1
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – eranreches Oct 22 '17 at 5:51
0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – jacksonf Oct 22 '17 at 5:56
  • $\begingroup$ the kernel is defined as the set of matrices which map to the $0$ vector, i.e. the additive identity in your vector space. $\endgroup$ – qbert Oct 22 '17 at 5:57
  • 2
    $\begingroup$ Got it! Sorry, mixing up my groups and vector space properties. $\endgroup$ – jacksonf Oct 22 '17 at 6:03
  • $\begingroup$ @jacksonf no problem! happy to help $\endgroup$ – qbert Oct 22 '17 at 6:09
1
$\begingroup$

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

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 $\ker T$

$\endgroup$
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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! $\endgroup$ – jerboa88 Oct 11 '19 at 17:22

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.