If $\frac{\mbox{Tr}(AB)}{\mbox{Tr}(B)} = 1$ for every $B \succeq 0$, then $A=I$

Question

Prove that if $\frac{\mbox{Tr}(AB)}{\mbox{Tr}(B)} = 1$ for every positive semidefinite matrix $B$, then $A$ is the identity.

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A professor of mine recently used this property in lecture to show that a given matrix is the identity, but this property is not obvious to me. Could someone show me a proof of this?

The best I could come up with (shared below) is that if $A$ is not the identity, there exists a vector $v$ of unit length such that $v^*Av\neq 1$, so we can define $B = vv^*$, so that $\operatorname{Tr}(B) = 1$ and $\operatorname{Tr}(AB)/Tr(B) = v^*Av \neq 1$, and this proves the statement by proving the contrapositive.

Does this proof require $A$ to be Hermitian? Are there any counterexamples where $A$ is not Hermitian? I would be grateful if anyone could offer a more rigorous proof.

Answer 1

I am assuming here that $A$ is Hermitian. Then, one can write $A-I=\sum_i\lambda_iu_iu_i^*$, where the $\lambda_i$'s are the eigenvalues and $u_i$'s are the (orthogonal) eigenvectors.

Then we have that

$$\mathrm{trace}((A-I)B)=\sum_i\lambda_iu_i^*Bu_i=0$$

If we assume that $A=I$, then we have that all the $\lambda_i$'s are zero, and the above expression holds for all $B$'s.

To prove the necessity, pick $B=u_ku_k^*$, $k=1,\ldots$. This yields

$$\mathrm{trace}((A-I)B)=\lambda_k=0$$

where we have used the fact that the eigenvectors or orthogonal. This implies that the above expression holds if and only if $\lambda_k=0$. Repeating for all $k$'s then implies that all the eigenvalues $\lambda_i$ must be zero and that $A$ must be the identity matrix.

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Update. In fact, a similar argument can be applied when $A$ is not Hermitian by using the singular value decomposition

$$A-I=\sum_i\sigma_iu_iv_i^*$$

where the matrices $U=[u_1\ \ldots\ u_n]$ and $V=[v_1\ \ldots\ v_n]$ are unitary (i.e. $U^*U=V^*V=I$). This yields

$$\mathrm{trace}((A-I)B)=\sum_i\sigma_iv_i^*Bu_i=0.$$

The rest of the proof follows from the same lines (using now the test matrices $B=v_ku_k^*$ or $B=u_ku_k^*$ for the symmetric PSD case with the extra assumption that $u_k^*v_k\ne0$.) and using the fact that if all the singular values of a matrix are zero, then this matrix must be the zero matrix.

Answer 2

I found a nice "physics-y" way to prove this. If A is NOT the identity, then there must exist some vector v of unit length such that the inner product v*Av =/= 1.

We can then use this vector v to construct the matrix B = vv*, which we know to be Hermitian and positive semidefinite, since it is just a projection onto v.

Since trace is invariant under change of basis, we have Tr(B) = v*v = 1.

We also have, by the same logic, Tr(AB) = v*Av =/= 1, and thus Tr(AB)/Tr(B) =/= 1.

Therefore, if A is not the identity, there exists a positive semidefinite Hermitian matrix B such that Tr(AB)/Tr(B) =/= 1.

The contrapositive of this statement proves the theorem: If Tr(AB)/Tr(B) = 1 for every positive semidefinite matrix B, then A must be the identity. QED

Answer 3

This answer expands on my comment. It is not necessary for $A$ to be Hermitian.

We note, as I did in my comment, that $$\operatorname{Tr}(AB) = \operatorname{Tr}(B) \iff \operatorname{Tr}(AB - B) = 0 \iff \operatorname{Tr}((A - I)B) = 0.$$ If we let $C = A - I$, then we have that $\operatorname{Tr}(CB) = 0$ for all $B$, and we wish to show $C = 0$.

Here's where a little peek at the big picture helps. The fact is, $$\langle A, B \rangle = \operatorname{Tr}(AB^*)$$ forms an inner product on the $n \times n$ complex matrices. So, to say that $\operatorname{Tr}(CB) = 0$ for all $B$, is equivalent to saying $\operatorname{Tr}(CB^*) = 0$ for all $B$, i.e. $\langle C, B \rangle = 0$ for all $B$. This is a classic exercise in the study of inner product spaces: the only such $C$ is $C = 0$. The proof is simple: consider $B = C$. Then $\langle C, C \rangle = 0$, and one of the axioms of inner product spaces implies $C = 0$.

If we don't take it for granted that $\langle \cdot, \cdot \rangle$ is an inner product, then we'll need to prove the relevant property: that $$\operatorname{Tr}(CC^*) = 0 \implies C = 0.$$ By definition, $$(CC^*)_{ii} = \sum_{j=1}^n (C)_{ij}(C^*)_{ji} = (C)_{ij}\overline{(C)}_{ij} = \sum_{j=1}^n |(C)_{ij}|^2.$$ Thus, $$\operatorname{Tr}(CC^*) = \sum_{i=1}^n (CC^*)_{ii} = \sum_{i=1}^n\sum_{j=1}^n |(C)_{ij}|^2.$$ This is a sum of non-negative terms. If $C \neq 0$, i.e. any $(C)_{ij} \neq 0$, then one of the terms will be strictly positive, making $\operatorname{Tr}(CC^*) > 0$. Thus, contrapositively, if $\operatorname{Tr}(CC^*) = 0$, then $C = 0$.

So, given that $\operatorname{Tr}(CB) = 0$ for all $B$, consider $B = C^*$ (which is $A^* - I$, as per my comment). We then get $\operatorname{Tr}(CC^*) = 0$, so $C = 0$, as we just showed. But, since $C = A - I$, this means $A = I$.

Answer 4

Another (equivalent) view as the one in @TheoBendit's answer but using rotational property of the trace : If $\mathrm{Tr}(CB)=0$ for all $B$, then for any $i$ and $j$

\begin{align*} \mathrm{Tr}(Ce_ie_j^T)&=\mathrm{Tr}(e_j^T C e_i)\\ &=e_j^T C e_i \end{align*} therefore $C=0$.