Find the basis for kernel of a matrix transformation $Let \ ψ\ :{Mat }_{ 2x2 }(ℝ)\ →\ { Mat }_{ 2x2 }(ℝ)\ be\ defined\ by$
$ ψ : \pmatrix{a&b\\c&d}→\pmatrix{a+b&a-c\\a+c&b-c}$ . 
Find basis for ker ψ
I'm not sure how to do it for a matrix, I've set $\pmatrix{a+b&a-c\\a+c&b-c}$ to the zero matrix, and obtained 4 equations, but not sure how proceed from there.
 A: Easy Way
Assuming that $\text{Mat}_{2\times 2}(\mathbb{R})$ are being though of as additive groups, the kernel is made up of all matrices that are send to the zero matrix. So you need $a+b=0$, $a-c=0$, $a+c=0$ and $b-c=0$. Well, $a+b=0$ if and only if $a=-b$. Let's substitute that into the remaining three equations: $-b-c=0$, $-b+c=0$ and $b-c=0$. Well, $-b-c=0$ if and only if $b=-c$. Let's put that into the two remaining equations: $-(-c)+c=0$ and $-(-c)-c=0$. Well, $2c=0$ if and only if $c=0$. Hence $a=-b$, $b=-c$ and $c=0$ meaning that $a=b=c=0$. 
You're free to choose $d$. So the matrix $\left[\begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array}\right]$ spans the kernel.
A: HINT First identify $\text{Mat}_{2 \times 2}$ with $\mathbb{R}^4$, with basis matrices $$\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, ...,$$ and write $\psi$ as a linear transformation $\mathbb{R}^4 \to \mathbb{R}^4$ with matrix $M$.
Then read the examples in this Wikipedia article, which give a method for finding a basis once you have a matrix.
