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I would like to find all matrices that commute with matrix $ A =\begin{pmatrix}1 & -1 \\ 0 & 1 \end{pmatrix}$

Proposed solution

$\begin{pmatrix}a&b \\c &d\end{pmatrix}\begin{pmatrix}1&-1 \\0 &1\end{pmatrix} = \begin{pmatrix}1& -1\\ 0&1\end{pmatrix}\begin{pmatrix}a&b \\ c&d\end{pmatrix} = \begin{pmatrix}a& -a+b\\ c&-c+d\end{pmatrix}=\begin{pmatrix}a-c& b-d\\ c&d\end{pmatrix}$

Unclear about the following

$a=a-c$

$-a+b =b-d$

$c=c$

$-c+d=d$

So any matrix of the form $\begin{pmatrix}d & 0 \\ 0 & a\end{pmatrix}$

Please could someone review and correct if needs be

Thanks

  • 0
    Why do you want $b$ to be $0$?2012-12-12

1 Answers 1

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Somewhat more generally: your matrix is $I - N$ where $N = \pmatrix{0 & 1\cr 0 & 0\cr}$. Every $2 \times 2$ matrix commutes with $I$, so the question is what commutes with $N$. We can write $N = u v^T$ where $u$ and $v$ are nonzero (column) vectors, in this case $\pmatrix{1\cr 0\cr}$ and $\pmatrix{0\cr 1\cr}$. Now $B(u v^T) = (Bu) v^T$ is a matrix whose rows are all scalar multiples of $v^T$ and whose columns are all scalar multiples of $Bu$, while $(uv^T) B = u (v^T B)$ is a matrix whose rows are scalar multiples of $v^T B$ and whose columns are all scalar multiples of $u$. Thus for some scalar $\lambda$, we must have $Bu = \lambda u$ and $v^T B = \lambda v^T$, i.e. $u$ and $v^T$ are right and left eigenvectors of $B$ for the same eigenvalue, and this condition is clearly sufficient as well as necessary. Now if $B$ is such a matrix, $C = B-\lambda I$ satisfies the same conditions with eigenvalue $0$. So the matrices that commutes with $A=I-u v^T$ are those of the form $B=\lambda I + C$ where $Cu = 0$ and $v^T C = 0$, i.e. all rows of $C$ are orthogonal to $u$ and all columns of $C$ are orthogonal to $v$.