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We have a matrix $A : 7x7$ with a minimal polynom: $$ m(t) = (t - \lambda)^3 $$

Also, we know that $A$ is not a Nilpotent matrix.

The rank of $A-\lambda I$ = 3

We need to find jordan forms for $A,A^2$.

For $A$ its ok:

We have: $$ m(t) = (t - \lambda)^3 $$

Therefore, the charactaristic polynomial: $$ p(t) = (t - \lambda)^7 $$

And $$ dim V_{\lambda} = 3 $$

Therefore: $$ G = diag\{J_{3}(\lambda), J_{2}(\lambda), \lambda, \lambda\} $$

NOW THE PROBLEM

Now i need to find jordan form for $A^2$.

The solution says: $$ P^{-1}A^2P = G^2 = diag\{J_{3}(\lambda)^2, J_{2}(\lambda)^2, \lambda^2, \lambda^2 \} = G' $$

And then he says that the charactaristic and minimal polynomial of $G'$ are: $$ p(t) = (t - \lambda^2)^7, m(t) = (t - \lambda^2)^3 $$

And because $A$ is not nilpotaent than $\lambda \neq 0$ and we conclude: $$ rank(G'-\lambda^2 I) = 3 $$

How did he conclude all of that? we dont know that?

Its like he conclude that $A^2$ has the same eigenvalues as $A$ just squared.

But why? how?

Thank you.

Alon
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1 Answers1

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Hint: If $Ax=\lambda x$ then $A^2 x=A(Ax)=A(\lambda x)=\lambda Ax=\lambda^2 x$.

Updated: Clearly $(A^2-\lambda^2)^3=(A-\lambda)^3(A+\lambda)^3=0$. Now the multiplicity of $\lambda^2$ in the minimal polynomial determines the size of the largest Jordan block, which we know is of size $3$. It follows that $(t-\lambda^2)^3$ is the minimal polynomial of $A^2$.

  • RIght, thanks, and how do you conclude about the minimal polynom that its with the same power for example? – Alon Apr 26 '20 at 04:13
  • The minimal poly needs to divide the characteristic poly which doesn't seem to happen with the given information. –  Apr 26 '20 at 04:32
  • I did a mistake in writing and now fixed it, the idea is it for $A^2$ with the same degree as the original $A$ and i dont understand why – Alon Apr 26 '20 at 04:43
  • Can you quote the problem verbatim? It seems to me there is some mistake. –  Apr 26 '20 at 05:02
  • Its in hebrew... – Alon Apr 26 '20 at 05:29
  • OK (I don't know Hebrew). Can you recheck that the minimal polynomial for $A^2$ is $m(t)=(t-\lambda)^3$. Assuming the solution is right, since the roots of the minimal poly and the characteristic poly have to be the same this means $\lambda=0,1$. Now $0$ implies that the matrix is nilpotent, which is impermissible so we it would suffice us to establish $\lambda=1$. But I can't seem to deduce that. –  Apr 26 '20 at 05:48
  • You are right, i fixed it its with $\lambda^2$ – Alon Apr 26 '20 at 05:52