# Jordan Normal Form

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?

• 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. – Berci Nov 25 '13 at 1:04

## 1 Answer

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.

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