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Let A be an $n\times n$ matrix with complex entries which is not a diagonal matrix. Pick out the cases when A is diagonalizable. (a) A is idempotent. (b) A is nilpotent. (c) A is unitary.

I think (c) is true. and (b) is false .not sure about (a), though order $2$ idempotent matrices are diagonalizable since it has two distinct eigenvalues $0,1$.but what is the general case .

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i think c is true. and b is false.not sure about a.though order 2 idempotent matrices are diagonalizable since it has two distinct eigenvalues 0,1.but what is the general case – poton Aug 7 '12 at 5:19
up vote 3 down vote accepted

(a) : $A$ is idempotent hence $A=A^{2}$ i.e. $A^{2}-A=0$ hence $x^{2}-x=x(x-1)$ is a polynomial $P$ s.t $P(A)=0$. Since the minimal polynomial of $A$ divides the polynomial we conclude that it is one of the following $x,x-1,x(x-1)$ . In all cases $A$ is diagonalize since the minimal polynomial splits into a product of coprime linear factors.

(b) : $A$ is nilpotent does not guarantee $A$ is diagonalize, as a counter example take $A=\begin{pmatrix}0 & 1\\ 0 & 0 \end{pmatrix}$ ($A$ is a $2\times2$ jordan block corresponding to the value $0$ , i.e. $A=J_{2}(0)$ ).

(c) : $A$ is unitary and in particular it is normal hence diagonalize

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In fact the only nilpotent $n \times n$ matrix that is diagonalizable is the $0$ matrix. – Robert Israel Aug 7 '12 at 5:28
Indeed, since it can be proven that the only eigenvalue is $0$ hence the Jordan form is a direct sum of matrices of the form $J_k(0)$ and is diagonizable iff $k=1$ for all the matrices – Belgi Aug 7 '12 at 5:32

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