Suppose you have a 4x4 matrix with the characteristic polynomial equal to the minimal polynomial $m_F(x)=C_F(x)=(x-3)^2(x+2)^2$. Find the Jordan form.
Is this the correct solution? $$ M=\left[\begin{array}{cccc} 3 & 1 & 0 & 0\\ 0 & 3 & 0 & 0\\ 0 & 0 & -2 & 1\\ 0 & 0 & 0 &-2 \end{array}\right] $$
Find the Jordan form of A $$ A=\left[\begin{array}{cccc} 0 & 0 & 1 & 0& 0\\ 0 & 0 & 0 & 1& 0\\ 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 &0\\ 0& 0 & 0 & 0 & 0 \end{array}\right] $$
To find the Jordan we need to find:
(a) The eigenvalues
since its an upper triangular matrix, then the elements on the diagonal matrix are the eigenvalues. Hence $C_A(\alpha)=\alpha^5\implies \alpha=0$
(b) Find the minimal polynomial $m_A(\alpha)$
We note that $m_A(A)=A^3=0$. hence, $m_A(alpha)=\alpha^3$ (I know the observation is correct, but not sure how to get the minimal polynomial)
(c)$d=dim(E_{alpha=0})$
$dim(null(A-0I))=dim(Null(A))=3 Now, we conclude that we need 3 jordan block and the first block is of size 3x3, hence:
so, $A=J^3\oplus J^1\oplus J^1$
IS my solution correct. I find forming jordan forms confusing .
Find the Jordan form of B $$ B=\left[\begin{array}{cccc} 0 & 0 & 0& 1\\ 0 & 0 & 0& 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 &0 \end{array}\right] $$
To find the Jordan we need to find:
(a) The eigenvalues
since its an upper triangular matrix, then the elements on the diagonal matrix are the eigenvalues. Hence $C_B(\alpha)=\alpha^4\implies \alpha=0$
(b) Find the minimal polynomial $m_B(\alpha)$
We note that $m_B(B)=B^2=0$. hence, $m_B(alpha)=\alpha^2$
(c)$d=dim(E_{alpha=0})$
$dim(null(B-0I))=dim(Null(B))=1 Now, we conclude that we need 1 jordan block and the first block is of size 2x2, but that can't be?
Your first answer is correct.
For the second, as a quick check look for permutations of columns and rows (same permutation) and see if it results in a nice Jordan form. In this case, with $\pi = (1,3,5,2,4)$ let $P$ be the corresponding permutation matrix $P e_k = e_{\pi_k}$, then $P^T A P = \begin{bmatrix} 0 & 1 & 0 & 0& 0\\ 0 & 0 & 1 & 0& 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 &1\\ 0& 0 & 0 & 0 & 0 \end{bmatrix}$, so we see that there are two blocks of sizes 2,3.
For the third question, take $\pi = (1,4,3,2)$ and let $Q$ be the corresponding permutation matrix, then $Q^T B Q = \begin{bmatrix} 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0& 0 & 0 & 0 \end{bmatrix}$, so we see that there is one Jordan block of size 2 and the rest are of size one.