# 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!

• The kernel of a linear transformation is the set of all vectors (=matrices A in your case) on which T vanish. Can you find all A's such that XA=0? – eranreches Oct 22 '17 at 5:51

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?

• Why am is the zero matrix the identity for the kernel? Since the operation is matrix multiplication, I was thinking I would be using the identity matrix. – jacksonf Oct 22 '17 at 5:56
• the kernel is defined as the set of matrices which map to the $0$ vector, i.e. the additive identity in your vector space. – qbert Oct 22 '17 at 5:57
• Got it! Sorry, mixing up my groups and vector space properties. – jacksonf Oct 22 '17 at 6:03
• @jacksonf no problem! happy to help – qbert Oct 22 '17 at 6:09

$A\in \ker T\implies TA=XA=0\implies a+c=b+d=0$

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 $\ker T$

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: $$\beta = \Bigg\{ \begin{pmatrix} 1&0\\-1&0\end{pmatrix},\begin{pmatrix} 0&1\\0&-1\end{pmatrix} \Bigg\}$$

• For someone not too familiar with the concept of kernels and images, this answer explains what is happening very well compared to the accepted answer. Thank you! – jerboa88 Oct 11 '19 at 17:22