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To perform spectral decomposition, do we require a matrix to be positive definite ?

Why ?

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Don't see what this has to do with statistics. The answer depends on whether you would like to guarantee real-valued spectrum, for example. If you are ok with having complex eigenvalues, matrix need not be positive definite. – gt6989b Mar 14 '13 at 19:54
Every diagonalizable matrix $A$ has a spectral decomposition $A=\sum \lambda_jp_j$. – 1015 Mar 14 '13 at 19:55
Every normal matrix can be diagonalized (so it doesn't have to be positive definite). – copper.hat Mar 14 '13 at 19:56
Ok thanks. Cos my friend said it has to be positive definite, which confused me – user1769197 Mar 14 '13 at 20:53

Positive definiteness is sufficient, but not necessary, for a spectral decomposition. Here, I'm taking the meaning of "spectral decomposition" of a matrix $A$ to mean an expression $$ A = \sum_{i} \lambda_i P_i $$ where $P_i$ is an orthogonal projection onto the corresponding eigenspace. Such a decomposition exists if and only if $A$ is a normal operator, and every positive-definite matrix is normal.

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Hum... I see now what people call spectral decomposition: unitarily diagonalizable. – 1015 Mar 14 '13 at 20:20

The answer depends on which properties of the resulting spectrum you would like to guarantee. It's perfectly possible to perform spectral decomposition on a large class on non-positive-definite matrices. Consider, e.g. the matrix $\left( 0 \quad 1\\-1 \quad 0 \right)$, which has eigenvalues $\pm \sqrt{-1}$.

Usually to guarantee real-valued spectrum, or some nice properties, we consider positive definite matrices, or work with $A^T A$, which is positive definite, instead of $A$ itself...

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