In general for a matrix $A$, we have $P^{-1}AP = D$
$P$ can be computed easily as they are formed from eigenvectors.
But sometimes we want $U^{-1}AU = D$, where $U$ is unitary. Why do we want $U$ when we have $P$ which can be computed more easily?
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Depends on whether your matrix is in $\mathbb{C}$ or $\mathbb{R}$. In $\mathbb{C}$ a matrix of eigenvectors is unitary, while in $\mathbb{R}$ it is orthogonal. – Hamed Oct 04 '15 at 03:35
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because we can use the spectral theorem:
a matrix is unitarily diagonalizable if and only if it is normal (AA*=A*A)
so it is more easy to check if a matrix is unitarily diagonalizable. and we prefer to find U, because the basis is orthonormal and we know that such U exist if the matrix normal.
D. A.
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but a matrix is also diagonalizable if and only if it is normal. Why do we want unitary? – Ka Wa Yip Oct 04 '15 at 03:46
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