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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?

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  • $\begingroup$ 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). $\endgroup$ – Minus One-Twelfth Jul 27 at 1:06
  • $\begingroup$ 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? $\endgroup$ – MEG Jul 27 at 15:16
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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.

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