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??


1 Answer 1


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 $Av \neq 0.$

  • $\begingroup$ Are you sure these are only two ?What about mine? But how to find $v$ $\endgroup$
    – Learnmore
    May 23, 2015 at 13:36
  • 1
    $\begingroup$ @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 $1.$ $\endgroup$
    – abel
    May 23, 2015 at 13:42
  • $\begingroup$ But you did not say how to find $v$ $\endgroup$
    – Learnmore
    May 23, 2015 at 13:43
  • $\begingroup$ @learnmore, see my edit. $\endgroup$
    – abel
    May 23, 2015 at 13:57
  • $\begingroup$ but what made you think that such a vector exists ?Is it very normal to find that i cant see or requires some effort $\endgroup$
    – Learnmore
    May 23, 2015 at 15:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.