MATH 323 Lecture 15

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Review

$m\times n$ matrix $A$

• Row space is the span of row vectors and a subset of ${\mathbb{ℝ}}^n$
• Column space is span of column vectors and subset of ${\mathbb{ℝ}}^m$

$\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}(A)=\dim (\text{row space of }A)$

Theorem 3.6.1: Two row equivalent matrices have the same row space.

Theorem 3.6.2: The system $A\overrightarrow{x}=\overrightarrow{b}$ has a solution iff $\overrightarrow{b}$ is contained in the column space of $A$.

Theorem 3.6.3: Let $A$ be a $m\times n$ matrix.

• The linear system $A\overrightarrow{x}=\overrightarrow{b}$ is consistent for every $\overrightarrow{b}\in {\mathbb{ℝ}}^n$ iff the column vectors of $A$ span ${\mathbb{ℝ}}^n$.
• The system $A\overrightarrow{x}=\overrightarrow{b}$ has at most one solution for every $\overrightarrow{b}\in {\mathbb{ℝ}}^n$ iff the column vectors of $A$ are linearly independent.

Properties of Row and Column Space

$\mathrm{S}\mathrm{p}\mathrm{a}\mathrm{n}\{\text{columns of }A\}={\mathbb{ℝ}}^m⟹n\ge m$

If Columns of $A$ are linearly independent, then $n\le m$

If columns of $A$ form a basis for ${\mathbb{ℝ}}^m$, then $n=m$

Corollary: the following are equivalent for a nonsingular square matrix

• $|A|=|A^T|\ne 0$
• columns form a basis for ${\mathbb{ℝ}}^n$
• rows form a basis for ${\mathbb{ℝ}}^n$

Given $A\to U$, where $U$ is in row echelon form; the column space of $A$ is NOT equal to the column space of $U$.

$\dim (\{\text{column space of }U\})=\dim (\{\text{column space of }A\})$

• columns containing the lead variables form basis for the column space of $U$
• above corresponds to columns of $A$ that form basis for $A$'s column space

Rank-Nullity Theorem

If $A$ is a $m\times n$ matrix, then $\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}(A)+\mathrm{n}\mathrm{u}\mathrm{l}\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{y}(A)=n$

$\mathrm{n}\mathrm{u}\mathrm{l}\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{y}(A)=\dim (N(A))$

$A$ can be converted to row-echelon form matrix $U$, so $A\overrightarrow{x}=\overrightarrow{0}⟺U\overrightarrow{x}=\overrightarrow{0}$.

$\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}(A)=\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}(U)=\text{num. nonzero rows}=r$

Thus we have $n-r$ and $\dim N(A)$ equal the number of free variables: $r+(n-r)=n$.

Example

$A=\left[\begin{array}{cccc}1& 2& -1& 1\\ 2& 4& -3& 0\\ 1& 2& 1& 5\end{array}\right]⟶U=\left[\begin{array}{cccc}1& 2& 0& 3\\ 0& 0& 1& 2\\ 0& 0& 0& 0\end{array}\right]$

Thus (1, 2, 0, 3) and (0, 0, 1, 2) form a basis for the row space of $A$

$\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}(A)=2$, and there are $4-2=2$ free variables.

$\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\\ x_4\end{array}\right]=\left[\begin{array}{c}-2\alpha -3\beta \\ \alpha \\ -2\beta \\ \beta \end{array}\right]=\alpha \left[\begin{array}{c}-2\\ 1\\ 0\\ 0\end{array}\right]+\beta \left[\begin{array}{c}-3\\ 0\\ -2\\ 1\end{array}\right]$

The vectors (-2, 1, 0, 0) and (-3, 0, -2, 1) form a basis for the null space of $A$, thus $\dim N(A)=2$

Theorem 3.6.6

If $A$ is a $m\times n$ matrix, the dimension of the row space of $A$ equals the dimension of the column space of $A$.

Proof

$\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{k}(A)=r=\text{num. lead vars}=\text{num. nonzero rows}$

Let $U_L$ be the matrix obtained from $U$ by deleting columns corresponding to free vars, and let $A_L$ be obtained by deleting the same columns from $A$. Both matrices are of size $r\times r$

$A_L\sim U_L$, so if $A_L\overrightarrow{x}=\overrightarrow{0}$, then $U_L\overrightarrow{x}=\overrightarrow{0}$ and $\overrightarrow{x}=\overrightarrow{0}$ because columns of $U_L$ are linearly independent. Therefore the columns of $A_L$ are linearly independent.

$A_L$ has $r$ columns, so the dimension of the column space of $A$ ≥ $r$ (and $r$ is also the dimension of the row space of $A$

the column space of

$A^T$

has the same dimension as the row space of

$A$

≥ the row space of

$A^T$

has the same dimension as the column space of

$A$

. By antisymmetry, the dimensions of the column and row spaces must be equal.

Q.E.D.

Example

$A=\left[\begin{array}{ccccc}1& -2& 1& 1& 2\\ -1& 3& 0& 2& -2\\ 0& 1& 1& 3& 4\\ 1& 2& 5& 13& 5\end{array}\right]⟶U=\left[\begin{array}{ccccc}1& -2& 1& 1& 2\\ 0& 1& 1& 3& 0\\ 0& 0& 0& 0& 1\\ 0& 0& 0& 0& 0\end{array}\right]$.

Thus ${\overrightarrow{u}}_1,{\overrightarrow{u}}_2,{\overrightarrow{u}}_5$ form a basis for the column space of $U$, and ${\overrightarrow{a}}_1,{\overrightarrow{a}}_2,{\overrightarrow{a}}_5$ form a basis for the column space of $A$.

$U(1,:),U(2,:),U(3,:)$ form a basis for the row spaces of $U$ and $A$ (since they are equivalent).

The nullity of $A$ is thus the number of columns − num. lead variables = 5 − 3 = 2.

Linear Transformation

Let $V,W$ be vector spaces.

$L:V\to W$ is a linear transformation if for all $v,v_1,v_2\in V$ and for all $\alpha ,\beta \in \mathbb{ℝ}$,

1. $L(v_1+v_2)=L(v_1)+L(v_2)$
2. $L(\alpha v)=\alpha L(v)$
3. (combination of 1 and 2) $L(\alpha v_1+\beta v_2)=\alpha L(v_1)+\beta L(v_2)$

Therefore, if $v\in V$, then $L(v)\in W$, where $L(v)$ is the image of $v$.

Let $L:V\to V$ be a linear operator on $V$.

For example:

1. $L(\overrightarrow{x})=2\overrightarrow{x}$.
2. $L(\overrightarrow{x})=x_1{\overrightarrow{e}}_1$ (projection onto $x$-axis)
3. $L(\overrightarrow{x})=⟨x_1,-x_2⟩$ (reflect vector about $x$-axis)
4. $L(\overrightarrow{x})=⟨-x_2,x_1⟩$ (rotate by 90° CCW)