array}\right]t$$