Lecture 10: Rank of a matrix

10.1 Definition of rank

#####Definition: Rankofthematrix

Description: Let $A$ be an $m\times n$ matrix. If at least one minor of order $k$ is nonzero and every minor of order $k+1$ is zero, then $r(A)=k$. For an $n\times n$ matrix: $$r\left(A\right)=n\Leftrightarrow\left|A\right|\neq0\Leftrightarrow A\text{ is invertible}.$$

Explanation

Minor:
A minor is a determinant formed by selecting the same number of rows and columns.
Minor of order $k$:
For $\left(\begin{matrix}1&2&3\\4&5&6\end{matrix}\right)$, one minor of order $2$ is $\left|\begin{matrix}1&3\\4&6\end{matrix}\right|$.
This matrix has three minors of order $2$.
Meaning: if at least one minor of order $k$ is nonzero and every minor of order $k+1$ is zero, then the largest linearly independent set of rows or columns has size $k$.
Meaning: r(A)=k
There are k linearly independent row vectors and k linearly independent column vectors.
Any k+1 vectors are related;
At this point, r(A)=k is the number of linearly independent vectors that make up the matrix;

Supplement: The geometric meaning of rank

Meaning:
Rank, representing the dimension of the transformed space after the transformation of the A matrix;
The space stretched by the transformed basis vector is the total possible transformation result -> That is: the dimension of the transformed basis vector and the space it stretches;
Example:
When a matrix $A_{32}$ appears with 32, it means mapping a two-dimensional figure into three-dimensional space; Because A has two basis vectors, and these two basis vectors are now three-dimensional [x,y,z];
When a matrix $A_{32}$ appears with 23, it means mapping a three-dimensional figure into a two-dimensional space; Because A has three basis vectors, and these three basis vectors are now two-dimensional [x,y] basis vectors;

10.2 Finding the Rank of a Matrix

Method: Find the rank of the matrix

Transform A into a row-stepped matrix using elementary row transformations, where the number of nonzero rows is the rank of A;

Method: Find the rank of the matrix - Example

Question: If $\begin{bmatrix}1&2&-1&1\\2&0&t&0\\0&-4&5&-2\end{bmatrix}$ has rank 2, find t
Analysis:
$A\rightarrow\left(\begin{matrix}1&2&-1&1\\0&-4&t+2&-2\\0&-4&5&-2\end{matrix}\right)\rightarrow\left(\begin{matrix}1&2&-1\\0&-4&t+2&-2\\0&0&3-t&0\end{matrix}\right)$
At this point, it is already a rowed ladder matrix;
And since the matrix rank is 2, all the following rows are 0, so 3-t=0
So t=3

10.3 Several Important Expressions Related to Rank

Concept: Let A be a m*n matrix, and B a matrix that satisfies the requirements for matrix operations, then:

(1) $0\leq r(A)\leq\min\{m,n\}$ (by definition).
Every matrix has nonnegative rank, and $r(A)=0$ iff $A=O$.
The order of a nonzero minor cannot exceed either the number of rows or the number of columns, so $r(A)\leq\min\{m,n\}$.
(2) $r(kA)=r(A)$ for $k\neq0$.
(3) $r(AB)\leq\min\{r(A),r(B)\}$.
Matrix multiplication cannot increase rank.
(4) $r(A+B)\leq r(A)+r(B)$.
（5）$r\left(A^{*}\right)=\begin{cases}n,&r\left(A\right)=n,\\1,&r\left(A\right)=n-1,\\0,&r\left(A\right)<n-1,\end{cases}$
(6) Let A = m*n a matrix, P、Q be m-th and n-order invertible matrices respectively, then r(A)=r(PA)=r(AQ)=r(PAQ)
In the process, all intermediate matrices are equivalent: $r\left(\begin{matrix}E_{r}&0\\0&0\end{matrix}\right)$
(7) If $A_{m\times n}B_{n\times s}=O$, then $r(A)+r(B)\leq n$.
n is the number of columns in A;
(8) Gram matrix: $r\left(A\right)=r\left(A^{T}\right)=r\left(A^{T}A\right)=r\left(AA^{T}\right)$
$r\left(A^{T}A\right)=r\left(AA^{T}\right)$ must have a solution, and it must be the best approximate solution;
For any matrix, the transpose rank equals its $r\left(A^{T}A\right)=r\left(AA^{T}\right)$;