Basis for Linear Transformation with Matrix Multiplication Let $V$ be the vector space of $ 2 \times 2 $ real matrices.
Find a basis for the kernel of the linear transformation $T :V \rightarrow V$ given by $T(A)= XA $.
$X$, being the matrix below.
\begin{bmatrix}1&1\\1&1\end{bmatrix}
The matrix $X$ isn't invertible, so I don't know what the kernel should be... And when I do, how do I find the basis for it?
*Sorry about the formatting, would appreciate an edit! 
 A: You are looking for the set of all vectors $x \in V$ such that $T(x) = 0$, which is the definition of the null space (kernel). So then you want to look for the elements in $V$ such that 
$$T(X) = 0 $$
First we observe the transformation
$$T\left(\begin{pmatrix} a&b\\c&d\end{pmatrix}\right) = \begin{pmatrix} 1&1\\1&1\end{pmatrix}\begin{pmatrix} a&b\\c&d\end{pmatrix} = \begin{pmatrix} a+c&b+d\\a+c&b+d\end{pmatrix}   $$ 
And in this case we are looking for it to be the zero matrix:
$$T\left(\begin{pmatrix} a&b\\c&d\end{pmatrix}\right) = \begin{pmatrix} 0&0\\0&0\end{pmatrix} \iff a+c =0 \ \ \ \text{and } \ \ b+d=0$$ 
So, because $c=-a$ and $d = -b$, we can write it as
$$\begin{pmatrix} a&b\\-a&-b\end{pmatrix} = a\begin{pmatrix} 1&0\\-1&0\end{pmatrix} + b\begin{pmatrix} 0&1\\0&-1\end{pmatrix}$$
And now we can write a basis $\beta $ for the kernel of $T$ as follows:
$$\beta = \Bigg\{ \begin{pmatrix} 1&0\\-1&0\end{pmatrix},\begin{pmatrix} 0&1\\0&-1\end{pmatrix}  \Bigg\}$$
A: $A\in \ker T\implies TA=XA=0\implies a+c=b+d=0$
if $A=$ \begin{bmatrix} a&b\\c&d\end{bmatrix}
Hence $a=-c,b=-d\implies $ \begin{bmatrix} 1&0\\-1&0\end{bmatrix} and \begin{bmatrix} 0&-1\\0&1\end{bmatrix} form a basis of $\ker T$
A: You are in search of matrices 
$$
\begin{bmatrix}
a&b\\c&d
\end{bmatrix}
$$
with 
$$
\begin{bmatrix}
1&1\\1&1
\end{bmatrix}\begin{bmatrix}
a&b\\c&d
\end{bmatrix}=
\begin{bmatrix}
0&0\\0&0
\end{bmatrix}=\vec{0}\in V
$$
can you see what conditions such a relation imposes on the matrices making up your kernel?
