I'm in an abstract linear algebra class.
In the solution to a question asking me to find the eigenvectors of a normal $2 \times 2$ matrix in $C$, I am told that for a $2 \times 2$ matrix with nonzero first row $(a, b)$, the null space has to be spanned by the vector $(-b, a)$.
Now, I understand that if we take $v = (-b, a)$, then $Av = w$ will give us some $w$ where the first row has value $0$, since we get the first row from $(a * -b) + (b * a)$. But for $(-b, a)$ to span the null space, doesn't it also need to ensure that the second row of $w$ will be zero?
For a $2 \times 2$ matrix $A$ there are two possibilities: either $A$ is invertible which means that the null space is trivial, or $A$ is not invertible which means that the rows of $A$ are linearly dependent.
So if $A$ is not invertible and the first row is non-zero, the second row is a multiple of the first row. For such a matrix, we find that if the first entry of $Av$ is zero, then the second entry of $Av$ must also be zero.