Previously we saw how to get echelon form, pivot/free column vectors of a matrix say

\(A\)

.
Now let's find

\vec{x}

that solves

A\vec{x}=\vec{0}

Solve $A\vec{x}=0$

(say)

A=\begin{bmatrix} \fbox{1} & 2 & 2 & 2\\ 0 & 0 & \fbox{2} & 4\\ 0 & 0 & 0 & 0\\ \end{bmatrix}

Note that we can get free columns by a linear combination of pivot columns.
So we are free to scale our free columns.
So consider

A'=\begin{bmatrix} \fbox{1} & \lambda_1*2 & 2 & \lambda_2*2\\ 0 & \lambda_1*0 & \fbox{2} & \lambda_2*4\\ 0 & \lambda_1*0 & 0 & \lambda_2*0\\ \end{bmatrix}

and

\lambda_i\in\mathbb{R}\ \forall i=\{1,2\}

Because free columns are the linear combination of pivot columns, we can say that the Null space we get by $A\vec{x}=\vec{0}$ is the same Null space we get by $A'\vec{x}=\vec{0}$

So now we will find solution for

A'\vec{x}=\vec{0}

A'\vec{x}= \begin{bmatrix} \fbox{1} & \lambda_1*2 & 2 & \lambda_2*2\\ 0 & \lambda_1*0 & \fbox{2} & \lambda_2*4\\ 0 & \lambda_1*0 & 0 & \lambda_2*0\\ \end{bmatrix} \begin{bmatrix} x_1\\x_2\\x_3\\x_4\\ \end{bmatrix} = \begin{bmatrix} 0\\0\\0\\ \end{bmatrix}

Let's see

\(2\)

approaches to achieve it.

$\text{Approach } 1$

So we have system of equation:
$\begin{matrix} x_1 & + & 2\lambda_1x_2 & + & 2x_3 & + & 2\lambda_2x_4 & = & 0 \\ 0 & + & 0 & + & 2x_3 & + & 4\lambda_2x_4 & = & 0 \\ 0 & + & 0 & + & 0 & + & 0 & = & 0 \\ \end{matrix}$
We are free to choose $\lambda_1$ and $\lambda_2$ so let $\lambda_1=\frac{1}{x_2}$ , $\lambda_2=0$
Then our system of equation becomes:
$\begin{matrix} x_1 & + & 2 & + & 2x_3 & + & 0 & = & 0 \\ 0 & + & 0 & + & 2x_3 & + & 0 & = & 0 \\ 0 & + & 0 & + & 0 & + & 0 & = & 0 \\ \end{matrix}$
$\Rightarrow x= \begin{bmatrix} -2\\1\\0\\0\\ \end{bmatrix}$
Now let's set $\lambda_1,\lambda_2$ some different value to get $$x_2,x_4$$
let $\lambda_1=0$ , $\lambda_2=\frac{1}{x_4}$
So we have system of equation:
$\begin{matrix} x_1 & + & 0 & + & 2x_3 & + & 2 & = & 0 \\ 0 & + & 0 & + & 2x_3 & + & 4 & = & 0 \\ 0 & + & 0 & + & 0 & + & 0 & = & 0 \\ \end{matrix}$
$\Rightarrow x= \begin{bmatrix} 2\\0\\-2\\1\\ \end{bmatrix}$

$\text{Approach } 2$

$A'\vec{x}= \begin{bmatrix} \fbox{1} & \lambda_1*2 & 2 & \lambda_2*2\\ 0 & \lambda_1*0 & \fbox{2} & \lambda_2*4\\ 0 & \lambda_1*0 & 0 & \lambda_2*0\\ \end{bmatrix} \begin{bmatrix} x_1\\x_2\\x_3\\x_4\\ \end{bmatrix} = \begin{bmatrix} 0\\0\\0\\ \end{bmatrix}$
We can write it as.
$A'\vec{x}= \begin{bmatrix} \fbox{1} & 2 & 2 & 2\\ 0 & 0 & \fbox{2} & 4\\ 0 & 0 & 0 & 0\\ \end{bmatrix} \begin{bmatrix} x_1\\ \lambda_1*x_2\\x_3\\ \lambda_2*x_4\\ \end{bmatrix} = \begin{bmatrix} 0\\0\\0\\ \end{bmatrix}$
We are free to choose $\lambda_1$ and $\lambda_2$ so let $\lambda_1=\frac{1}{x_2}$ , $\lambda_2=0$
By solving it we get $\Rightarrow x= \begin{bmatrix} -2\\1\\0\\0\\ \end{bmatrix}$
By choosing $\lambda_1=0$ , $\lambda_2=\frac{1}{x_4}$
We get $\Rightarrow x= \begin{bmatrix} 2\\0\\-2\\1\\ \end{bmatrix}$

So we have two special solution.

They are special solution because we crafted those solution in such a way that, in one solution we didn't consider free vector $$c_4$$ but consider $$c_2$$ , and in another solution we didn't consider $$c_2$$ but consider $$c_4$$ .

We can get all the possible

\vec{x}

that solves

A'\vec{x}=0

by taking linear combinations of those possible solutions.
So

x=\alpha \begin{bmatrix} -2\\1\\0\\0\\ \end{bmatrix} + \beta \begin{bmatrix} 2\\0\\-2\\1\\ \end{bmatrix};\quad

\alpha,\beta\in\mathbb{R}

Now we got the Null Space of matrix

\(A\)

it's a plane passing through

\begin{bmatrix} -2\\1\\0\\0\\ \end{bmatrix}

and

\begin{bmatrix} 2\\0\\-2\\1\\ \end{bmatrix}

Null Space is the linear combination of all special solutions.

But how many special solutions are there?

Say there are

\(k\)

dependent column vectors in our matrix

\(A\)

, then as we discussed we craft these special solution such that we consider one dependent vector at a time.
So # special solution = # dependent column vector
remember Rank(A) = # pivot columns
Say that shape of our matrix is

m\times n

, so we have

\(n\)

column vectors in

\mathbb{R}^m

, and say Rank of the matrix is

\(r\)

, then

# special solution = # dependent column vector = $$n-r$$

Then the Null space of

\(A\)

is linear combinations of these

\(n-r\)

special solution.

For a matrix $A\in\mathbb{R}^{m\times n}$ with $\text{Rank}(A)=r$ , Null Space is a vector space in $\mathbb{R}^{n-r}$

Solve \(A\vec{x}=0\)

\(\text{Approach } 1\)

\(\text{Approach } 2\)

But how many special solutions are there?

Solve $A\vec{x}=0$

$\text{Approach } 1$

$\text{Approach } 2$