# diagonalizing a matrix $A$: can $P$ be bigger than $A$?

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?

• 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? Apr 30 '13 at 16:44
• 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. Apr 30 '13 at 16:48
• @rschwieb so true, sorry my brain is fried... Apr 30 '13 at 16:52

If $A$ is an $n \times n$ matrix then the diagonalization of $A$ will be an $n \times n$ matrix.
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 $n \times n$.