$\forall P \succ 0, \langle X,P \rangle = \langle Y,P \rangle$, implies $X=Y$?

Question

For n by n matrices X,Y,P, if $\forall P \succ 0, \langle X,P \rangle = \langle Y,P \rangle$, does $X=Y$, where the inner product is the Forbenius inner product?

This question arises from some optimization work I'm doing and in the case of my question $Y=X^{T}$. Since all positive definite matrices are not a subspace, we cannot really pick a basis. I tried to use some special matrix P to isolate elements of X but it doesn't go really well. Maybe we can do something with the trace?

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My original question is about when X and Y are not necessarily symmetric. In that case, X does not have to be Y (as described in the first comment), so what if we require X and Y to be symmetric?

Answer 1

Here is a perspective on this problem that you might find helpful. If we let $M = X - Y$, then the following statements are equivalent: $$ \forall P \succ 0, \langle X,P\rangle = \langle Y,P\rangle\\ \forall P \succ 0, \langle X,P\rangle - \langle Y,P\rangle = 0\\ \forall P \succ 0, \langle X-Y,P\rangle = 0\\ \forall P \succ 0, \langle M,P\rangle = 0. $$ Thus, the question amounts to the following: for a matrix $M$, does $\langle M,P \rangle = 0$ for all $M$ imply that $M = 0$ (or, what can we say about $M$ under this condition)?

As the comments on your question reflect, it is not necessarily the case that $M = 0$. This is because the set of positive definite matrices does not span the space of all matrices, it spans only the set of the symmetric matrices. That is, $$ \operatorname{span}\{P:P \succ 0\} = \{A:A = A^T\} =: \mathcal S. $$ The fact that $\langle M,P \rangle = 0$ tells us that $M$ is within the orthogonal complement of this span. That is, $M \in \mathcal S^\perp$. As it turns out, this implies that $M^T = -M$, which is to say that $M$ is skew-symmetric.

To reframe things in terms of the original question $\langle X,P \rangle = \langle Y,P\rangle$ will hold for all positive definite $P$ if and only if the matrix $X-Y$ is skew-symmetric.

Answer 2

If $Y=X^T$ take $P=X^TX$ and note that $\langle X^T,P\rangle = \langle X,P\rangle $ implies $$ \|X^T\|^{3/2}=\|X\|^{3/2}. $$ Therefore $X^T=X$.

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Below is a 'background' consisting of a series of numbered facts to understand the above solution.

Fact 1. For all $U,V\in\mathbb{R}^{n\times n}$ we have $$ \langle U, V \rangle \overset{_{\rm def}}{=}\mathop{\rm trace}(U^TV), \mbox{ and } \|X\|=\sqrt[2\,\,]{\langle X, X\rangle }. $$ Fact 2. For all $X\in \mathbb{R}^{n\times n}$ we have $X^{T}X \succ 0 $ because for all $v\in\mathbb{R}^{n}$ we have $(X\cdot v)^T(X\cdot v)>0 $ and $v^T(X^TX)v=(X\cdot v)^T(X\cdot v)$.

Fact 3. For all $X\in \mathbb{R}^{n\times n}$ we have $X^{T}X$ symmetric because $(X^{T}X)^{T}=X^T(X^T)^T=X^TX$.

Fact 4. By the Spectral Theorem for Symmetric Operators ( finite dimensional case) if $A$ is symmetric on $\mathbb{R}^{n\times n}$, then there exists an orthonormal basis of $\mathbb{R}^{n}$ consisting of eigenvectors $v_1,\ldots,v_n$ of $A$ and each eigenvalue $\lambda_{1},\ldots,\lambda_{n}$ is real. So we have the

Fact 5. If $A\succ 0$ and $A=A^T$ $$ {\rm trace}(A)=\lambda_1+\ldots+\lambda_{n} $$