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According to the Wikipedia article on the common spatial pattern algorithm, one can find the following matrices by joint diagonalization of a pair of covariance matrices $R_1$ and $R_2$:

$$ P = [\mathbf{p_1}, \cdots , \mathbf{p_n}] $$ $$ D = \mbox{diag} \{\lambda_1, \cdots ,\lambda_n\} $$

such that $\mathbf{P^{-1}R_1P=D}$ and $\mathbf{P^{-1}R_2P=I_n}$, where $\mathbf{I_n}$ is the identity matrix of rank $n$.

Now, I know that there are "plenty" of non-diagonalizable matrices, so I'm inclined to wonder what guarantees the above is possible, and when.

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I read the article and I'm baffled too... Isn't $P^{-1}R_2P = I$ the same as $R_2 = I$? –  Tunococ May 27 '13 at 2:39
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Then I'd say it is jointly diagonalizable if $R_2 = I$, and you only need to diagonalize $R_1$ because no matter what $P$ and $D$ are, $P^{-1}R_2P = I$ is guaranteed. I feel that the Wikipedia page you're referring to is not very reliable. –  Tunococ May 27 '13 at 2:48
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If $P^{-1}R_2P$ is diagonal, it must contain eigenvalues of $R_2$, so you at least require $R_2$ to have only one distinct eigenvalue. –  Tunococ May 27 '13 at 5:46
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I concur with Tunococ. The Wikipedia article you cited does not sound right. If $P^{-1}R_2P=I_n$, then $R_2$ is necessarily equal to $I_n$, but by assumption, $R_2$ is just an arbitrary covariance matrix. In general, it is not the identity matrix. –  user1551 May 27 '13 at 10:39
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In general, two square matrices $\,A,B\,$ are simultaneously diagonalizable (which seems to be more or less the same as jointly diagonalizable) iff there are diagonalizable and $\,AB=BA\,$ ... –  DonAntonio May 27 '13 at 21:25

1 Answer 1

The most likely explanation is that they are referring to joint approximate diagonalization by an orthogonal matrix the minimizes the sum of the Frobenius norm of the off-diagonal terms. The algorithm used is often called JADE and a quick web search for JADE and common spatial pattern picks up many promising hits.

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