Pick out which are diagonalisable:

1.any $n\times n$ unitary matrix over $\mathbb C$

2.any $n\times n$ hermitian matrix over $\mathbb C$

3.any $n\times n$ upper triangular matrix over $\mathbb C$

4.any $n\times n$ matrix over $\mathbb C$ having eigen values real

For 3 and 4 i think it is not diagonalisable since I can easily find matrices whose minimal polynomial does not split into distinct linear factors .I am not sure about 1 and 2


Hint: for 1 and 2, remember the spectral theorem and what it says about normal matrices. You're correct about 3 and 4.

See this source for the spectral theorem that I'm working with here. The statement we want is that a matrix is normal (that is, $A^*A = AA^*$) if and only if it is unitarily diagonalizable (that is, there is a unitary $U$ such that $A = UDU^* = UDU^{-1}$, where $D$ is diagonal).

This version of the spectral theorem is the fastest approach since unitary and Hermitian matrices are both necessarily normal.

  • 1
    $\begingroup$ i believe comment to the question is the place for a hint. $\endgroup$ – abel Jan 10 '15 at 16:05
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    $\begingroup$ @abel If OP understands the hint, then there really isn't anything else to say, and so no one would post an answer. On the other hand, if this answer is insufficient, he can ask for clarification of the hint. $\endgroup$ – Omnomnomnom Jan 10 '15 at 16:44
  • $\begingroup$ See my latest edit for a source and brief explanation. You can also find more in the linear algebra text by Horn and Johnson. $\endgroup$ – Omnomnomnom Jan 11 '15 at 14:08

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