I have the following matrix and I need to find its Jordan form.
$$A=\begin{pmatrix} -3 & 2 & 0 & 0 & 0 & 0 \\ -2 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 & 0 & 0\\ 0 & 0 & 1 & -1 & 0 & 0\\ 0 & 0 & 1 & 1 & 3 & 0\\ 0 & 0 & 0 & 0 & 0 & 3\\ \end{pmatrix}$$
So I first need to find the characteristic polynomial of each block
For $B_1: P_{B_1}(\lambda)=(\lambda +1)^2$
For $B_2: P_{B_2}(\lambda)=(\lambda +1)^2(\lambda-3)$
For $B_3: P_{B_3}(\lambda)=(\lambda-3)$
a. for $B_1,B_2$ it is said that there are "product of linear elements"? although $(\lambda +1)^2=\lambda^2+2\lambda+1$?
Next we have to find the minimal polynomial, we know that each root of characteristic polynomial need to be in the minimal polynomial so
For $B_1: M_{B_1}(\lambda)=(\lambda +1)^2$ has $(A+I)^2\neq 0$
For $B_2: P_{B_2}(\lambda)=(\lambda +1)^2(\lambda-3 )$ has $(A+I)^2(A-3*I)\neq 0$
For $B_3: P_{B_3}(\lambda)=(\lambda-3)$
So the characteristic polynomial of $A$ is: $P_{A}(\lambda)=(\lambda +1)^4(\lambda-3)^2$
How do I find the minimal polynomial when I know that it is equal to the LCM?
The characteristic polynomial of the matrix is the product of those of the blocks, which is (λ + 1)⁴(λ – 3)².
For the minimal polynomial, the LCM of the exponents of (λ + 1) for the blocks is 2, and that of the exponents of (λ – 3) is 1, so altogether, the minimal polynomial for the matrix is (λ + 1)²(λ – 3).