# Possible Jordan Canonical Forms: Intuition

As I was reviewing linear algebra before I head off to grad school in the fall, I came across a question about Jordan Canonical Forms. It reads: "Suppose that A is a square complex matrix with characteristic polynomial $$c_A(x) = (x−1)^4(x+ 3)^5$$. Assume also that $$A−I$$ has nullity 4 and $$A+3I$$ has nullity 1, where $$I$$ is the identity matrix of the same size as $$A$$. Find, with justification, all possible Jordan canonical forms of $$A$$, and give the minimal polynomial for each."

I believe that there will be 2 Jordan Blocks, for each eigenvalue of $$A$$. Since the rank of the null space of the linear operator $$A-I$$ is 4, then the dimension of the eigenspace $$E_1$$ will be four. So there will be four linearly independent eigenvectors with eigenvalue 1. Thus there will one 4 by 4 Jordan Block with no ones in the super-diagonal. Since the dimension of the null space of $$A+3I$$ has rank 1, then there is only one linearly independent eigenvector with eigenvalue -3. If $$K_{-3}$$ is the generalized eigenspace, then $$dim(K_{-3})=5$$. Does this implies that there are 5 linearly independent generalized eigenvectors with corresponding to -3? Is so, then there will be one 5 by 5 Jordan block for the eigenvalue -3 with ones in the superdiagonal. So we get $$\begin{bmatrix} 1& 0 &0 &0 &0 &0 &0 &0 &0 \\ 0& 1 &0 &0 &0 &0 &0 &0 &0 \\ 0& 0 &1 &0 &0 &0 &0 &0 &0 \\ 0& 0 &0 &1 &0 &0 &0 &0 &0 \\ 0& 0 &0 &0 &-3 &1 &0 &0 &0 \\ 0& 0 &0 &0 &0 &-3 &1 &0 &0 \\ 0& 0 &0 &0 &0 &0 &-3 &1 &0 \\ 0& 0 &0 &0 &0 &0 &0 &-3 &1 \\ 0& 0 &0 &0 &0 &0 &0 &0 &-3 \end{bmatrix}$$

How do I determine the minimal polynomial? Are there any other options for the Jordan blocks corresponding to the eigenvalue -3? Is it possible for a different linear transformation with the same characteristic polynomial to have the dimension of the null space of $$A-3I$$ to be 2,3 or 4? Is so, how does this affect the Jordan form and why?

• Once you know the Jordan form, you can figure out the minimal polynomial by using the fact that the power for $(x-\lambda)$ in the minimal polynomial will be the size of the largest Jordan block for that eigenvalue. By the way, the four $1$'s in the Jordan form are actually four $1\times 1$ Jordan blocks (rather than a single $4\times 4$ Jordan block, since a single $4\times 4$ Jordan block means there has to be $1$'s in the superdiagonal). Jul 27, 2019 at 1:06
• The the minimal polynomial will be $(x-1)(x+3)^5$ because there are five Jordan blocks for the eigenvalue of 1 and one Jordan block for the eigenvalue of 3. So $(x-1)(x+3)^5$ will be the largest invariant factor. Can we determine what the other invariant factors are?
– MEG
Jul 27, 2019 at 15:16

We have a matrix $$A$$ whose characteristic polynomial is $$\chi_A(t)=(t-1)^4(t+3)^5$$ This immediately tells us that $$A$$ is a $$9\times 9$$ matrix whose eigenvalues are $$1$$ and $$-3$$. Moreover, the algebraic multiplicities of these eigenvalues are \begin{align*} \operatorname{am}_A(1) &= 4 & \operatorname{am}_A(-3)=5 \end{align*} The eigenspaces of $$A$$ are \begin{align*} E_1 &= \operatorname{Null}(1\cdot I_9-A) & E_{-3} &= \operatorname{Null}(-3\cdot I_9-A) \end{align*} The geometric multiplicities of the eigenvalues are the dimensions of these eigenspaces, which are \begin{align*} \operatorname{gm}_A(1) &= 4 & \operatorname{gm}_A(-3) &= 1 \end{align*} The algebraic and geometric multiplicities of the eigenvalues of $$A$$ give us partial information about the Jordan canonical form $$J$$ of $$A$$. For each eigenvalue $$\lambda$$, we have \begin{align*} \operatorname{am}_A(\lambda) &= \text{number of \lambda's on diagonal of J} & \operatorname{gm}_A(\lambda) &= \text{number of Jordan blocks in J corresponding to \lambda} \end{align*} So, in our example, $$J$$ will have
• four Jordan blocks corresponding to $$\lambda=1$$ whose sizes sum to four
• one Jordan block corresponding to $$\lambda=-3$$ whose size is five
So, it turns out that there is only one possible Jordan form $$J = \left[\begin{array}{r|r|r|r|rrrrr} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ \hline 0 & 0 & 0 & 0 & -3 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & -3 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & -3 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & -3 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -3 \end{array}\right]$$ Finally, the minimal polynomial is $$\mu_A(t)=(t-1)(t+3)^5$$ since the largest Jordan block corresponding to $$\lambda=1$$ has size one and the largest Jordan block corresponding to $$\lambda=-3$$ has size five.