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I have this question:

Let $Z\in M_{2\times2}(\mathbb{R})$ be defined as

$$Z = \left( \begin{align} 1 &&1\\1 &&1 \end{align} \right)$$

and consider $T: M_{2\times2}(\mathbb{R}) \to M_{2\times2}(\mathbb{R})$ such that $$T(X) = ZX$$

Determine a basis for $\ker(T)$

Now I have transformation matrix with respect to the order basis as:

$$\left[ \begin{align} 1&&0&&1&&0\\0&&1&&0&&1\\1&&0&&1&&0\\0&&1&&0&&1 \end{align} \right]$$

Not sure if that is helpful for this, but I don't know how to work it out.

Is $\ker(T) = \emptyset$ or just the identity? I feel like there is no $\ker$.

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    $\begingroup$ Do you know how to find the basis for the nullspace of a matrix? $\endgroup$ – Alex Wertheim Jun 25 '14 at 2:41
  • $\begingroup$ @AWertheim No I don't, should I just check wiki for nullspace basis? $\endgroup$ – Katie Jun 25 '14 at 2:42
  • $\begingroup$ Katie: that would be a good place to start, I think. :) This video may also be helpful: youtube.com/watch?v=qjkafy42CWg $\endgroup$ – Alex Wertheim Jun 25 '14 at 2:44
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    $\begingroup$ the kernel is two-dimensional: $$ \left( \begin{align} a &&b\\-a &&-b \end{align} \right) $$ $\endgroup$ – yoyo Jun 25 '14 at 3:26
  • $\begingroup$ @yoyo If you post an answer, I'll be glad to remove mine; I felt it necessary to add a correct answer along with the (accepted) wrong one. $\endgroup$ – egreg Jun 25 '14 at 15:56
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The kernel of $T$ is $$ \ker T=\{A\in M_{2\times2}(\mathbb{R}):T(A)=0\} $$ that is $ZA=0$. You can find it using the representing matrix with respect to the basis $\{E_1,E_2,E_3,E_4\}$ where $$ E_1=\begin{bmatrix}1&0\\0&0\end{bmatrix}\quad E_2=\begin{bmatrix}0&1\\0&0\end{bmatrix}\quad E_3=\begin{bmatrix}0&0\\1&0\end{bmatrix}\quad E_4=\begin{bmatrix}0&0\\0&1\end{bmatrix} $$ which is the matrix you wrote. Now row reduce it: $$ \begin{bmatrix} 1&0&1&0\\0&1&0&1\\1&0&1&0\\0&1&0&1 \end{bmatrix} \to \begin{bmatrix} 1&0&1&0\\0&1&0&1\\0&0&0&0\\0&0&0&0 \end{bmatrix} $$ which means that the solutions of the homogeneous system are generated by the vectors $$ \begin{bmatrix}-1\\0\\1\\0\end{bmatrix} \quad\text{and}\quad \begin{bmatrix}0\\-1\\0\\1\end{bmatrix} $$ Taking into account the meaning of the representing matrix, we can say that $$ \{-E_1+E_3,-E_2+E_4\}= \left\{ \begin{bmatrix}-1&0\\1&0\end{bmatrix}, \begin{bmatrix}0&-1\\0&1\end{bmatrix} \right\} $$ is a basis of the kernel of $T$.

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