1
$\begingroup$

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

  • 2
    It seems to me there is a sign error $-a+b=b-d$, not $b+d$ as you wrote.2012-12-11
  • 2
    The first thing to do is to check to see whether your answer works, that is, whether it commutes with the given matrix. You don't need anyone's help to do that.2012-12-11
  • 2
    Any matrix commutes with the identity... If your answer doesn't contain $I_2$, it cannot be right.2012-12-11
  • 1
    Your second equation implies $a = d$.2012-12-11
  • 3
    Any matrix commutes with itself, so if your answer doesn't contain $A$, it can't be right.2012-12-11
  • 0
    typos amended ..thanks all for the comments2012-12-11
  • 0
    Why do you want $b$ to be $0$?2012-12-12

1 Answers 1

0

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 $v$.