# How to find basis for $\ker(L)$ and $\operatorname{ran}(L)$?

Let $V=\mathbb R^{2 \times 2}$ be the vector space of $2\times 2$ matrices and let $L\colon V\to V$ be defined by $L(X)= AX$, where $$A = \begin{bmatrix}8&-4\\2&-1\end{bmatrix}.$$ Find a basis for $\ker(L)$. Find a basis for $\operatorname{ran}(L)$.

The answer is supposed to be in the form of two $2\times2$ matrices for each but I'm not sure how to find the basis for a two by two matrix. I row reduced to find the basis of $\ker(L) = \begin{bmatrix}1/2\\1\end{bmatrix}$ and the basis of $\operatorname{ran}(L) = \begin{bmatrix}8\\2\end{bmatrix}$ but I'm not sure how to proceed after that.

• Hint: $$\pmatrix{8 & -4 \\ 2 & -1}\pmatrix{a & b \\ c & d} = \pmatrix{8a-4c & 8b -4d \\ 2a -c & 2b-d} \leftrightarrow \pmatrix{8 & 0 & -4 & 0 \\ 0 & 8 & 0 & -4 \\ 2 & 0 & -1 & 0 \\ 0 & 2 & 0 & -1}\pmatrix{a \\ b \\ c \\ d}=\pmatrix{8a-4c \\ 8b-4d \\ 2a-c \\ 2b-d}$$ – user137731 Nov 21 '15 at 20:23
• Thank you! I understand how to find the basis for 2x2 now – Sally K Nov 21 '15 at 20:34
• @Bye_World: whoops, misread the question! – Alex R. Nov 21 '15 at 20:44

AS noted in the comment of Bye_World we have: $$L(M)= \begin{bmatrix} 8&-4\\ 2&-1 \end{bmatrix} \begin{bmatrix} a&b\\ c&d \end{bmatrix} = \begin{bmatrix} 4(2a-c)&4(2b-d)\\ 2a-c&2b-d \end{bmatrix}$$ so, for the kernel of $L$ we have: $L(M)=0 \Rightarrow 2a=c \;\land \; 2b=d$ , i.e. the matrices : $$K= \begin{bmatrix} a&b\\ 2a&2b \end{bmatrix}= a\begin{bmatrix} 1&0\\ 2&0 \end{bmatrix} +b\begin{bmatrix} 0&1\\ 0&2 \end{bmatrix}$$ And, since the two matrices at RHS are linearly independent they are a basis for $Ker (L)$.
For the Rank of $L$ we can do the same notice that a matrix $R$ in the range is of the form: $$R=\begin{bmatrix} 4k&4h\\ k&h \end{bmatrix}=k \begin{bmatrix} 4&0\\ 1&0 \end{bmatrix}+ h\begin{bmatrix} 0&4\\ 0&1 \end{bmatrix}$$