Does there exists a vector v such that $Av\neq 0$ but $A^{2}v=0$

Question

Let $A$ be a $4\times4$ matrix over $\mathbb C$ such that $\operatorname{rank}(A)=2$ and $A^{3}=A^{2}\neq0$.

Suppose that $A$ is not diagonalizable.

My question is , "Does there exists a vector $v$ such that $Av\neq0$ but $A^{2}v=0$?"

I think not. Since $\operatorname{rank}A=2$ so $A$ is not invertible. So for any vector $v$ if we find $A^{2}v=0$ then we can not multiply both sides by $A^{-1}$.

But I am not sure about it.

Are there any general rule or any theorem to solve this problem or it can be done only setting example?

Some related questions: http://math.stackexchange.com/questions/1311834/characteristic-polynomial-of-a-complex-4-by-4-matrix http://math.stackexchange.com/questions/1086438/true-or-false-statement-about-a-non-diagonal-matrix http://math.stackexchange.com/questions/1295229/show-that-there-exists-a-vector-v-such-that-av-neq-0-but-a2v-0 — Martin Sleziak, Jun 14 '16 at 13:51

score 4 · Accepted Answer · answered Oct 24 '14 at 06:45

4

The question asks you to show that $A$ is nilpotent on some subspace.

Now, from the assumptions you know that the minimal polynomial divides $$ t^3-t^2 = t^2(t-1). $$ Hence $0$ and $1$ are the only possible eigenvalues. Moreover, the associated Jordan blocks have maximal dimension $2$ (eigenvalue $0$) and $1$ (eigenvalue $1$). It remains to determine size and count of these blocks.

Since $A$ is not diagonalizable, it must have eigenvalue $0$ with multiplicity $\ge 2$.

Since $rank(A)=2$, the null space of $A$ is two-dimensional. Since $A$ is not diagonalizable, this implies that multiplicity of $0$ is $\ge 3$. Hence the Jordan form of $A$ contains a Jordan block of size $2$, hence there is a vector $v$ such that $Av\ne0$ but $A^2v=0$.

answered Oct 24 '14 at 06:45

daw

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there was also the condition $A^3=A^2$... – daw Oct 24 '14 at 06:59
For a $n\times n$ matrix $A$ if $A$ contains a Jordan block of size $m(\lt n)$ then there is a vector $v$ such that $Av\neq0$ but $A^{m}v=0$. Is it true as generalization? – Empty Oct 24 '14 at 11:13
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You even have $A^{m-1}v\ne 0$. See nilpotent mappings. – daw Oct 24 '14 at 11:27

copper.hat · Answer 2 · 2014-10-24T07:23:42.620

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Let $A=\begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$.

Then $A^n \neq 0$ for all $n$, $A e_2 =e_1$, but $A^2 e_2 = 0$. $A$ is not diagonalizable (it is in Jordan form).

Modulo permutations, this is the only Jordan form the matrix can have.

$A^3 =A^2$ means any Jordan block of size $>1$ must have eigenvalue zero and size at most 2. Since $A$ is not diagonalisable, it must have at least one Jordan block of at size at least 2. If there were two Jordan blocks of size 2, then we would have $A^2 = 0$, hence there is exactly one Jordan block of size 2 and hence the remaining blocks have size 1. One of these remaining blocks must be zero, otherwise the rank would be greater than 2, and the remaining block must be non-zero. Since $A^3 = A^2$, the remaining block has eigenvalue 1.

It follows that there always exists some $v$ such that $Av \neq 0$, but $A^2 v= 0$.

edited Oct 24 '14 at 07:23

answered Oct 24 '14 at 06:45

copper.hat

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For that A what is the vector v? – Empty Oct 24 '14 at 06:47
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$v=e_2 = (0,1,0,0)^T$. – copper.hat Oct 24 '14 at 06:48
Is the choice of $e_{2}$ arbitrary here or it can be choose from the canonical form? – Empty Oct 24 '14 at 07:49
I don't understand what you are asking. The choice is not arbitrary, it is chosen so that $Av \neq 0$ and $A^2v=0$. However, it is not unique, we could have chosen $v=e_1+e_2$ for example. – copper.hat Oct 24 '14 at 07:53
I want to say that the choice of $v$ is only by inspection such that the conditions are satisfied or $v$ follows from the Jordan canonical form – Empty Oct 24 '14 at 08:29
Sorry, I really don't understand what you are saying/asking. – copper.hat Oct 24 '14 at 08:39

score 0 · Answer 3 · answered Oct 24 '14 at 06:41

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Hint: You are looking for an element $v \in ker(A^2) - ker(A)$ or you could also look for an $w \in im(A) \cap ker(A) - 0$. Think about why such an element would yield a solution to your problem.

And $A^{-1}$ is only defined for matrices with full rank.

answered Oct 24 '14 at 06:41

Daniel Valenzuela

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Does there exists a vector v such that $Av\neq 0$ but $A^{2}v=0$

3 Answers3