Why is $y=1$ for this basis? If we have:
$ \begin{bmatrix}
        -2 & 0 \\
        0 & 0 \\
        \end{bmatrix}\begin{bmatrix}
        x \\
       y \\
        \end{bmatrix} =\begin{bmatrix}
        0 \\
        0 \\
        \end{bmatrix}$
This will result in the basis of \begin{bmatrix} 0 \\
       1 \\
        \end{bmatrix} as $x=0$ and $y=t$ for any t in the real numbers
Why is this?  When I multiply the matrix out, I get $ (-2)*x+0*y= 0$ and $0*x+0*y =0$ this would seem to imply that both x and y are equal to 0, so why is y=t, and thus the basis have a 1?  Should it not be \begin{bmatrix}
        0 \\
         0 \\
        \end{bmatrix}this basis for the null space? Since both x and y should equal 0?  Thank you for your help!
 A: The first equation gives you that $x=0$, but says nothing about $y$, since zero times anything is zero, so any value of $y$ will make the equation true. The second equation tells you nothing at all, for the same reason; any value of $x$ and $y$ will make that one true. So you have to have $x=0$, but you have no restrictions on $y$, meaning you can actually pick anything you like. It's customary to choose 1 in that case, but in fact $\begin{bmatrix}0 \\ s\end{bmatrix}$ is just as good for any $s\neq 0$.
A: $-2x + 0y = 0$ only implies that $x = 0$. It leaves y free (ex: plug in $x = 0, y = 100$). That is why $\begin{bmatrix}
0 \\
1 \\
\end{bmatrix}$ is a basis for the null space of the transformation, because x has to be 0 and y can be anything.
Note: $\begin{bmatrix}
0 \\
0 \\
\end{bmatrix}$
can never be a basis for anything.
A: Firs observe that $$ (-2)x + 0 y = 0 \iff -2x = 0 \iff x = 0$$ Now look at the other equation$$0x + 0y = 0$$ and note that for all $x,y \in \mathbb{R}$ this is true, so this gives us no information. The only restriction for a vector to be in the null space is for $x$ to be zero, so we can write it  in a general form:
$$\begin{bmatrix}0 \\ y \end{bmatrix} = y \begin{bmatrix}0 \\ 1 \end{bmatrix}$$
so we can conclude that  $\begin{bmatrix}0 \\ 1 \end{bmatrix}$ is an element of the basis for the null space. Also, notice how $\begin{bmatrix}0 \\ 0 \end{bmatrix} = 0 \cdot \begin{bmatrix}0 \\ 1 \end{bmatrix}$.
