# Let $0 \preceq A \preceq I$ can we relate $Tr( (I-A) B+A C)$ to $Tr(I-A)Tr(B)+Tr(A) Tr(C)$

Let $0 \preceq A \preceq I$ and let $B$ and $C$ be two symmetric positive definite matrices.

Can we related \begin{align} {\rm Tr}( (I-A) B+A C) \end{align}

to

\begin{align} Tr(I-A)Tr(B)+Tr(A) Tr(C) \end{align}

via some inequality?

Note, that in the case $A=aI$ for any $a\in [0,1]$ there is an equality.

Edit: Note, from a very nice answer below we have:

\begin{align} \lambda_{\min}(I-A){\rm Tr}(B)+\lambda_{\min}(A){\rm Tr}(C) \le {\rm Tr}( (I-A) B+A C) \le \lambda_{\max}(I-A){\rm Tr}(B)+\lambda_{\max}(A){\rm Tr}(C). \end{align}

• Just an observation, though it may not be useful to you. If $0\preceq A\preceq I$, then $0\preceq I-A\preceq I$. Therefore, if $B$ and $C$ are nonnegative matrices, $0\leq\operatorname{Tr}((I-A)B)\leq\operatorname{Tr}(B)$ and $0\leq\operatorname{Tr}(AC)\leq\operatorname{Tr}(C)$. It follows that in this case, $$0\leq\operatorname{Tr}((I-A)B+AC)\leq\operatorname{Tr}(B+C)$$ and $$0\leq\operatorname{Tr}(I-A)\operatorname{Tr}(B)+\operatorname{Tr}(A)(C)\leq\operatorname{Tr}(B+C)$$ (both quantities are nonnegative and bounded above by $\operatorname{Tr}(B+C)$). – parsiad Apr 19 '17 at 13:52

## 1 Answer

Note that the function $$\langle A,B \rangle = Tr(A^TB)$$ defines an inner product over the real matrices. As such, $\|A\|_2^2 = Tr(A^TA)$ is the norm induced by this inner product. We have $$Tr((I-A)B + AC) = Tr((I-A)B) + Tr(AC) =\\ \langle I-A,B \rangle + \langle A,C \rangle \leq\\ \|I - A\|\,\|B\| + \|A\|\,\|C\| = \\ \sqrt{Tr[(I - A)^2]Tr(B^2) + Tr(A^2)Tr(C^2)}$$ We can do a bit better using the inequality $|\langle A,B \rangle| \leq \sigma_1(A)\sum \sigma_i(B)$. In the symmetric positive semidefinite case, amounts to $|\langle A,B \rangle| \leq \rho(A) Tr(B)$, where $\rho$ denotes the spectral radius. We now have $$\langle I-A,B \rangle + \langle A,C \rangle \leq\\ \rho(I - A)Tr(B) + \rho(A) Tr(C)$$ This reduces to your result in the case that $A = aI$ for $a \in [0,1]$.

The reverse inequality works perfectly well for our purposes: note that $A \succeq \lambda_{min}(A)I$. So, $C^{1/2}AC^{1/2} \succeq \lambda_{min}(A)C^{1/2}C^{1/2}$. We therefore have $$Tr((I-A)B + AC) = Tr((I-A)B) + Tr(AC) =\\ Tr(B^{1/2}(I-A)B^{1/2}) + Tr(C^{1/2}AC) \geq\\ Tr(\lambda_{min}(I - A)B^{1/2}B^{1/2}) + Tr( \lambda_{min}(A)C^{1/2}C^{1/2}) = \\ \lambda_{min}(I-A)Tr(B) + \lambda_{min}(A) Tr(C)$$ in retrospect, a similar inequality could have worked in reverse.

• I don't have a reference for these inequalities of the top of my head. If someone could find such a reference, I would appreciate it. I'm using the Hölder's inequalities for Schatten p-norms. – Omnomnomnom Apr 19 '17 at 13:58
• What is this second inequality called that you are using? $|<A ,B>| \le \sigma_1 \sum \sigma_i(B)$. – Boby Apr 19 '17 at 14:00
• I don't know. In Pedersen's "Analysis Now", it's lemma 3.4.10, and that's the infinite dimensional version. – Omnomnomnom Apr 19 '17 at 14:06
• See also proposition IV.2.4 from Bhatia's Matrix Analysis – Omnomnomnom Apr 19 '17 at 14:11
• @Boby I'm going to post my own question about that. – Omnomnomnom Apr 19 '17 at 14:13