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I'm having difficulty solving a linear algebra problem:
Let $A,B,C,D$ be real $n \times n$ matrices. Show that there is a non-zero $n \times n$ matrix $X$ such that $AXB$ and $CXD$ are both symmetric.

There is an accompanying hint:
Show that the set of all matrices $X$ for which $AXB$ is symmetric is a vector space, and compute its dimension.

I introduced some notation: let $S(A,B)$ denote the set of all $n \times n$ real matrices such that $AXB$ is symmetric. It's easy to prove that for any $A,B$ real $n \times n$ matrices $S(A,B)$ is a subspace of $\cal {M}_{n \times n} (\mathbb{R})$-it's closed under addition, scalar multiplication, and contains the zero matrix. Now I'm not entirely sure where to go. I'd like to be able to show that the space $S(A,B) \cap S (C,D)$ has dimension greater than $0$, but I haven't made any progress thus far.

Any help would be greatly appreciated!

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2 Answers 2

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Hint: The product $AXB$ has entries that can be expressed as linear combinations of the entries of $X$, where the coefficients come from the entries of $A$ and $B$. The symmetry condition imposes equations on the entries, so we have a linear system of equations with unknowns the off-diagonal entries of $X$. How many equations are there? You should be able to use this to compute the dimension, and it will be big enough to get what you want.

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Let $T$ map every matrix $X$ to $AXB$.

  1. Show that $T$ is a linear transformation.

  2. What is the preimage of the space of symmetric matrices under $T$?

  3. What is the dimension of the space of symmetric matrices?

  4. What can you deduce from 1 and 2 about the dimension of $S(A,B)$?

  5. Use 4 to prove the intersection is nontrivial.

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  • $\begingroup$ @MattSamuel: Who said anything about nonzero? $S(A,B)$ is the space of matrices $X$ such that $AXB$ is symmetric, not such that $AXB$ is nonzero symmetric. I think it works out in the end - every addition to $T$'s kernel adds a vector that maps to 0, and takes away at most a vector that maps to a nonzero symmetric matrix. I think I have a proof but I'm not 100% sure about it, can you provide a counterexample? $\endgroup$
    – Meni Rosenfeld
    Jun 17, 2015 at 19:28
  • $\begingroup$ I believe I see the point you are driving at, but am still having difficulty: let $S$ denote the space of all symmetric $n \times n$ real matrices. If we can prove that $T |_{S(A,B)}$ gives an isomorphism from $S(A,B)$ to $S$ then we'll be set-the latter space has dimension $\frac{n^2 + n}{2}$, then a counting argument shows the intersection is non-trivial. I can't see how to prove that $T|_{S(A,B)}$ is surjective, and I believe that for the argument to work this is necessary.. $\endgroup$
    – aherring
    Jun 18, 2015 at 20:27
  • $\begingroup$ @aherring: $T$ can have a nontrivial kernel, which $S(A,B)$ contains, so $T|_{S(A,B)}$ needn't be injective, and our plan isn't to show that it is an isomorphism. Rather, we want to prove $\dim T^{-1}(S) \ge \dim S$ (it can be strictly larger, obvious example is $A=B=0$. Decompose the dimension of the domain of $T|_{S(A,B)}$, and use theorem about dimension of intersection to bound $\dim(S)\cap Im T$. It should all work out unless I made a mistake somewhere. PS I originally thought that in general $\dim T^{-1}(S)\ge\dim S$ is an obvious well-known theorem, but even if not, we can prove it. $\endgroup$
    – Meni Rosenfeld
    Jun 18, 2015 at 20:57

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