Row space and rank of a matrix

Andrew Thangaraj

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 \(F^m\)
what about the rows?

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 \(\{0\}\).