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Null Space - Solve Ax⃗=0⃗A\vec x = \vec 0

Previously we saw how to get echelon form, pivot/free column vectors of a matrix say AA.
Now let's find x⃗\vec{x} that solves Ax⃗=0⃗A\vec{x}=\vec{0}

(say) A=[122200240000]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′=[1λ1∗22λ2∗20λ1∗02λ2∗40λ1∗00λ2∗0] and λi∈R ∀i={1,2}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} \text{ and } \lambda_i\in\mathbb{R}\ \forall i=\{1,2\}
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Because free columns are the linear combination of pivot columns, we can say that the Null space we get by Ax⃗=0⃗A\vec {x}=\vec{0} is the same Null space we get by A′x⃗=0⃗A'\vec{x}=\vec{0}

So now we will find solution for A′x⃗=0⃗A'\vec{x}=\vec{0}

A′x⃗=[1λ1∗22λ2∗20λ1∗02λ2∗40λ1∗00λ2∗0][x1x2x3x4]=[000]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 22 approaches to achieve it.

Approach 1​

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So we have system of equation:

x1+2λ1x2+2x3+2λ2x4=00+0+2x3+4λ2x4=00+0+0+0=0\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 λ1\lambda_1 and λ2\lambda_2 so let λ1=1x2\lambda_1=\frac{1}{x_2}, λ2=0\lambda_2=0
Then our system of equation becomes:

x1+2+2x3+0=00+0+2x3+0=00+0+0+0=0\begin{matrix} x_1 & + & 2 & + & 2x_3 & + & 0 & = & 0 \\ 0 & + & 0 & + & 2x_3 & + & 0 & = & 0 \\ 0 & + & 0 & + & 0 & + & 0 & = & 0 \\ \end{matrix}
⇒x=[−2100]\Rightarrow x= \begin{bmatrix} -2\\1\\0\\0\\ \end{bmatrix}

Now let's set λ1,λ2\lambda_1,\lambda_2 some different value to get x2,x4x_2,x_4
let λ1=0\lambda_1=0, λ2=1x4\lambda_2=\frac{1}{x_4}
So we have system of equation:

x1+0+2x3+2=00+0+2x3+4=00+0+0+0=0\begin{matrix} x_1 & + & 0 & + & 2x_3 & + & 2 & = & 0 \\ 0 & + & 0 & + & 2x_3 & + & 4 & = & 0 \\ 0 & + & 0 & + & 0 & + & 0 & = & 0 \\ \end{matrix}
⇒x=[20−21]\Rightarrow x= \begin{bmatrix} 2\\0\\-2\\1\\ \end{bmatrix}

Approach 2​

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A′x⃗=[1λ1∗22λ2∗20λ1∗02λ2∗40λ1∗00λ2∗0][x1x2x3x4]=[000]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′x⃗=[122200240000][x1λ1∗x2x3λ2∗x4]=[000]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 λ1\lambda_1 and λ2\lambda_2 so let λ1=1x2\lambda_1=\frac{1}{x_2}, λ2=0\lambda_2=0
By solving it we get

⇒x=[−2100]\Rightarrow x= \begin{bmatrix} -2\\1\\0\\0\\ \end{bmatrix}

By choosing λ1=0\lambda_1=0, λ2=1x4\lambda_2=\frac{1}{x_4}

We get ⇒x=[20−21]\Rightarrow x= \begin{bmatrix} 2\\0\\-2\\1\\ \end{bmatrix}

So we have two special solution.

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They are special solution because we crafted those solution in such a way that, in one solution we didn't consider free vector c4c_4 but consider c2c_2, and in another solution we didn't consider c2c_2 but consider c4c_4.

We can get all the possible x⃗\vec{x} that solves A′x⃗=0A'\vec{x}=0 by taking linear combinations of those possible solutions.
So,

x=α[−2100]+β[20−21];α,β∈Rx=\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 AA it's a plane passing through [−2100]\begin{bmatrix} -2\\1\\0\\0\\ \end{bmatrix} and [20−21]\begin{bmatrix} 2\\0\\-2\\1\\ \end{bmatrix}

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Null Space is the linear combination of all special solutions.

But how many special solutions are there?​

Say there are kk dependent column vectors in our matrix AA, 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×nm\times n, so we have nn column vectors in Rm\mathbb{R}^m, and say Rank of the matrix is rr, then

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# special solution = # dependent column vector = n−rn-r

Then the Null space of AA is linear combinations of these n−rn-r special solution.

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