# Possibly non-trivial inequality about traces and eigenvalues

The problem:

Let $A, B$ be real matrices of dimension $n$ and $\alpha_1 \geq \dots \geq \alpha_n$ and $\beta_1 \geq \dots \geq \beta_n \geq 0$ its eigenvalues, respectively ($B$ is symmetric positive semidefinite). Also, suppose that $\sum_{i=1}^n\beta_i=1$ and that the eigenvalues are all real.

Is it true that $\mbox{trace}(A^TB)\geq \alpha_n$?

My attempt:

Let $A=PJP^T$ be the Jordan representation of $A$ and $B=QDQ^T$ the spectral decomposition of $B$, that is, $P$ and $Q$ are orthonormal matrices of eigenvector bases, $J$ is the Jordan form of $A$ and $D=diag(\beta_1,\dots,\beta_n)$.

Then $$trace(A^TB)=trace(PJ^TP^TQDQ^T)=trace(\underbrace{(Q^TP)J^T(Q^TP)^T}_{this:=S}D)=\sum_{i=1}^n S_{ii}\beta_i.$$ And I can't continue.

My second (and wrong) attempt:

My wrong attempt is similar, but the underlined step is probably wrong. I know it is in general not true, but I am not sure about this particular case. $$trace(A^TB)=\underline{trace(PJ^TP^TQDQ^T)=trace(J^tPP^TQQ^TD)}=trace(J^TD)=\sum_{i=1}^n \alpha_i\beta_i\geq \alpha_n.$$

For instance, if this is true, then the lesser eigenvalue of $A$ is the solution of a semidefinite program: $$\alpha_n=\min_B\{trace(A^TB)\mid trace(B)=1,B\succeq 0\}.$$

Thank you!

Let \begin{align}A&=\begin{bmatrix}0&2\\0&0\end{bmatrix}&B=\frac{1}{2}\begin{bmatrix}1&-1\\-1&1\end{bmatrix}\end{align}\text{.} Under your conjecture one would have $\mathrm{Tr}\,A^{\mathsf{T}}B\geq 0$. Instead, $$\mathrm{Tr}\,A^{\mathsf{T}}B=-1\text{.}$$
• The eigenvalues of $A$ and $B$ are $\{0,0\}$ and $\{0,1\}$, respectively... – K B Dave Mar 2 '18 at 19:56
• Is it true when $A$ is symmetric? – Koto Mar 2 '18 at 19:59
• Yes, because by subtracting $I\alpha_n$ we can assume without loss of generality that $A$ is psd, in which case $\mathrm{Tr}\, AB=\mathrm{Tr}\, B^{1/2}AB^{1/2}\geq 0$. – K B Dave Mar 2 '18 at 20:01
• $A$ symmetric implies $A-\alpha_n I$ is psd, so set $A\leftarrow A-\alpha_n I$; what needs to be shown now is $\mathrm{Tr} AB\geq 0$. $B$ is psd, so its psd square root $B^{1/2}$ exists. by cyclicity of the trace, $\mathrm{Tr}AB^{1/2}B^{1/2}=\mathrm{Tr}B^{1/2}AB^{1/2}$, so the question is reduced to demonstrating the inequality $\mathrm{Tr}B^{1/2}AB^{1/2}\geq 0$. But $A$ psd implies $B^{1/2}AB^{1/2}$ psd, and traces of psd matrices are nonnegative. – K B Dave Mar 2 '18 at 20:13