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If $$X = \left[ \begin{array}{ccc} 3 & 4 & 1\\ 4 & 1 & 3\\ 1 & 3 & 4\end{array} \right]$$

find the possible matrix $Y$ such that: $$XY - YX = I$$

The method my professor gave us was that if we observe the diagonal elements of $XY$, they will always be equal to the diagonal elements of $YX$.

$\therefore Trace(XY - YX)$
$=Trace(XY) - Trace(YX)$
$=0$

If $XY - YX = I$,
$\implies Trace(XY - YX) = Trace(I)$

The trace of an Identity matrix of the same order would be $1+1+1=3$.

$\because 0 \neq 3 \implies Y$ doesn't exist.

But, I decided to cross-check. I multiplied 3 pairs of matrices and none of their diagonal elements were same as their commutative counter-parts.

I'm confused. Is what my professor said correct? And did I mess up my calculations? Or does what he said not hold? If not, what would be the method to solve this problem, because I'm stumped. (assuming a matrix Y and solving it to get 9 equations isn't the way, I'm guessing).

mikhailcazi
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    try again, $tr(XY)=tr(YX)$ is definitely true. the diagonal elements of $XY, YX$ wont be the same, but the sum of the diagonal elements will be – yoyo Nov 08 '13 at 15:14

2 Answers2

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What your professor said, what he used, and what is true, is that the sum of the diagonal entries of $XY$ is the same as that of $YX$.

If you pay attention to the argument you wrote, you are never using the single entries of $XY$ and $YX$, but only the sum of their diagonals.

Martin Argerami
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Define the matrices $$ A = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n}\\ a_{21} & a_{22} & \cdots & a_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{bmatrix} \qquad B = \begin{bmatrix} b_{11} & b_{12} & \cdots & b_{1m}\\ b_{21} & b_{22} & \cdots & b_{2m}\\ \vdots & \vdots & \ddots & \vdots\\ b_{n1} & b_{n2} & \cdots & b_{nm} \end{bmatrix} $$ The $j^{th}$ diagonal entry of $AB$ is given by $$ \sum_{i=1}^n a_{ji}b_{ij} $$ Thus, the trace of $AB$ (the sum of all such diagonal entries) is given by $$ \sum_{j=1}^m \left[\sum_{i=1}^n a_{ji}b_{ij}\right] $$ Similarly, we may find the trace of $BA$ as $$ \sum_{i=1}^n \left[\sum_{j=1}^m a_{ji}b_{ij} \right] $$ Since these summations are equal (why?), we find $\operatorname{trace}(AB)=\operatorname{trace}(BA)$

Ben Grossmann
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