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So this problem appeared on a list called "Matrix Problems", meant to be like training for competition-level maths. The whole problem goes like this: Let $A$ and $B$ be $n\times n$ matrices with complex entries. Prove there exists an invertible matrix $S$ and another matrix $T$ such that $B=S(A+T)S^{-1} -T$ if and only if $\text{tr}(A)=\text{tr}(B)$.

I didn't have any trouble solving the left-to-right part of the implication since it is easy to see that:

$B + T = S(A+T)S^{-1}$ $\implies$ $\text{tr}(B+T) = \text{tr}(A+T)$ $\implies$ $\text{tr}(B)=\text{tr}(A)$ .

But supposing traces are equal and then proving there exists another matrix that makes $B+T$ similar to $A+T$ is probably not so trivial. Of course, similar matrices have the same traces, but the converse does not always hold. I hope the idea here is to define $T$ in such a way that it "fixes" other requirements for the matrices to be similar, but I don't really remember any theorem that can make me be sure of that. I also don't know if the "complex entries" could help here, maybe to reach a Jordan form or something, since the trace, in this case, equals the sum of eigenvalues, but I couldn't find anything with this. Any solution or idea is appreciated.

Nothing special
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Kendall Rodriguez
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1 Answers1

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Let $P$ be an invertible matrix and define \begin{align*} B'&=P(B-A)P^{-1},\\ S'&=PSP^{-1},\\ T'&=P(A+T)P^{-1}. \end{align*} The equation $B=S(A+T)S^{-1}-T$ is then equivalent to $$ B'=S'T'S'^{-1}-T'.\tag{1} $$ Since $B-A$ has a zero trace, we may choose $P$ such that $B'$ has a zero diagonal. Now take $S'$ to be any diagonal matrix with $n$ different nonzero elements. Equation $(1)$ is solved by $t_{ij}'=\left(\dfrac{s_{ii}'}{s_{jj}'}-1\right)^{-1}b_{ij}'$ for all $i\ne j$. (The diagonal elements of $T'$ can be arbitrary.) Hence the original equation is solved by $S=P^{-1}S'P$ and $T=P^{-1}T'P-A$.

user1551
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