4
$\begingroup$

I'm asked to find the Jordan Normal form of $A\in M_5(\mathbb{C}^{5x5})$ with the characteristic polynomial: $p(A)=(\lambda-1)^3(\lambda+1)^2$ and minimum polynomial $m(A)=(\lambda-1)^2(\lambda+1)$

I got so far:

$$m_A(x)=(x-1)^2(x+1)\;\;\;:\;\;\;\;\begin{pmatrix}1&1&0&0&0\\ 0&1&0&0&0\\ 0&0&1&0&0\\ 0&0&0&-1&0\\ 0&0&0&0&-1\\ \end{pmatrix}$$

The above matrix i got it by substitution of $m(A)$ in a $A\in M_5(\mathbb{C}^{5x5})$ matrix.

And i think it's correct(if it isn't please let me know) but i don't know why does it works?

Can you please give me a hint of why does this works?

$\endgroup$
  • 2
    $\begingroup$ Exactly how did you substitute and what? The result is correct. In this particular case, $m(A)$ and $p(A)$ were enough to conclude the Jordan normal form, i.e. to determine $A$ up to similarity. $\endgroup$ – Berci Nov 25 '13 at 1:04
4
$\begingroup$

Knowing the Jordan normal form theorem, you can just start out from the Jordan normal form. Assume that it belongs to a basis $e_1,..,e_5$. The characteristic polynomial tells you the diagonal elements, and in this case the minimal polynomial could determine the sizes of the blocks. Because $(\lambda+1)$ in $m(A)$ has exponent $1$, there cannot be a $2\times 2$ block around diagonal $-1$. Similarly, there cannot be a $3\times 3$ block around diagonal $1$, but if only $1\times 1$ blocks it contained, $(\lambda-1)$ in $m(A)$ would also have exponent $1$. So, there must be a $2\times 2$ and a $1\times 1$ block of $1$.

We can divide the space into $U:={\rm span}(e_1,e_2,e_3)$ and $V:={\rm span}(e_4,e_5)$, these are $A$-invariant, and we have $$m(A|_U)=(\lambda-1)^2\,,\quad\quad m(A|_V)=(\lambda+1)\,.$$ This second one tells use that $A|_V+I_V=0$, that is, $A|_V=-I_V$ (where $I_V$ is the identity of subspace $V$).

Since the nilpotent $N=\pmatrix{0&1&0\\0&0&1\\0&0&0}$ has $N^2\ne 0$ but $N^3=0$, we have that $m(\pmatrix{a&1&0\\0&a&1\\0&0&a})=(\lambda-a)^3$. This justifies the reasoning text above.

$\endgroup$
  • $\begingroup$ Thank you for your answer is exactly what i asked $\endgroup$ – user111034 Nov 25 '13 at 3:03

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.