0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Marc van Leeuwen May 18 '16 at 16:20
  • $\begingroup$ Yes Rn is positive definite throughout. $\endgroup$ – The Dude May 19 '16 at 17:00
1
$\begingroup$

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 $P^{-1}=P^T$.

$\endgroup$
  • $\begingroup$ A clear explanation! Thanks! $\endgroup$ – The Dude May 29 '16 at 20:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.