The relationship between a matrix and its coefficient matrix decomposed in Pauli matrix

Question

For a dimension-$4$ Hermitian matrix $A$, denote pauli matrices $\{I,X,Y,Z\}$ as $\{\sigma_0,\sigma_1,\sigma_2,\sigma_3\}$ respectively. The pauli matrices form a basis of the matrix space if we take the trace as inner product.

$A$ can be written as $A=\sum_{ij}c_{ij}\sigma_i\otimes\sigma_j$. Here $\otimes$ denotes the Kronecker product. The coefficients $\{c_{ij}\}$ form a $4\times 4$ matrix $C=(c_{ij})_{ij}$.

My question is: is it possible to analyze the relationship between the eigenvalues of $A$ and singular values of $C$ (since $C$ may not be Hermitian)? Or more specifically, given the eigenvalues of $A$, can I predict any properties of $C$?

I tried some examples: $A=I\otimes I-X\otimes X+Y\otimes Y+Z\otimes Z$, the eigenvalues of matrix $C$ are $\{-1,1,1,1\}$, and eigenvalues of $A$ are $\{0,0,0,4\}$, and it seems difficult to find relations.

Any related problems, references or ideas are welcome.

Answer 1

Let $S_{ij}=\sigma _{i}\otimes \sigma _{j}.$ These matrices are unitary and Hermitian (with eigenvalues $\{-1,-1,1,1\}$, except for $S_{00}=I_{4}$ with unit eigenvalues). They are orthogonal with respect to the Frobenius inner product $\left\langle X,Y\right\rangle = \mathrm{tr} (X^{\ast }Y),$ (c.f. this post), which means that we can premultiply $A$ by $S_{ij}^{\ast }$ and take the trace to find $ c_{ij}=\frac{1}{4}\mathrm{tr}\left( S_{ij}^{\ast }A\right) $, as noted by Carlo Beenakker. So trivially the eigenvalues of $A$ are related to $c_{00}$ by

$$ c_{00}=\frac{1}{4}\mathrm{tr}\left( I_{4}A\right) =\frac{1}{4}\mathrm{tr}\left( A\right) =\frac{1}{4}\sum_{i=1}^{4}\lambda _{i} $$

More interesting is that

$$ \sum_{i=1}^{4}\lambda _{i}^{2}=4\sum_{i=1}^{4}\sigma _{i}^{2} $$

where $\sigma _{i}$ are the singular values of $C$. To see this, first note for convenience that since both $S_{ij}$ and $A$ are Hermitian, the $c_{ij}$ are real:

$$ 4c_{ij}=\left\langle S_{ij},A\right\rangle =\mathrm{tr}(S_{ij}^{\ast }A)=\mathrm{tr} (S_{ij}A^{\ast })=\mathrm{tr}(A^{\ast }S_{ij})=\left\langle A,S_{ij}\right\rangle =\overline{\left\langle S_{ij},A\right\rangle }=4 \overline{c_{ij}} $$

Now consider

$$ \mathrm{tr}\left( A^{\ast }A\right) =\mathrm{tr}\left( A^{\ast }\sum_{ij}c_{ij}S_{ij}\right) =\mathrm{tr}\left( \sum_{ij}c_{ij}A^{\ast }S_{ij}\right) =\sum_{ij}c_{ij}% \mathrm{tr}\left( A^{\ast }S_{ij}\right) =4\sum_{ij}c_{ij}^{2} $$

The quantity $\mathrm{tr}\left( A^{\ast }A\right) $ is the square of the Frobenius norm of $A$ and so

$$ \sum_{ij}|a_{ij}|^{2}=\left\Vert A\right\Vert ^{2}=\mathrm{tr}\left( A^{\ast }A\right) =4\sum_{ij}c_{ij}^{2} $$

I interpret this result is a Parseval equality. Since $A$ is normal, we have $ \sum_{ij}|a_{ij}|^{2}=\sum_{k}|\lambda _{k}|^{2}$ [H & J, Thm 2.5.4][1]. For $C$ (non-normal in general), there is a similar relationship involving singular values, $\sum_{ij}|c_{ij}|^{2}= \sum_{k}\sigma _{k}{}^{2}$ [H&J, Ch.7, Prob 4][1], and these relationships combine with the above equation to give the relationship between the eigenvalues of $A$ and singular values of $C$ given earlier.

In terms of the eigenvalues $\mu_i$ of $C$ we have only an inequality

$$ \sum_{i=1}^{4}\lambda _{i}^{2}\geq 4\sum_{i=1}^{4}|\mu_{i}|^{2} $$

with equality if and only if $C$ is normal [H&J Ch 5 Prob 26][1].

[1]: R.A.Horn and C.R. Johnson, Matrix Analysis, Cambridge University Press, 1985, pp. 101, 421 and 316.