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I am trying to prove (if possible), for given $A_{n\times n}$ and $B_{n\times n}$, there exists a $C_{n\times n}$ satisfying $$A\pmb{X}_1 + B\pmb{X}_2 \stackrel{D}{=} C\pmb{X},$$ where $X_1, ~X_2$, and $X$ are independently and identically distributed Cauchy random variables with location vector $\pmb{0}$ and scale matrix identity, $I$.

I planned to proceed using the characteristic function rather than using convolution: \begin{align} \Phi_{A\pmb{X}_1 + B\pmb{X}_2}(\pmb{t}) &= E\left(e^{i\pmb{t}^T(A\pmb{X}_1 + B\pmb{X}_2)}\right)\\ &=E\left(e^{i\pmb{t}^T(A\pmb{X}_1)}\right)E\left(e^{i\pmb{t}^T(B\pmb{X}_2)}\right)\\ &=E\left(e^{i(A^{T}\pmb{t})^T(\pmb{X}_1)}\right)E\left(e^{i(B^{T}\pmb{t})^T(\pmb{X}_2)}\right)\\ &=e^{-\sqrt{ \pmb{t}^TAA^{T}\pmb{t}}}e^{-\sqrt{ \pmb{t}^TBB^{T}\pmb{t}}}, \end{align} and \begin{align} \Phi_{C\pmb{X}}(\pmb{t}) &= e^{-\sqrt{ \pmb{t}^TCC^{T}\pmb{t}}}; \end{align}

thus comparing RHS and LHS will reduce the problem to finding a $C$, for given $A$ and $B$, satisfying $$ \sqrt{\pmb{t}^TAA^{T}\pmb{t}} + \sqrt{\pmb{t}^TBB^{T}\pmb{t}} = \sqrt{\pmb{t}^TCC^{T}\pmb{t}}, $$ for all possible choices of $\pmb{t}$.

After this step, the idea of taking different choices of $\pmb{t}$ to find an explicit solution of $C$ can be implemented, but I could not find $C$ explicitly in terms of $A$ and $B$ following this idea. Any suggestions will be greatly appreciated.

Note: $A$, $B$, and $C$ are real square matrices and $A^{T}$ indicates the transpose of the matrix $A$.

Shanks
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  • Such a $C$ does not exist in general unless $A$ and $B$ commute. Use this fact to construct explicit counterexamples for $n=2.$ – whuber Jun 03 '18 at 14:26
  • Can you find $C$ explicitly if $A$ and $B$ commute? If that is the case, you can assume so. – Shanks Jun 03 '18 at 14:29
  • When $A$ and $B$ commute, upon simultaneous diagonalization the problem reduces to a set of one-dimensional problems. I leave it to you to find the easy solution to those! – whuber Jun 03 '18 at 14:30
  • If $A$ and $B$ are simultaneously diagonalizable, then $A$ and $B$ commute. The converse is not necessarily true. – Shanks Jun 03 '18 at 14:44
  • I would be interested in seeing a counterexample. – whuber Jun 03 '18 at 16:06
  • Please visit this [Wikipedia Page](https://en.wikipedia.org/wiki/Commuting_matrices). – Shanks Jun 03 '18 at 16:32
  • Thank you, but that's not a counterexample because your matrices $AA^\prime$ are special: they represent (positive-semidefinite) quadratic forms and are symmetric. – whuber Jun 03 '18 at 20:17
  • In your first comment, you said that the commutativity of $A$ and $B$ is needed. $A$ and $B$ are any real matrices. They are not necessarily symmetric or positive-semidefinite. – Shanks Jun 03 '18 at 20:40
  • I apologize for any confusion that might have ensued by referrig to $A$ and $B$ when I meant $AA^\prime$ and $BB^\prime$. The latter, of course, are what are involved in your question and so only the properties of the latter are going to be of any relevance to its solution. – whuber Jun 03 '18 at 22:06

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