# Simultaneous Diagonalization of A and B via $\Sigma = A^{-1}B$

I am reading the paper "A Generalized Subspace Approach for Enhancing Speech Corrupted by Colored Noise" by Yi Hu and Philipos C. Loizou. In the paper, they claim that given two matrices $R_{n}$ and $R_{x}$, both real symmetric $n \times n$ matrices, there exists a matix $V$ which can simultaneously diagonalize both of them in the following way:

$$V^{T} R_{x} V = \Lambda_{x}$$ $$V^{T} R_{n} V = I_{n}$$

Where $\Lambda_{x}$ and $V$ are the eigenvalue and eigenvector matrices of $\Sigma = R_{n}^{-1} R_{x}$. Ie:

$$\Sigma V = V \Lambda_{x}$$.

Then they claim that $\Lambda_{x}$ is real given that $R_{n}$ is positive definite. To back up their claims, they cite two books, which I don't have access to. From the references I have available, I can't seem to find a proof for this.

Does anyone have any ideas about how to show both of these things?

• You should state your hypotheses more clearly. Notably is $R_n$ positive definite throughout, or does it only become so in the before-last paragraph? – Marc van Leeuwen May 18 '16 at 16:20
• Yes Rn is positive definite throughout. – The Dude May 19 '16 at 17:00

I will suppose that $R_n$ was assumed positive definite, as otherwise $V^TR_nV=I_n$ is impossible.
I believe this is just the spectral theorem, which says that every symmetric matrix $S$ admits an orthogonal matrix $P$ such that $P^TSP$ is diagonal, slightly disguised. Note that if we associate to$~S$ the symmetric bilinear form $f_S:(v,w)\mapsto v\cdot Sw=v^T\!Sw$, then this theorem says that there exists a basis for $\Bbb R^n$ (defined by the columns of $P$) that is both orthonormal for the standard inner product, and orthogonal (but not orthonormal) for$~f_S$. That matches the two equations in the question (in reverse order).
So here it goes. The first step is to find a matrix $Q$ such that $Q^TR_nQ=I_n$. One can (for instance) first find an orthogonal $Q'$ such that $(Q')^TR_nQ'$ is diagonal; since $R_n$ is positive definite, the diagonal entries $d_i$ of $(Q')^TR_nQ'$ are positive (they are $Q_i^TR_nQ_i$ where $Q_i$ is column $i$ of $Q$). Then with $C$ the diagonal matrix with diagonal entries $d_i^{-1/2}$ take $Q=Q'C$; then $Q^TR_nQ =C^T(Q')^TR_nQ'C =I_n$. Note that $Q$ is not orthogonal; the orthogonality of $Q'$ which the spectral theorem gave us is irrelevant.
The second step is to apply the spectral theorem to $Q^TR_xQ$, which is a symmetric real matrix. So there is an orthogonal (this time it is relevant) $P$ such that $P^TQ^TR_xQP$ is diagonal; call it $\Lambda_x$. Since $P$ is orthogonal one has $P^T=P^{-1}$, and $P^TQ^TR_nQP=P^{-1}I_nP=I_n$. So one can take $V=QP$ in the question.
The relation $R_n^{-1}R_xV=V\Lambda_x$ is not very transparent to me, but it does appear to check out. Using $V=QP$ write it as $R_xQP=R_nQP\Lambda_x$ (which must be proved). We saw that $R_nQ$ is the inverse of $Q^T$, so this will follow from $Q^TR_xQP=P\Lambda_x$ or from $P^{-1}Q^TR_xQP$, which follows from the definition of $\Lambda_x$ since $P^{-1}=P^T$.