1
$\begingroup$

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.

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

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: $$ R=\begin{bmatrix} 4k&4h\\ k&h \end{bmatrix}=k \begin{bmatrix} 4&0\\ 1&0 \end{bmatrix}+ h\begin{bmatrix} 0&4\\ 0&1 \end{bmatrix} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.