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If I take a diagonalizable block upper triangular matrix whose diagonal blocks are of finite order, is it true that away from the leading block diagonal, the matrix is zero?

I think the statement is true (Jordan-Chevalley/Levi decomposition sort of thing), but I'm really not an expert in the field and I don't want to reinvent the wheel.

Thanks!

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    $\begingroup$ What you wrote is probably not what you have in mind. Any upper triangular matrix with its diagonal entries all distinct (finite order or not) is diagonalizable. $\endgroup$
    – abx
    Commented Dec 13, 2013 at 13:43

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I assume that you are either working over an algebraically closed field, or that by "diagonalizable", you mean "diagonalizable, possibly after extending the field". To go ( a little, but actually not really any way) beyond the example abx gives, the matrix is diagonalizable I and only if its minimum polynomial has no repeated factors. In particular, it is diagonalizable if the characteristic polynomial has no repeated factors. The characteristic polynomial only depends on the "main diagonal blocks". So, for any set of main diagonal blocks which give a multiplicity free characteristic polynomial, you can supplement it with any blocks you like above the main block diagonal, and still obtain a matrix which is diagonalizable. It is questionable whether this question is research level.

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