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Given that $A=\begin{pmatrix}1 &2\\ 0 &-1\end{pmatrix}$ solve $$XA = AX.$$

Multiplying with inverses from the different sides doesn't seem to do the job.

Parseval
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    This is a small enough problem that you could just set $X = \begin{pmatrix}a&b\c&d\end{pmatrix}$, calculate the entries of $XA$ and $AX$, and solve the resulting system of equations. – JimmyK4542 Oct 21 '17 at 22:00
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    The more general picture is given by Sylvester equation (https://en.wikipedia.org/wiki/Sylvester_equation). But I don't advise you to use this sledgehammer. The advise of @JimmyK4542 is a good one. – Jean Marie Oct 21 '17 at 22:05
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    @JeanMarie there is a way with intermediate difficulty, see my answer here or http://math.stackexchange.com/questions/92480/given-a-matrix-is-there-always-another-matrix-which-commutes-with-it/92832#92832 – Will Jagy Oct 22 '17 at 00:08

2 Answers2

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Let $$X = \begin{pmatrix} a & b \\ c & d\end{pmatrix}$$ Now, let's consider what conditions $X$ must satisfy such that it commutes: $$\begin{pmatrix} 1 & 2 \\ 0 & -1\end{pmatrix}\begin{pmatrix} a & b \\ c & d\end{pmatrix} = \begin{pmatrix} a & b \\ c & d\end{pmatrix}\begin{pmatrix} 1 & 2 \\ 0 & -1\end{pmatrix}$$ This multplies out to be: $$\begin{pmatrix}a +2c & b+2d \\ -c & -d\end{pmatrix} = \begin{pmatrix}a & 2a-b \\c & 2c-d\end{pmatrix}$$ So, we now have 4 equations that must be satisfied. Some of them, like $c = -c$, easily give us a conclusion (that $c = 0$). After this, we see that $a = a$, and $-d = -d$, so these can be anything. The only other real condition is that $b+2d = 2a-b\implies b = a-d$. So, we have that: $$X = \begin{pmatrix}a & a-d \\ 0 & d\end{pmatrix}$$ Should work for any choice of $a$ and $d$.

Mark Schultz-Wu
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As the two (evident) eigenvalues are distinct, it follows that the characteristic polynomial and the minimal polynomial of $A$ coincide. Therefore, the only matrices that commute with $A$ are polynomials in $A$ itself, that is $$ sI + t A = \left( \begin{array}{cc} s + t & 2t \\ 0 & s-t \end{array} \right) $$ It is not necessary to use terms $A^2$ or $A^3$ because of Cayley-Hamilton

See http://math.stackexchange.com/questions/92480/given-a-matrix-is-there-always-another-matrix-which-commutes-with-it/92832#92832 for some background

Will Jagy
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