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 $\mathbb {R} ^{m}$

$\mathrm {rank} (A)=\dim({\mbox{row space of }}A)$

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

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

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

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

Properties of Row and Column Space

$\mathrm {Span} \{{\mbox{columns of }}A\}=\mathbb {R} ^{m}\implies n\geq m$

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

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

Corollary: the following are equivalent for a nonsingular square matrix

• $|A|=|A^{T}|\neq 0$
• 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 $U$.

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

• 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 $\mathrm {rank} (A)+\mathrm {nullity} (A)=n$

$\mathrm {nullity} (A)=\dim(N(A))$

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

$\mathrm {rank} (A)=\mathrm {rank} (U)={\mbox{num. nonzero rows}}=r$

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

Example

$A={\begin{bmatrix}1&2&-1&1\\2&4&-3&0\\1&2&1&5\end{bmatrix}}\longrightarrow U={\begin{bmatrix}1&2&0&3\\0&0&1&2\\0&0&0&0\end{bmatrix}}$

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

$\mathrm {rank} (A)=2$, and there are $4-2=2$ free variables.

${\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\\x_{4}\end{bmatrix}}={\begin{bmatrix}-2\alpha -3\beta \\\alpha \\-2\beta \\\beta \end{bmatrix}}=\alpha {\begin{bmatrix}-2\\1\\0\\0\end{bmatrix}}+\beta {\begin{bmatrix}-3\\0\\-2\\1\end{bmatrix}}$

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

$\mathrm {rank} (A)=r={\mbox{num. lead vars}}={\mbox{num. nonzero rows}}$

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 $A_{L}{\vec {x}}={\vec {0}}$, then $U_{L}{\vec {x}}={\vec {0}}$ and ${\vec {x}}={\vec {0}}$ 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

$A={\begin{bmatrix}1&-2&1&1&2\\-1&3&0&2&-2\\0&1&1&3&4\\1&2&5&13&5\end{bmatrix}}\quad \longrightarrow \quad U={\begin{bmatrix}1&-2&1&1&2\\0&1&1&3&0\\0&0&0&0&1\\0&0&0&0&0\end{bmatrix}}$.

Thus ${\vec {u}}_{1},{\vec {u}}_{2},{\vec {u}}_{5}$ form a basis for the column space of $U$, and ${\vec {a}}_{1},{\vec {a}}_{2},{\vec {a}}_{5}$ 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. $L({\vec {x}})=2{\vec {x}}$.
2. $L({\vec {x}})=x_{1}{\vec {e}}_{1}$ (projection onto $x$-axis)
3. $L({\vec {x}})=\left\langle x_{1},-x_{2}\right\rangle$ (reflect vector about $x$-axis)
4. $L({\vec {x}})=\left\langle -x_{2},x_{1}\right\rangle$ (rotate by 90° CCW)