This is hardly "beginning algebra" but

The "kernel" of a linear transformation, A, is the set of vectors, v, such that Av= 0.

It can be shown that the kernel of a linear transformation is always a subspace of the original vector space and that, if A is "invertible", its kernel is the trivial subspace consisting of the vector only.

So to find the kernel of T, mapping \(\displaystyle \begin{pmatrix} x & y \\ z & t \end{pmatrix}\) to (x+ t, y- z, t) we must solve x+ t= 0, y- z= 0, and t= 0. Since t= 0, x+ t= x= 0. y- z= 0 means that y= z so we can write any matrix satisfying this as \(\displaystyle \begin{pmatrix} 0 & y \\ y & 0 \end{pmatrix}\).

That is, the kernel is the one-dimensional subspace of all 4 by 4 matrices that are multiples of \(\displaystyle \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\).