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For square diagonal $L$ and $M$, and invertible $A$ and $B$, if

$$X = A L A^{-1},$$ and $$Y = B M B^{-1}$$

are equal ($X=Y$), is there a way to express $A$ explicitly in terms of $B$, $L$, $M$?

Seems easy enough; brain is just not working today...

(P.S. Thanks for answers on earlier question on this problem!)

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I gave an explicit expression for $L$ in terms of $A,B,M$ in your previous question. The logic is the same. There is no requirement that $L,M$ be diagonal, just that they be invertible. –  Ross Millikan Jun 27 '13 at 21:35
Can you clarify 'the logic is the same'? The problem I am having is that, unlike L, the A and its inverse appear split in the RHS, which makes manipulation to find A a bit more tricky, in my opinion. If I move inv(A) over to the LHS as .A, it looks very similar to an eigenproblem -- except for the order. Thanks for all the hints and efforts thus far! –  Lucozade Jun 27 '13 at 22:26
You need to recognize the combinations $(AB)^{-1}=B^{-1}A^{-1}$ that appear, then it is the same style of unpacking, as David shows. –  Ross Millikan Jun 27 '13 at 22:54

2 Answers 2

Since $L$ and $M$ are each diagonalizations of the same matrix $X=Y$, we know that $L$ and $M$ have the same diagonal entries. That is, there is some permutation matrix $P$ such that $L = P M P^{-1}$. Thus, the first equality becomes $$ X = APMP^{-1}A^{-1} = (AP)M(AP)^{-1} = Y = BMB^{-1} $$ That is, $B=AP$ for some permutation matrix $P$.

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If $A$ and $B$ are rectangular, then they can't possibly have inverses, so the premise doesn't hold. However, assuming that $A$ and $B$ were square, then:

$X = ALA^{-1} = BMB^{-1} = Y$ is what we're working with.


$ALA^{-1} = BMB^{-1}$

$ALA^{-1}A = AL = BMB^{-1}A$

$ALL^{-1} = A = BMB^{-1}AL^{-1}$ This only works if all the elements on the diagonal are non-zero. If so, then $L$ is invertible.

$A = BMB^{-1}AL^{-1}$, which is the closest you're going to get. Again, this assumes that that $A$ and $B$ are both square, because they can't have inverses if they're rectangular, and $L$ must be non-zero for every diagonal entry.

Granted, this isn't explicit, but I believe it is the closest you will get.

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