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Let $A\in\mathcal{M}_4(\mathbb C)$ such that $\operatorname{rank}(A)=2$ and $A^{3}=A^2$ $\neq0$. Suppose that $A$ is not diagonalizable. Then
1. One of the Jordan blocks of the Jordan cannonical form of $A$ is $$A= \begin{bmatrix} 0 & 1 \\ 0 & 0 \\ \end{bmatrix} $$
2. $A^2=A\neq0$
3. There exists a vector $v$ such that $Av\neq0$ but $A^2v=0$.
4. The characteristic polynomial of $A$ is $x^4-x^3$.

Please someone give me some useful link on complex matrix its rank and diagonalizability.I want to solve this question on my own ,please tell me what theory I should read to solve it

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Since $A^3=A^2$ then the polynomial $P=x^3-x^2=x^2(x-1)$ annihilates $A$ and since $A$ isn't diagonalisable then the polynomial $x(x-1)$ with simple roots doesn't annihilate $A$ so $P$ is the minimal polynomial of $A$ because $A^2\ne0$ and then $A$ is similar to $$\begin{pmatrix}0&0&0&0\\0&0&1&0\\0&0&0&0\\0&0&0&1\end{pmatrix}$$ because $\operatorname{rank}A=2$ and $1$ is a simple roots of the minimal polynomial. Can you take it from here?

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  • $\begingroup$ What is the meaning of annihilates? $\endgroup$ – antara Aug 17 '14 at 15:59
  • $\begingroup$ plz give me the link about the theory part you have used $\endgroup$ – antara Aug 17 '14 at 16:00
  • $\begingroup$ The polynomial $P$ annihilates the matrix $A$ means $P(A)=0$. See this $\endgroup$ – user63181 Aug 17 '14 at 17:46
  • $\begingroup$ Sir it is given that a finite dimensional vector space over a field F is diagonalizable if and only if its minimal polynomial factors completely over F into distinct linear factors what's the proof of it? $\endgroup$ – antara Aug 18 '14 at 13:18
  • $\begingroup$ Sir according to you A isn't diagonalisable then the polynomial x(x−1) with simple roots doesn't annihilate A , please explain this $\endgroup$ – antara Aug 18 '14 at 17:30

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