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Question. Let $A$ be a $3\times 3$ matrix. If $A^3=0$ and $A^2 \neq 0$, prove that $A^2v=0$ for some $v\in\mathbb{R}^3\setminus0$.

To generalize the solution I'm defining $A$ as: \begin{pmatrix} a_1 & a_2 & a_3\\ a_4 & a_5 & a_6\\ a_7 & a_8 & a_9 \end{pmatrix}

and have calculated $A^2$ to be:

\begin{pmatrix} a_1^2+a_2a_4+a_3a_7 & a_1 a_2+a_2 a_5+a_3 a_8 & a_1 a_3+a_2 a_6+a_3 a_9\\ a_4 a_1+a_5 a_4+a_6 a_7 & a_4 a_2+a_5^2+a_6 a_8 & a_4 a_3+a_5 a_6+a_6 a_9\\ a_7 a_1+a_8 a_4+a_9 a_7 & a_7 a_2+a_8 a_5+a_9 a_8 & a_7 a_3+a_8 a_6+a_9^2 \end{pmatrix}

Just the thought of having to calculate $A^3$ gives me a headache... I've read something about diagnolization, but I couldn't apply it here.

I assume there must be some characteristic of a matrix $A$ for which $A^3=0$ I couldn't think about.

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If $A^2\ne0$ how do you want to prove that $A^2 v=0$ for all $v$? this contradicts the assumption! – user63181 Apr 5 '14 at 10:33
Arggg.. so it does.. thanks. Edited: Prove that ∃v(A²v=0), v∈R³ – TastySpaceApple Apr 5 '14 at 10:51
In this case this is simply the meaning of $A^2\ne 0$ i.e. there exists $v$ such that $A^2v\ne0$. Isn't it? – user63181 Apr 5 '14 at 10:55
I believe you mean nonzero $v$, otherwise is trivial. – DanielV Apr 5 '14 at 11:20
up vote 7 down vote accepted

$$\det A^3 = 0 \Rightarrow \det A = 0 \Rightarrow \det A^2 = 0$$

Because $A^2$ is not invertible it has to have nontrivial kernel. Therefore there is $v$ that $A^2 v = 0$

You don't really need to know that $A^2 \neq 0$.

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The question is about the rank of the matrix. The assumption $A^3=0$ tells you that $A^3$ has rank zero, in particular its determinant is zero. Now, by Binet's theorem, $\det A^3= \det A \det A^2=(\det A)^3$, and the first term being zero tessl you that both $A$ and $A^2$ have not maximal rank (null determinant). Now, having a null determinant is a necessari and sufficient condition for the existence of a nonzero $v$ such that $A^2 v=0$ (notice that if $v$ can be zero you can always choose one).

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$$A^2 \begin{bmatrix} a_1 & a_2 & a_3 \\ a_4 & a_5 & a_6 \\ a_7 & a_8 & a_9 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix}$$

Therefore $$A^2 \begin{bmatrix} a_1 \\ a_4 \\ a_7 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \\ \end{bmatrix}$$

At least one of the columns of $A$ isn't zero if $A^2$ isn't zero.

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Very nice ..... +1 – user1551 Apr 6 '14 at 9:15

$A^{3}=0$ hence for any $v\in\mathbb{R}^{3}$ $A^{3}v=0$ thus $A^{2}(Av)=0$.

That is: Any vector of the form $Av$ is a solution $x$ for $A^{2}x=0$.

Since $A\neq0$ then there is some $x\in\mathbb{R}^{3}$ s.t $Ax\neq0$. This gives a non trivial solution to $A^{2}x=0$ (hence also shows it is not of full rank)

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