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can you have a P bigger than the original A matrix? in other words after I found the eigenvalues I then found all the eigenvectors so when I constructed the P vector turns out to be bigger than my original A. is that good?

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    $\begingroup$ Close to incomprehensible. Is $P$ the matrix you are diagonlizing $A$ with? If so, then no. How could you multiply $A$ and $P$ if their dimensions don't match? $\endgroup$
    – rschwieb
    Apr 30 '13 at 16:44
  • $\begingroup$ The matrix $P$ can only have the same size as $A$. Note that when you find the eigenvectors, you need to look for a basis for the eigenspace, not for all eigenvectors. $\endgroup$
    – N. S.
    Apr 30 '13 at 16:48
  • $\begingroup$ @rschwieb so true, sorry my brain is fried... $\endgroup$ Apr 30 '13 at 16:52
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If $A$ is an $n \times n$ matrix then the diagonalization of $A$ will be an $n \times n$ matrix.

So no that's not good.

Remember to diagonalize means to find an $n \times n$ invertible matrix $P$ so that $D = PAP^{-1}$ is diagonal. This is a product of $n \times n$ matrices so it must itself be $n \times n$.

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