Rank of matrices

Consider a $m\times n$ matrix $A$ of rank $r$.
So we have $n$ column vectors in $\mathbb{R}^m$, and there are $(n-r)$ column vectors which we can get by the linear combination of other $r$ column vectors.
We describe rank as number of pivots of a matrix.

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# pivots can't be $\gt$ # rows ($m$), so $r\leq m$
# pivots can't be $\gt$ # columns ($n$), so $r\leq n$

Full column rank matrix

Consider a $m\times n$ matrix $A$ of rank $r$ and $m\gt n$.
Here $\text{Rank}(A)=r=n$ it means that all of the column vectors are independent, # pivots $=n$ $\Rightarrow$ # free column vectors $=0$.

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Can we always get a solution, can we always find $x$ such that $A\vec{x}=\vec{b}$ for every $\vec{b}$?
Here in $A$ we have $n$ independent vectors in $\mathbb{R}^m$ and $m\gt n$, to fill a $m$ dimensional space we alteast need $m$ independent vectors, but we had $n$ independent vectors and $n\lt m$ so we don't have enough vectors to fill $\mathbb{R}^m$.
Linear combination of $n$ independent vectors can't give us a space in $\mathbb{R}^m$
So we can't get every vector $\vec{b}$ by the linear combinations of those independent vectors.
So answer is No we can't get every $\vec{b}$.

We have no free vectors here so Null Space of $A$ is just $\vec{0}$, so $A\vec{x}=\vec{0}$ has only one solution that is $\vec{x}=\vec{0}$
Complete solution of $A\vec{x}=\vec{b}$ is $\vec{x}=\vec{x}_p+\vec{x}_n=\vec{x}_p+\vec{0}=\vec{x}_p$

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So complete solution is just a single vector $\vec{x}_p$ if it exists.
# solution = $\left\{\begin{matrix} 1 & \text{ if there is a solution} \\ 0 & \text{ otherwise} \\ \end{matrix}\right.$

Example:

$A = \begin{bmatrix} 1 & 4\\ 2 & 3\\ 3 & 2\\ 4 & 1\\ \end{bmatrix}$
Can we always get a solution, can we always find $x$ such that $A\vec{x}=\vec{b}$ for every $\vec{b}$?
Here in $A$ we have $2$ vectors in $\mathbb{R}^4$.
and linear combination of $2$ vector can give us a space in $\mathbb{R}^2$ so we can only get those $\vec{b}$ which are in the linear combination(plane) of those two vectors.

possible $\vec{b}$ is

$$\vec{b}= \alpha\begin{bmatrix} 1\\2\\3\\4\\ \end{bmatrix} +\beta\begin{bmatrix} 4\\3\\2\\1\\ \end{bmatrix};\quad \alpha,\beta\in\mathbb{R}$$

So answer is No we can't get every $\vec{b}$

Row reduced echelon form is

$$\begin{bmatrix} \fbox{1} & 0 \\ 0 & \fbox{1} \\ 0 & 0 \\ 0 & 0 \\ \end{bmatrix} = \begin{bmatrix} I_{2\times2}\\0 \end{bmatrix}$$

So # of pivots are $2$

Full row rank matrix

Consider a $m\times n$ matrix $A$ of rank $r$ and $m\lt n$.
Here $\text{Rank}(A)=r=m$ now all of the column vectors are not independent, # pivots $=m$ $\Rightarrow$ # free column vectors $=n-r$.

note

Can we always get a solution, can we always find $x$ such that $A\vec{x}=\vec{b}$ for every $\vec{b}$?
Here in $A$ we have $m$ independent vectors in $\mathbb{R}^m$, so we have enough vectors to fill $\mathbb{R}^m$.
Linear combination of $m$ independent vectors can give us a space in $\mathbb{R}^m$
So we can get every vector $\vec{b}$ by the linear combinations of those independent vectors.
So answer is Yes we can get every $\vec{b}$.

Here we have $n-r$ free vectors so Null Space of $A$ is in $\mathbb{R}^{n-r}$.

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# solutions are $\infty$

Example:

$A = \begin{bmatrix} 1 & 2 & 3 & 4 \\ 4 & 3 & 2 & 1 \\ \end{bmatrix}$
Can we always get a solution, can we always find $x$ such that $A\vec{x}=\vec{b}$ for every $\vec{b}$?
Here in $A$ we have $4$ vectors in $\mathbb{R}^2$ and out of those $2$ vectors are independent.
And linear combination of $2$ independent vector can give us a space in $\mathbb{R}^2$. So we covered the whole $2$-dimensional space.
So answer is Yes we can get every $\vec{b}$.

Row reduced echelon form is

$$\begin{bmatrix} \fbox{1} & 0 & -1 & -2 \\ 0 & \fbox{1} & 2 & 3 \\ \end{bmatrix} = \begin{bmatrix} I_{2\times2} & F_{2\times2} \end{bmatrix}$$

So # of pivots are $2$

Full rank matrix

Consider a $m\times n$ matrix $A$ of rank $r$ and $m = n$.
Here $\text{Rank}(A)=r=n$ it means that all of the column vectors are independent, # pivots $=n$ $\Rightarrow$ # free column vectors $=0$.

note

Can we always get a solution, can we always find $x$ such that $A\vec{x}=\vec{b}$ for every $\vec{b}$?
Here in $A$ we have $n$ independent vectors in $\mathbb{R}^n$, so we have enough vectors to fill $\mathbb{R}^n$.
Linear combination of $n$ independent vectors can give us a space in $\mathbb{R}^n$
So we can get every vector $\vec{b}$ by the linear combinations of those independent vectors.
So answer is Yes we can get every $\vec{b}$.

We have no free vectors here so Null Space of $A$ is just $\vec{0}$, so $A\vec{x}=\vec{0}$ has only one solution that is $\vec{x}=\vec{0}$

info

# solutions = $1$
row reduced echelon form of $A$ is $I_{n\times n}$