The Hilbert—Schmidt inner product

By definition, $(A^\dagger B)_{jj}=\sum_i(A^\dagger)_{ji}B_{ij}=\sum_iA_{ij}^*B_{ij}$, and summing over $j$ gives $\tr(A^\dagger B)=\sum_{i,j}A_{ij}^*B_{ij}$. This is exactly the usual Hermitian inner product on $\C^{n^2}$ when the $n^2$ entries of a matrix are arranged into a single vector. It is therefore linear in $B$, satisfies Hermitian symmetry $\langle B,A\rangle_{\rm HS}=\langle A,B\rangle_{\rm HS}^*$, and is positive definite: $\langle A,A\rangle_{\rm HS}=\sum_{i,j}|A_{ij}|^2\geq0$, with equality if and only if all entries vanish. The associated norm, $\norm A_{\rm HS}=\sqrt{\sum_{i,j}|A_{ij}|^2}$, is called the Hilbert—Schmidt norm, or Frobenius norm. The space $M_n(\C)$ equipped with this inner product is a Hilbert space of dimension $n^2$.

From the previous question, $\langle E_{kl},E_{k'l'}\rangle_{\rm HS}=\sum_{i,j}(E_{kl})_{ij}(E_{k'l'})_{ij}=\delta_{kk'}\delta_{ll'}$, since each matrix has a nonzero entry in only one position. These $n^2$ orthonormal matrices form a basis, and the expansion of a matrix in this basis is simply $A=\sum_{k,l}A_{kl}E_{kl}$, the formula $\hat A=\sum_{k,l}A_{kl}\ket{e_k}\bra{e_l}$ in Dirac notation. For a unitary matrix, $\norm U_{\rm HS}^2=\tr(U^\dagger U)=\tr\mathbf{1}=n$.

Since the Pauli matrices are self-adjoint, $\langle\sigma_j,\sigma_k\rangle_{\rm HS}=\tr(\sigma_j\sigma_k)$. For $j=k$, this gives $\tr\mathbf{1}=2$. For $j\neq k$, the product $\sigma_j\sigma_k$ is $\pm i\sigma_l$, whose trace vanishes. Finally, $\langle\mathbf{1},\sigma_j\rangle_{\rm HS}=\tr\sigma_j=0$ and $\langle\mathbf{1},\mathbf{1}\rangle_{\rm HS}=2$. The four matrices $\mathbf{1},\sigma_x,\sigma_y,\sigma_z$ are therefore orthogonal and have norm $\sqrt2$; dividing them by $\sqrt2$ gives four orthonormal matrices in a space of dimension $4$: this is an orthonormal basis.

The Cauchy—Schwarz inequality reads $|\tr(A^\dagger B)|^2\leq\tr(A^\dagger A)\,\tr(B^\dagger B)$. Taking $A=\mathbf{1}$ and renaming $B$ as $A$, we obtain $|\tr A|^2\leq\tr(\mathbf{1})\,\tr(A^\dagger A)=n\,\tr(A^\dagger A)$. Equality holds if and only if $A$ is proportional to the identity. For example, for a Hermitian matrix with eigenvalues $\lambda_k$, this inequality reads $\bigl(\sum_k\lambda_k\bigr)^2\leq n\sum_k\lambda_k^2$: the square of the mean of the eigenvalues is less than or equal to the mean of their squares.