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Show that if $A$ is a strictly upper triangular nonzero matrix, then $A$ cannot be diagonalizable.

I have only shown that $A$ is nilpotent, but I don't know if that implies that $A$ isn't diagonalizable.

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    $\begingroup$ Hint: Show that the only possible eigenvalue of a nilpotent matrix is zero. $\endgroup$ – Kenny Wong Apr 3 '17 at 21:02
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Hint:

Well, you are close, as a nilpotent matrix is diagonalizable iff it is the zero matrix . Why? Because in a diagonal matrix the elements on its main diagonal are the matrix's eigenvalues....but a nilpotent matrix has one unique eigenvalue: zero.

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Hint: The eigenvalues of such a matrix are all equal to the same value; what is that value? In that case, what would the diagonal matrix look like?

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Suppose $A$ is a strictly upper triangular non-zero matrix. You have shown that $A$ is nilpotent. Let us show that the only eigenvalues of $A$ are $0$.

Note that $A^n = 0$ for some $n$. Then the minimal polynomial of $A$ is $x^k$ from some $k$. Since the eigenvalues of $A$ are precisely the roots of the minimal polynomial, this shows that the only eigenvalues of $A$ are $0$.

But if $A$ were diagonalizable, the only option for its diagonalization would then be the zero matrix. That is, there is an invertible matrix $B$ such that:

$$0 = BAB^{-1}.$$

Then $A = B^{-1}0B = 0$.


A little more directly, you can see the eigenvalues of $A$ are all $0$ because $A-xI$ is upper triangular with $x$'s down the diagonal. Therefore the characteristic polynomial is $x^n$ for some $n$.

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