# Joint convexity of trace of matrices

Let $$\Gamma_{m\times m}$$ be a diagonal matrix with positive diagonal entries and $$\mathbf{A}_{m\times m}$$ be an arbitrary matrix. Then, is the following trace function jointly convex on $$\Gamma_{m\times m}$$ and $$\mathbf{A}_{m\times m}$$? $$\mathrm{Tr}[\mathbf{A}\Gamma^{-2}\mathbf{A}^{\mathrm{H}}\Gamma^{2}],$$ where superscript $$(.)^{\mathrm{H}}$$ stands for hermitian operator.

This is false.

First, $$\mathrm{Tr}(AA^{\mathrm{H}}) = \|A\|_F^2$$ is the square of the Frobenius norm. Then the trace has this nice circular property $$\mathrm{Tr}(ABC)= \mathrm{Tr}(CAB)$$, by which

$$\mathrm{Tr}[\mathbf{A}\Gamma^{-2}\mathbf{A}^{\mathrm{H}}\Gamma^{2}] = \mathrm{Tr}[\Gamma\mathbf{A}\Gamma^{-1}(\Gamma\mathbf{A}\Gamma^{-1})^{\mathrm{H}}] = \|\Gamma A\Gamma^{-1}\|_F^2.$$

Now to conclude you would take $$A,B \in \mathbb{C}^{n\times n}$$ and $$\Gamma,\Phi$$ diagonal, Hermitian positive definite and a $$\lambda \in (0,1)$$ and look at the convex combination $$\lambda(A,\Gamma) + (1-\lambda)(B,\Phi)$$ and the evaluation of your function. This leads to $$(*) \quad \quad\|(\lambda\Gamma + (1-\lambda)\Phi) (\lambda A + (1-\lambda)B) (\lambda\Gamma + (1-\lambda)\Phi)^{-1}\|_F^2$$

With this in mind it is easy to construct a counterexample. Let $$A=\begin{bmatrix} 0 & 1 \\ 0 & 0\end{bmatrix}, \quad \Gamma=\begin{bmatrix} 1 & 0 \\ 0 & \varepsilon \end{bmatrix},\quad B=\begin{bmatrix} 0 & 0 \\ 1 & 0\end{bmatrix}, \quad \Phi=\begin{bmatrix} \varepsilon & 0 \\ 0 & 1\end{bmatrix}.$$

Then $$\|\Gamma A\Gamma^{-1}\|_F^2 = \varepsilon^{-2} = \|\Phi B\Phi^{-1}\|_F^2$$ and for $$\lambda =1/2$$ plugged into (*) (note that one factor $$1/2$$ cancels) $$\|1/2(\Gamma + \Phi) (A + B) (\Gamma + \Phi)^{-1}\|_F^2 = \frac{1}{4}\|A + B\|_F^2 = \frac{1}{2}.$$

Now choose any $$\varepsilon > \sqrt{2}$$ and you have a counterexample.