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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.

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    Your notation for $C$ is bizarre. I assume that you mean $C$ is a diagonal matrix with (diagonal) entries $1,2,\dots,N$. Is this correct? – 2017-02-21
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    C is a diagonal matrix with diagonal entries $C_1$, $C_2$...$C_N$, and they are all real positive numbers. – 2017-02-21
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    Aha! That makes sense now. Actually, I realize now your notation is not so unusual. – 2017-02-21
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    Sorry, there is a condition, $R_{i,j} = R_{j,i}$ – 2017-02-21

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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.

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    Let $\tilde{\mathbf{A}}=\mathbf{C}^{1/2}\mathbf{A}\mathbf{C}^{-1/2}$. Then,$(\tilde{\mathbf{A}})^T=(\mathbf{C}^{1/2}\mathbf{A}\mathbf{C}^{-1/2})^T = -(\mathbf{C}^{-1/2}\mathbf{G}\mathbf{C}^{-1/2})^T = -(\mathbf{C}^{-1/2})^T\mathbf{G}^T(\mathbf{C}^{-1/2})^T$. So $\tilde{\mathbf{A}}$ is symmetric. So, $\tilde{\mathbf{A}}$ is diagonalizable. Thus, A is similar to a diagonalizable matrix and it is also diagonalizable. Is this correct? Thank you. – 2017-02-21
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    That's exactly it. Just make sure that you say what you mean by $C^{1/2}$. – 2017-02-21