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2/ Complex => A has an eigenvector, by fundamental theorem of algebra
1/ Short proof that complex normal matrics (A A* = A* A) are diagonalizable
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3/ Work in a basis including the eigenvector as first element, so first column of A is zero, except in top-left corner
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Is it obvious that the normal property is basis-independent?
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4/ Looking at top left element of A A* = A* A, length of first column of A = length of first row
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5/ First column of A is zero off-diagonal, so the first row must be, too, by 4.
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6/ So A has block diagonal form. Recurse on sub-matrices. QED
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7/ Incidentally, the proof of the converse (diagonalizable matrices are normal) is a trivial calculation. So, normal iff diagonalizable
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8/ Key ideas: AA*=A*A implies A's row lengths = col lengths; work in a basis including an e-vector of A; conclude A is block-diag; recurse
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9/ The search for nerdier ways to tweet continues...
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This is more elementary, since I don't need to prove the Schur decomp
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