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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.

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I think this should be tagged as homework. Also I presume you meant you have no idea about (b) rather than about (a) (which you seem to have answered) so I've edited it. –  Clive Newstead Sep 16 '12 at 17:21
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Multiple choice question on 20 percent accept rate: it's (a) low; (b) very low; (c) extremely low. –  Gerry Myerson Dec 12 '12 at 2:15
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2 Answers 2

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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is there any result that "a matrix is diagonalizable iff minimal polynomial have distinct linear factor". if it is true then b is false. –  mintu Sep 16 '12 at 17:30
    
@mintu: See Hint 1 above. –  Clive Newstead Sep 16 '12 at 17:32
    
then x-1 is not the minimal polynomial and hence it is not diagonalizable. am i right? –  mintu Sep 16 '12 at 17:41
    
@mintu: Correct! –  Clive Newstead Sep 16 '12 at 17:44
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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?

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