Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

Why ?

share|improve this question
    
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
1  
Every diagonalizable matrix $A$ has a spectral decomposition $A=\sum \lambda_jp_j$. –  1015 Mar 14 '13 at 19:55
1  
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
add comment

2 Answers

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.

share|improve this answer
    
Hum... I see now what people call spectral decomposition: unitarily diagonalizable. –  1015 Mar 14 '13 at 20:20
add comment

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...

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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