Vtyjkyuk

Determine the null space of the matrix:$$beginbmatrix 1 & -1 \ 2 & 3 \ 1 & 1 endbmatrix$$

My try:
$$beginbmatrix 1 & -1 \ 2 & 3 \ 1 & 1 endbmatrix_R_2rightarrow R_2-2R_1\R_3rightarrow R_3-R_1$$
$$beginbmatrix 1 & -1 \ 0 & 5 \ 0 & 2 endbmatrix_R_3rightarrow 5R_3-2R_2$$
$$beginbmatrix 1 & -1 \ 0 & 5 \ 0 & 0 endbmatrix_R_2rightarrow fracR_25$$
$$beginbmatrix 1 & -1 \ 0 & 1 \ 0 & 0 endbmatrix$$

From this I got $$x-y=0implies x=y\y=0$$
$$(x,y,z)^T=(y,0,z)^T=y(1,0,0)^T+z(0,0,1)^T$$
So, $(1,0,0)^T$ and $(0,0,1)^T$ is the null space. Is this correct?

Recall that by definition the nullspace is the subspace of all vectors $vec x$ such that $Avec x=vec 0$ and in that case we have ony the trivial solution $(x_1,x_2)=(0,0)$ then $Null(A)=vec 0$.

Notably to solve $Ax=0$ we can proceed by RREF to obtain

$$beginbmatrix 1 & -1 \ 2 & 3 \ 1 & 1 endbmatrixto
beginbmatrix 1 & -1 \ 0 & 5 \ 0 & 2 endbmatrixto
beginbmatrix 1 & -1 \ 0 & 1 \ 0 & 0 endbmatrix$$

that is

$$beginbmatrix 1 & -1 \ 0 & 1 \ 0 & 0 endbmatrixbeginbmatrix x_1 \ x_2 endbmatrix=beginbmatrix 0 \ 0 endbmatrix$$

that is $x_1=x_2=0$.

Your method is correct to find that $x=y$ and $y=0$. But you have made an incorrect conclusion after that stage and you have made a mistake in the dimension of your vectors.

Firstly, see that your matrix acts on vectors in $mathbbR^2$ to form vectors in $mathbbR^3$ like so:

$$left( beginmatrix 1 & -1 \ 2 & 3 \ 1 & 1 endmatrix right) left( beginmatrix x \ y endmatrix right)=left( beginmatrix x-y\ 2x+3y\ x+y endmatrix right)
$$

So you're searching for vectors $(x,y)^T$ in your null space.

If $x=y$ and $y=0$ then $x=y=0$, showing that your null space is just $(0,0)^T$

Right Nullspace
If vector $alpha = [x,y]$ is in the right nullspace of A then
$$beginbmatrix
1 & -1 \
2 & 3 \
1 & 1 \
endbmatrix[x,y]^T = beginbmatrix 0 \ 0 \ 0
endbmatrix$$
This gives
$$x - y = 0$$
$$2x + 3y = 0$$
$$x + y = 0$$
First and third equation tells us that $x = y = -x$, i.e. the only solution is $[0,0]$. So, the right nullspace has dimension zero

Left Nullspace
If vector $alpha = [x,y,z]$ is in the left nullspace of A then
$$beginbmatrix
x & y & z
endbmatrix
beginbmatrix
1 & -1 \
2 & 3 \
1 & 1 \
endbmatrix = beginbmatrix
0 & 0
endbmatrix$$
So we get
$$x + 2y + z = 0$$
and
$$-x + 3y + z = 0$$
First equation tells us $$x = -(2y+z)$$
and second one says
$$x = 3y+z$$
So by equating $x=x$, we get
$$-2y-z = 3y+z$$
that is
$$5y = -2z$$
or $$y = - frac25z$$. Plugging this back into one of the two x equations, we get
$$x= 3y + z = 3(- frac25z) + z = frac-6+55z = -frac15z$$
So, the Null space takes the form
$$beginbmatrix
-frac15 \
- frac25 \
1
endbmatrix
z$$