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Find the basis of the kernel of the linear transformation $f: M_{2×2} (\mathbb R) \rightarrow M_{2×2}(\mathbb R)$ given $f(A) = AB − BA$ where $B = \begin{bmatrix}1&1\\0&1\end{bmatrix} $ and $M_{2×2} (\mathbb R)$ is the vectorial space of real $2×2$ square matrices.

My problem is probably that I don't understand the question. I know how to find the basis of the kernel of a linear transformation when $f$ is given as a vector with equations, which i can turn into a matrix, put the identity below, and then echelon it and take the vectors under the null columns. It's probably the same, but I'm a little lost as it's given in a matricial way.

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    $\begingroup$ Go back to basics. The kernel is the set of matrices $A$ such that $AB-BA=0$. This will give you a system of linear equations to solve that give you the general form of a matrix in the kernel. From there, you should be able to come up with a basis. Alternatively, “vectorize” the matrices and proceed as you would normally. $\endgroup$ – amd Apr 2 '17 at 0:52
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    $\begingroup$ Note, too, that $AB-BA=0$ is equivalent to $AB=BA$, so the kernel consists of all matrices that commute with $B$. $\endgroup$ – amd Apr 2 '17 at 0:56
  • $\begingroup$ why didn't you accept an answer? $\endgroup$ – Viktor Glombik Dec 16 '18 at 18:45
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As commented by amd, the kernel of $f$ is the set of matrices $A$ such that $f(A) = 0.$ So we compute $$\begin{align} f(A) & = AB - BA\\\\ & = \begin{bmatrix}a_{11} & a_{12}\\ a_{21} & a_{22}\end{bmatrix} \begin{bmatrix} 1 & 1\\ 0 & 1\end{bmatrix} - \begin{bmatrix} 1 & 1\\ 0 & 1\end{bmatrix} \begin{bmatrix}a_{11} & a_{12}\\ a_{21} & a_{22}\end{bmatrix}\\\\ & = \begin{bmatrix}a_{11} & a_{11} + a_{12}\\ a_{21} & a_{21} + a_{22}\end{bmatrix} - \begin{bmatrix}a_{11} + a_{21} & a_{12} + a_{22}\\ a_{21} & a_{22}\end{bmatrix}\\\\ & = \begin{bmatrix}-a_{21} & a_{11} - a_{22}\\ 0 & a_{21}\end{bmatrix}\end{align}$$ which we want equal to the zero matrix. Hence, we see that the kernel of $f$ consists of matrices such that $a_{21} = 0,$ and $a_{22} = a_{11},$ i.e., matrices of the form $$\begin{bmatrix}a_{11} & a_{12}\\ 0 & a_{11}\end{bmatrix},$$ which can be expressed as $$a_{11}\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix} + a_{12}\begin{bmatrix}0 & 1\\ 0 & 0\end{bmatrix}$$ where $a_{11}$ and $a_{12}$ here are arbitrary real numbers. Thus, a (not the as in the question) basis for the kernel of $f$ is $$\left\{\begin{bmatrix}1 & 0\\ 0 & 1\end{bmatrix}, \begin{bmatrix}0 & 1\\ 0 & 0\end{bmatrix} \right\}.$$

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If $A\in \ker (f)$ then $$f(A)=0\implies AB = BA$$

So if $$A = \begin{bmatrix}a & b\\ c & d\end{bmatrix}$$ we get

$$ \begin{bmatrix}a & b\\ c & d\end{bmatrix}\begin{bmatrix}1 & 1\\ 0 & 1\end{bmatrix}= \begin{bmatrix}1 & 1\\ 0 & 1\end{bmatrix}\begin{bmatrix}a & b\\ c & d\end{bmatrix}$$

so $$\begin{bmatrix}a & a+b\\ c & b+d\end{bmatrix} = \begin{bmatrix}a+c & b+d\\ c & d\end{bmatrix}$$

So $c =0$ and $b=0$ and $a=d$. So $$A = \begin{bmatrix}a & 0\\ 0 & a\end{bmatrix}= a\cdot I$$

So $\ker (f) = \{a\cdot I;\; a\in \mathbb{R}\}$ and thus def$ (f)=1$.

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  • $\begingroup$ Why do you obtain a different answer than @Maurice P ? $\endgroup$ – Viktor Glombik Dec 16 '18 at 18:47
  • $\begingroup$ I don't know, but you can trace every step in both solution to find a mistake. $\endgroup$ – Aqua Dec 16 '18 at 19:51

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