Is the scalar field $ X \mapsto \mbox{tr} \left( X A X^T B \right) $ always convex?

Question

If the matrices $A$ and $B$ are positive semidefinite, is the scalar field $$ X \mapsto \mbox{tr} \left( X A X^T B \right) $$ always convex? I get an error when I put this in CVX even though I double-checked that $A$ and $B$ do not have negative eigenvalues.

_{Related: Is the function $X \mapsto \mbox{trace} \left( X X^T \right)$ convex?}

What is $X?$ An arbitrary matrix? – Igor Rivin Dec 20 '13 at 03:56 — Igor Rivin, Dec 20 '13 at 03:56
Yes its an arbitrary matrix – Jessica Spelman Dec 20 '13 at 03:59 — Jessica Spelman, Dec 20 '13 at 03:59

score 1 · Answer 1 · answered Dec 20 '13 at 04:01

1

Yes, this is $trace(Y Y^T)$ where $Y = B^{1/2} X A^{1/2}$ is a linear function of $X$, so it is indeed convex.

answered Dec 20 '13 at 04:01

Robert Israel

448,999

Indeed; and this in turn means that $\mathop{\textrm{Tr}}(XAX^TB)=|B^{1/2}XA^{1/2}|_F^2$, which can be represented in CVX as square_pos(norm(sqrtm(B)*X*sqrtm(A),'Fro')). – Michael Grant Dec 31 '13 at 06:09

score 0 · Answer 2 · answered Dec 20 '13 at 04:01

If $B$ is positive semi-definite, it is equal to $CC^t,$ so your expression is equal to $\mathrm{trace}((C^t X) A (C^t X)^t).$ This is basically just the Frobenius norm of the vector $C^t X,$ which is a linear function of $X,$ and is convex (strictly if $A, B$ are positive definite, not strictly otherwise).

Is the scalar field $ X \mapsto \mbox{tr} \left( X A X^T B \right) $ always convex?

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