# Row space and rank of a matrix

Aug-Nov 2020

## Recap

• Vector space $V$ over a scalar field $F$
• $F$: real field $\mathbb{R}$ or complex field $\mathbb{C}$ in this course
• $m\times n$ matrix A represents a linear map $T:F^n\to F^m$
• null$(A)=$ null $T=\{v\in V:Tv=0\}$, colspace$(A)$ = range $T=\{Tv:v\in V\}$
• dim null $T$ + dim range $T$ = dim $V$
• Linear equation: $Ax=b$
• Solved using elementary row operations
• Solution (if it exists): $u+$ null$(A)$
• Linear map $T$ induces a one-to-one map $V/\text{null }T\to \text{range }T$

## Row space

$A$: $m\times n$ matrix over $F$

• represents a linear map $T:F^n\to F^m$

• column space is range $T$, subspace of $F^m$

rowspace $A=$ span$\{$rows of $A\}$ is a subspace of $F^n$.

• rank or column rank $A=$ dim colspace $A$

• what about dim rowspace $A$?

## Row rank equals column rank

Let $A$ be an $m\times n$ matrix. Then, dim rowspace $A$ $=$ dim colspace $A$. Popularly known as: row rank equals column rank

Proof

• Suppose column rank is $r$. Let colspace $A=$ span$\{v_1,\ldots,v_r\}$

• Write $j$-th column of $A$ as $c_{1j}v_1+\cdots+c_{rj}v_r$

$A=\begin{bmatrix} \vdots&\cdots&\vdots\\ v_1&\cdots&v_r\\ \vdots&\cdots&\vdots \end{bmatrix}\begin{bmatrix} c_{11}&\cdots&c_{1n}\\ \vdots&\cdots&\vdots\\ c_{r1}&\cdots&c_{rn} \end{bmatrix}$

• Each row of $A$ is a linear combination of $r$ vectors; so, row rank $A\le r=$ col rank $A$

• Use above result for $A^T$: row rank $A^T\le$ col rank $A^T$, or col rank $A\le$ row rank $A$

## Matrix, transpose, invertibility, rank

• $A:m\times n$
• $A^T$: represents a linear map from $F^m\to F^n$
• row space $A=$ col space $A^T$
• col space $A=$ row space $A^T$
• rank $A=$ rank $A^T$

Square, $m=n$

• $A$ invertible $\leftrightarrow$ columns are linearly independent $\leftrightarrow$ rows are linearly independent
• $A$ invertible $\leftrightarrow$ columns form a basis for $F^n$ $\leftrightarrow$ rows form a basis for $F^n$
• $A$ invertible $\leftrightarrow$ rank $A=n$
• $A$ invertible $\leftrightarrow$ $A^T$ invertible

Non-square, $m\ne n$

• rank $A\le m$, rank $A\le n$
• rank $A\le$ min$(m,n)$

## Rank and invertible operators

$T:V\to W$ linear map with rank $T=$ dim range $T$. Suppose $S:V\to V$ and $U:W\to W$ are invertible operators. Then,

rank $T=$ rank $TS=$ rank $UT=$ rank $UTS$

• Is $T=TS$? Is $T=UT$? Is $T=UTS$? Not equal, in general.

$A:m\times n$ matrix, $B:n\times n$ invertible matrix, $C:m\times m$ invertible matrix.

rank $A=$ rank $AB=$ rank $CA=$ rank $CAB$

• Elementary row operations do not change the row space

• Elementary row operations do not change the rank

• Elementary row operations can change the column space

• Elementary column operations: elementary row operations on $A^T$

## Result of elementary row operations

For any matrix $A$, there are elementary row operations $E_i$ such that $\left(\prod_i E_i\right)A=\begin{bmatrix} 0&\cdots&0&1&\ast&\cdots&\ast&\ast&\ast&\cdots&\ast&\ast&\cdots&\cdots\\ 0&\cdots&0&0& 0 &\cdots& 0 & 1 &\ast&\cdots&\ast&\ast&\cdots&\cdots\\ 0&\cdots&0&0& 0 &\cdots& 0 & 0 & 0 &\cdots& 0 & 1 &\cdots&\cdots\\ 0&\cdots&0&0& 0 &\cdots& 0 & 0 & 0 &\cdots& 0 & 0 &\cdots&\cdots\\ \vdots&\vdots&\vdots&\vdots& \vdots &\vdots& \vdots & \vdots & \vdots &\vdots& \vdots & \vdots &\vdots &\vdots\\ \end{bmatrix}$

rank $A$ $=$ number of nonzero pivots

dim null $A=n-$rank $A$

Basis of null $A$: found by solving $Ax=0$ using above form

## Echelon form of a matrix

Let $A$ be an $m\times n$ matrix. There are elementary row operations $E_i$ and column operations $F_j$ such that $\left(\prod_i E_i\right)A\left(\prod_j F_j\right)=\begin{bmatrix} 1&0&\cdots&0&0&\cdots&0\\ 0&1&\cdots&0&0&\cdots&0\\ \vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\vdots\\ 0&0&\cdots&1&0&\cdots&0\\ 0&0&\cdots&0&0&\cdots&0\\ \vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\ 0&0&\cdots&0&0&\cdots&0\\ \end{bmatrix},$ where the number of 1s equals the rank of $A$.

• Elementary row operations: get upper triangular form

• Do column swaps to get nonzero pivots together

• Do column operations to make values above and right of pivots zero

## Left null space of a matrix

Let $A$ be an $m\times n$ matrix. Then, left null $A=\{x\in F^m: xA=0\}$.

left null $A=$ null $A^T$

• Fundamental theorem on $A$ $n=\text{col rank }A+\text{nullity }A$

• Fundamental theorem on $A^T$ $m=\text{row rank }A+\text{left nullity }A$

• Since row rank equals column rank… $n=\text{row rank }A+\text{nullity }A$

• row space $A$, null $A$ are subspaces of $F^n$

## Row space and null space

If $F=\mathbb{R}$, the intersection of row space $A$ and null $A$ is equal to $\{0\}$.

Proof

• Suppose $v=(v_1,\ldots,v_n)\in$ row space $A$. Then, there exists $x=(x_1,\ldots,x_m)$ such that $v=xA.$

• Multiply both sides by $v^T$. $vv^T=xAv^T=0$

What about $F$ being the field of complex numbers? What is going on with $A^T$? Answers later…