Given $N\times N$ square matrix $\mathbf{G}=[G_{i,j}]_{N\times N}$.
$G_{ii}=\sum_{i=1}^N 1/R_{i,j}$ when on diagonal; otherwise, $G_{i,j} = -1/R_{i,j}$, $i,j\in \{1,\cdots,N\}$, $R_{i,j}=R_{j,i}>0$ and real.
$\mathbf{C}=diag\{C_1,\cdots,C_N\}$, $C_i>0$.
Prove $\mathbf{A}=-\mathbf{C}^{-1}\mathbf{G}$ is diagonalizable.
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we can see $\mathbf{G}$ is diagonal dominant.
Since $\mathbf{G}$ is real symmetric, there exists an orthogonal matrix $\mathbf{Q}$, $\mathbf{Q}^T = \mathbf{Q}^{-1}$, that diagonalize $\mathbf{G}$ as $\mathbf{\Lambda}= \mathbf{Q}^T\mathbf{G}\mathbf{Q}$, where $\mathbf{\Lambda}$ is diagonal, with $\mathbf{G}$'s eigenvalue on the diagonal.
$\mathbf{C}$ is a diagonal matrix.
$\mathbf{A}=-\mathbf{C}^{-1}\mathbf{G}=-\mathbf{C}^{-1}\mathbf{Q}\mathbf{\Lambda}\mathbf{Q}^T$
Does $\mathbf{C}^{-1}\mathbf{Q} = \mathbf{Q}\mathbf{C}^{-1}$? How can I prove $\mathbf{A}$ is diagonalizable?
THANK YOU.
Hint: Note that $A$ is similar to $$ C^{1/2}AC^{-1/2} = -C^{-1/2}GC^{-1/2} $$ which is symmetric. I will leave it to you to figure out what $C^{1/2}$ should mean here.