# Jordan form of different matrices

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?

• $Rank(A-0I)^0=5$, $Rank(A-0I)^1=3,$ $Rank(A-0I)^2=1,$ $Rank (A-0I)^3=0.$ Look at the differences, you get a "dot diagram" $2,2,1.$ This means your JCF is of type $J_3 \oplus J_2.$ – matt biesecker May 16 '15 at 3:59
• I don't know what is a dot diagram – user3382078 May 16 '15 at 4:03
• Ultimately, the JCF can be computed by examining the generalized eigenvectors. The rank differences 2,2,1 tell you have 2 eigenvectors of (rank 1), 2 generalized eigenvectors of rank 2, and 1 generalized eigenvector of rank 3. – matt biesecker May 16 '15 at 4:16
• $A^0=I$ (by convention) – matt biesecker May 16 '15 at 6:39
• 2 Gen. E-vect. of rank 1 means 2 Blocks Total. 2 Gen. E-vect. of rank 2 means 2 Blocks of size $\geq 2.$ 1 Gen. E-vect of rank 3 means 1 Block of size $\geq 3.$ Thus you have blocks of size 3,2. – matt biesecker May 16 '15 at 6:45

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.