i have been asked to construct a matrix A such that $A^2$ is not equal to '0' but, $A^3=0$. how should i proceed.

i can only understand that all the eigenvalues for A , $A^2$ and $A^3$ will be 'zero' but then how to proceed !

please don't give me the answer only. i want to know how to proceed !

I TRIED CAYLEY HAMILTON THEOREM . (if $\lambda $ is the eigenvalue of A)

if $A^3=0 =>(\lambda)^3 =0 => (\lambda)=0 =>(\lambda)^2=0 =>A^2=0 $ so this is not possible . is there any fallacy in my application cayley hamilton. please point me out.

  • $\begingroup$ The implication in the Cayley-Hamilton goes only so far - it tells us that the matrix satisfies its own characteristic equation. But it doesn't go further and tell us that the matrix satisfies every polynomial equation satisfied by its eigenvalues. $\endgroup$ – Mark Bennet Jun 21 '15 at 12:04
  • $\begingroup$ $A3=0=>(λ)^3=0=>(λ)=0=>(λ)^2=0$ . you mean , upto this is right ? then A^2=0 is not right ! right ? $\endgroup$ – saudade Jun 21 '15 at 12:06
  • $\begingroup$ I think that's it. $\endgroup$ – Mark Bennet Jun 21 '15 at 12:06

First the minimal polynomial of the matrix is $x^3$, hence the matrix has dimension $\ge3$.

With the Jordan canonical form, you can have this: $$A=\begin{bmatrix}0&1&0\\0&0&1\\0&0&0\end{bmatrix},\quad A^2=\begin{bmatrix}0&0&1\\0&0&0\\0&0&0\end{bmatrix},\quad A^3=0.$$ Any matrix that is similar to $A$ will have the same property.

  • $\begingroup$ thanks but , I TRIED CALEY HAMILTON THEOREM . (if $\lambda $ is the eigenvalue of A) if $A^3=0 =>(\lambda)^3 =0 => (\lambda)=0 =>(\lambda)^2=0 =>A^2=0 $ so what mistake did i do ? $\endgroup$ – saudade Jun 21 '15 at 11:59
  • 2
    $\begingroup$ You confused eigenvalue ($\lambda$, $0$ here) andcharacteristic polynomial which has eigenvalues as roots ($x^3$). Indeed, in the real numbers, $\lambda^3=0\Rightarrow \lambda=0$. But in the ring of $3\times 3$ matrices, this is no more true: a matrix can have a power equal to 0, without being $0$ (such a matrix is called a nilpotent matrix). This ring also has zero-divisors ($AB=0$ with $A,B\ne 0$), which is very different from what happens with real or complex numbers. $\endgroup$ – Bernard Jun 21 '15 at 12:12

The matrix A not equal to 0 but which when squad it gives 0 is, 0 1 0 0. Hint: The matrix is suppose to be in bracket.


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