What can I say about $\mathbf{X}$ if $\mathbf{y}^{\mathrm{T}} \mathbf{X} \mathbf{y} = \mathbf{z}^{\mathrm{T}} \mathbf{X} \mathbf{z}$?

37 Views Asked by At

Consider the $N$-dimensional positive semidefinite matrix $\mathbf{X}$ and two $N$-dimensional vectors $\mathbf{y}$ and $\mathbf{z}$ with $\|\mathbf{y} \|^{2} = \| \mathbf{z} \|^{2}$. If I know that $\mathbf{y}^{\mathrm{T}} \mathbf{X} \mathbf{y} = \mathbf{z}^{\mathrm{T}} \mathbf{X} \mathbf{z}$, what more can I say about $\mathbf{X}$? Ideally, I'd like to find $\mathbf{X}$ as a function of $\mathbf{y}$ and $\mathbf{z}$.

Tentative solution. Letting $\mathbf{Y} = \mathbf{y} \mathbf{y}^{\mathrm{T}}$ and $\mathbf{Z} = \mathbf{z} \mathbf{z}^{\mathrm{T}}$, the condition $\mathbf{y}^{\mathrm{T}} \mathbf{X} \mathbf{y} = \mathbf{z}^{\mathrm{T}} \mathbf{X} \mathbf{z}$ can be rewritten as $\mathrm{tr}(\mathbf{X} (\mathbf{Y} - \mathbf{Z}))=0$. Evidently, the matrix $(\mathbf{Y} - \mathbf{Z})$ has rank 2 and is not positive semidefinite; if it was positive semidefinite, I could say that $\mathbf{X}$ and $(\mathbf{Y} - \mathbf{Z})$ had the same eigenvectors and complementary eigenvalues, but this is not the case and thus I don't know how to proceed.