# $N$ is a matrix such that $N^3=0$

Given a $3\times 3$ matrix $N$ such that $N^3=0$, then which of the following are/is true?

1. $N$ has a non zero eigenvector

2. $N$ is similar to a diagonal matrix

3. $N$ has $3$ linearly independent eigenvector

4. $N$ is not similar to a diagonal matrix

Well, eigenvalues of $N$ are all zeroes and characteristic polynomial is $x^3=0$, clearly not diagonalizable. so only $1$ is true.

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One of 2 and 4 has to be true.... – David Wheeler Jun 13 '12 at 16:42
I think you need to worry about if $N$ is the zero matrix or not. – user641 Jun 13 '12 at 16:43
non-zero eigenvector is a pleonasm: eigenvectors are non-zero by definition – Marc van Leeuwen Jun 13 '12 at 17:02
The body and title of the question give different information. Do we only have $N^3=0$, or do we also have that $N$ is nilpotent of degree $3$ (i.e. $N^2\ne0$)? If so, this should be added to the body of the question. – joriki Jun 13 '12 at 17:15
@DavidWheeler Actually sometimes 2 is true and sometimes 4 is true, thus neither 2 nor 4 are true statements... – N. S. Jun 13 '12 at 18:06

"Clearly not diagonalizable" is not correct; if we know that $N^2\neq 0$, then you are correct (that would imply that the minimal polynomial of $N$ is also $x^3$, and since it is not square free then $N$ is not diagonalizable). But just from knowing that $N$ has characteristic polynomial $x^3$, we do not know whether $N$ is diagonalizable or not. It could be diagonalizable. Explicitly, $N$ is diagonalizable if and only if it is the zero matrix; prove it!
As noted, you cannot have 2 and 4 both false, since they are negations of each other and the excluded middle applies here. And 2 and 3 are logically equivalent for a $3\times 3$ matrix.
If the question explicitly states that $N^2\neq 0$, then you know that 2 is false. If the question does not explicitly state so, then you don't.