I was asking myself whether, if $Y\in M_n(\mathbb C)$ is such that $XY-YX=Y$ for some other $X\in M_n(\mathbb C)$, then the spectrum $\text{eig}(Y)$ must be $\{0\}$. I think that this is true for $n=1,2$ but I don’t know how to prove it in general. I only know that $\text{tr}(Y)=0$ and so the sum of the eigenvalues (counting multiplicity) has to be zero. Thanks!
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$XY=Y(X+I)$ then $XY^2=Y^2(X+2I)$ and $XY^k=Y^k(X+kI)$ for all $k\geq 0.$ Not sure how that helps. – Thomas Andrews May 14 '24 at 16:03
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You can prove it using the following identity for the matrix exponential $$ e^ABe^{-A}= I+[A,B]+\frac12[A,[A,B]]+\frac16[A,[A,[A,B]]]+\ldots\,, $$ which shows that $$ e^{-tX} Ye^{tX} = e^{-t}Y\,. $$ Then we can consider how the characteristic polynomial $p(\lambda)=\det(Y-\lambda I)$ behaves under a rescaling of its argument $$ p(e^t \lambda) = \det(Y-e^t\lambda I) = e^{nt}\det(e^{-t}Y-\lambda I)= e^{nt}\det(e^{-tx}[Y-\lambda I]e^{tX})= e^{nt}p(\lambda)\, , $$ which immediately implies that $p(\lambda)\propto \lambda^n$.
Ben
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1Inspired by @Ben's beautiful proof.
By the product rule for derivatives $\frac{d}{dt}e^{tX}Ye^{-tX}=e^{tX}(XY-YX)e^{-tX}=e^{tX}Ye^{-tX}.$ So $e^{tX}Ye^{-tX}$ and $e^{t}Y$ both satisfies the initial value problem $f'(t)=f(t),f(0)=Y.$ Thus $e^{tX}Ye^{-tX}=e^{t}Y.$ That is $Ye^{-tX}=e^{t}e^{-tX}Y.$ If $x$ is is eigenvector for $Y$ corresponding to the eigenvalue $w,$ then $Ye^{-tX}x=we^{-t}e^{tX}x.$ Hence, the numbers $we^{-t},-\infty<t<\infty$ are all eigenvalues for $Y.$ Since $Y$ only has a finite number of eigenvalues $w=0.$
– Steen82 May 14 '24 at 18:43
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My attempt for the two-by-two case seems to confirm the conjecture.
Since the trace of $Y$ is zero it is sufficient to consider the cases where the diagonal entries of $Y$ are $1,-1$ and $0,0$.
The $\textbf{first}$ case is $Y=\left(\begin{array}{cc} 1 & y\\ x & -1 \end{array}\right).$ Then the eigenvalues $w$ of $Y$ are determined by $1-w^{2}+xy=0,$ hence $w=\pm\sqrt{1+xy}.$
Now consider $X=\left(\begin{array}{cc} a & b\\ c & d \end{array}\right)$ and see what we can learn about $xy$ from $XY-YX=Y.$ The equation $XY-YX=Y$ simplifies to $$\left(\begin{array}{cc} bx-cy & ay-2b-dy\\ 2c+dx-ax & cy-bx \end{array}\right)=\left(\begin{array}{cc} 1 & y\\ x & -1 \end{array}\right).$$ The diagonal gives $$bx-cy=1$$ and the off-diagonals gives
$$ x =\frac{2c}{1+\left(a-d\right)}\mathrm{ \,\, and \,\, } y =\frac{-2b}{1-\left(a-d\right)}.$$ Plugging into the diagonal condition we get $$\frac{4bc}{1-\left(a-d\right)^{2}}=1.$$ Using this we get $$xy=\frac{2c}{1+\left(a-d\right)}\frac{-2b}{1-\left(a-d\right)}=\frac{-4bc}{1-\left(a-d\right)^{2}}=-1.$$ Hence the eigenvalues of $Y$ are $0,0.$
The $\textbf{second}$ case is $Y=\left(\begin{array}{cc} 0 & y\\ x & 0 \end{array}\right).$ Then the eigenvalues are $w=\pm\sqrt{xy}.$
Again consider $X=\left(\begin{array}{cc} a & b\\ c & d \end{array}\right).$ The equation $XY-YX=Y$ simplifies to $$\left(\begin{array}{cc} bx-cy & ay-dy\\ dx-ax & cy-bx \end{array}\right)=\left(\begin{array}{cc} 0 & y\\ x & 0 \end{array}\right).$$ The diagonal gives $bx-cy=0$ and the off-diagonals gives $$ 0 =\left(1+\left(a-d\right)\right)x \mathrm{\,\, and \,\, } 0 =-\left(1-\left(a-d\right)\right)y. $$ If $x$ and $y$ both are $\neq0,$ then $\left(1+\left(a-d\right)\right)=0$ and $\left(1-\left(a-d\right)\right)=0,$ hence $a-d=-1$ and $a-d=1.$ Consequently, at least one of $x$ and $y$ is zero. Again the eigenvalues of $Y$ are $0,0.$
Steen82
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