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Let $A$ be an $n \times n$ upper triangular matrix with complex entries. Pick out the true statement(s):

$(a)$ If $A \neq 0$, and if $a_{ii} = 0$, for all $1\leq i \leq n$, then $A^n = 0$.

$(b)$ If $A \neq I$ and if $a_{ii} = 1$ for all $1\leq i \leq n$, then $A$ is not diagonalizable.

$(c)$ If $A \neq 0$, then $A$ is invertible.

$(a)$ is true as eigenvalues of the upper triangular matrices are diagonal elements and here all the eigenvalues are $0$. Hence $x^n=0$ is the characteristic equation and by Cayley-Hamilton theorem $A^n=0$.

I have no idea about $(b)$.

$(c)$ is not true.

2 Answers 2

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For (b), consider the minimal polynomial of the matrix. Its characteristic polynomial is plainly $(x-1)^n$. So what must its minimal polynomial be if it is to be diagonalisable? Can its minimal polynomial be this?

Hint 1 (mouse-over to reveal):

A matrix is diagonalisable if and only if its minimal polynomial has distinct linear factors.

Hint 2:

The minimal polynomial of a matrix divides its characteristic polynomial.

Hint 3:

A matrix satisfies its own minimal polynomial.

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    @mintu: Correct!2012-09-16
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Part (b) is actually the easiest one among all three. Hint: if $A$ is diagonalizable, what would be the diagonal matrix it is similar to?