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Let $A$ be a $4\times 4$ matrix over $\mathbb C$ such that $rank A=2$ and $A^3=A^2\neq 0$.Suppose that $A$ is not diagonalisable. Then

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

My try:$\dim \operatorname{Im}(A)+\dim \ker (A)=4$ so $\dim(\ker A)=\dim (\operatorname{Im}A)=2$.

Now $A$ satisfies $x^3=x^2\implies x^2(x-1)=0$ thus $0,1$ are the only eigen values of $A$. Since $A$ has rank $2$ so geometric multiplicity of $0$ is $2$ but $A$ is not diagonalizable thus algebraic multiplicity of $0$ is $3$.

So the characteristic polynomial will be $x^4-x^3=0$

Obviously $A$ will have a Jordan block of size $2$ corresponding to 0

$$\ A \text{ will have Jordan form as } \pmatrix{1&0&0&0\\0&0&1&0\\0&0&0&0\\0&0&0&0} $$

How should I use these information to conclude my result??

Martin Sleziak
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Learnmore
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  • Your $A$ has rank $3$, one of the $1$s on the diagonal ought to be a $0$. – Daniel Fischer May 23 '15 at 10:50
  • Some related posts: http://math.stackexchange.com/questions/1086438/true-or-false-statement-about-a-non-diagonal-matrix http://math.stackexchange.com/questions/1311834/characteristic-polynomial-of-a-complex-4-by-4-matrix – Martin Sleziak Jun 14 '16 at 13:15

1 Answers1

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There are only two possible Jordan forms: $$\pmatrix{0&1&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&1}, \quad \pmatrix{0&0&0&0\\0&0&0&0\\0&0&1&1\\0&0&0&1}$$ only the first first one satisfies $A^2 = A^3.$

$\bf p.s.$

the nonzero vector $v$ you are looking for is in $\ker(A^2)\setminus \ker(A)$ we have $dim(\ker(A^2) = 3,dim(\ker(A) = 2.$

step 1. find a basis of $3$ vector for $ker(A^2)$ by row reducing $A$ if you must.

step 2. find the vector $v$ in the basis so that $Av \neq 0.$

Martin Sleziak
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abel
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  • Are you sure these are only two ?What about mine? But how to find $v$ – Learnmore May 23 '15 at 13:36
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    @learnmore your matrix is similar to the first one on my list. both have $2,1$ block corresponding to the zero eigenvalue and $1$ block corresponding to the eigenvalue $1.$ – abel May 23 '15 at 13:42
  • But you did not say how to find $v$ – Learnmore May 23 '15 at 13:43
  • @learnmore, see my edit. – abel May 23 '15 at 13:57
  • but what made you think that such a vector exists ?Is it very normal to find that i cant see or requires some effort – Learnmore May 23 '15 at 15:12
  • @learnmore, of course it requires effort on your part and a particular matrix $A.$ – abel May 23 '15 at 15:23
  • i.e my question sir!! how did you see it? – Learnmore May 23 '15 at 16:09
  • @learnmore, did you read my postscript. it gives you an algorithm to find a $v$ if you have a matrix satisfying the hypothesis. i don't know what you mean by see it? what is it? – abel May 23 '15 at 18:18
  • when we try to solve the problem we first make sure that it is solvable ;we try to guess why it is true and then solve it by giving concrete arguements by "see it" I mean why you thought that such $v$ should exist ;did you see that from rank of $A$ being 2 or some other hypothesis – Learnmore May 24 '15 at 05:03
  • @learnmore,yes. the reason such a $v$ exists is that $dim(\ker(A)) = 2$ and $dim(\ker(A^2)) = 3.$ that $\ker(A^2)$ has dimension one bigger than the dimension $\ker(A)$ makes it possible for such a $v$ to exist. – abel May 24 '15 at 11:02