For a hermitian matrix R and a diagonal one Q, is there any relationship between eigenvalues/eigenvectors of R and QRQ?
To be specific, assuming the eigenvalue decomposition of R is R=VDV*, then can the eigenvalues/eigenvectors of QRQ be represented by V, D & Q?
Thanks a lot!
In fact I've ask this question on math.stackexchange already, and was advocated to ask here. I do hope to get a proof for it.
I agree with the comments. However, there is a special case of interest, more general than commutation. Assume $R$ is a positive definite Hermitian matrix: this means that we can find a matrix $V$ such that $$ R=VDV^*,\quad V^*V=I,\quad\text{$D$ diagonal positive.} $$ Now if $S$ is another Hermitian matrix, the matrix $V^*SV$ is also Hermitian as well as $D^{-1/2}V^*SVD^{-1/2}$. Thus we can find $W$ such that $$ W^*D^{-1/2}V^*SVD^{-1/2}W=\Delta,\quad W^*W=I,\quad\text{$\Delta$ diagonal.} $$ We have thus $$ S=V D^{1/2}W\Delta W^* D^{1/2}V^*,\quad R=VDV^*=VD^{1/2}WW^*D^{1/2}V^* $$ which means that the $\boxed{\text{quadratic forms } R,S}$ are simultaneously diagonalizable: with the invertible matrix $P=W^*D^{1/2}V^*$, we have $$ \langle Rx,x\rangle=\langle P^*Px,x\rangle=\Vert Px\Vert^2,\quad \langle Sx,x\rangle=\langle P^*\Delta Px,x\rangle=\langle \Delta Px,Px\rangle,\quad\text{with $\Delta$ diagonal.} $$ In particular, if $(e_j)_{1\le j\le n}$ is the canonical basis of $\mathbb R^n$, $(f_j=P^{-1}e_j)_{1\le j\le n}$ is also a basis and for $x=\sum_j x_j f_j$, we have $$ \langle Rx,x\rangle=\sum_j x_j^2,\quad \langle Sx,x\rangle=\sum_j \Delta_j x_j^2,\quad \Delta=\text{diag}(\Delta_1,\dots,\Delta_n). $$
