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Suppose that $V$ is a $5$ dimensional complex vector space over $\mathbb{C}$. Moreover, let $T : V \rightarrow V$ be such that $\exists v \in V$ with $\{T^{i}(v)\}_{i=0}^{4}$ a spanning set of $V$, and $T$ has eigenvalues $\{1,2\}$. Determine the possible jordan canonical forms. I think if I can determine the characteristic polynomial of $T$, then I should be able to determine the possible Jordan Canonical Forms. I know that $\exists a_{0} , \dots , a_{5} \in \mathbb{C} : \sum_{i=0}^{5} a_{i} T^{i}(v) = 0$, but I don't know how to deduce anything about the characteristic polynomial from this.

user100101212
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2 Answers2

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Hint: There exists a $v$ such that $\{v,Tv,\dots,T^4v\}$ are linearly independent. It follows that the matrices $I,T,\dots,T^4$ are linearly independent. Consider the minimal polynomial of $T$.

Ben Grossmann
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  • If the matrices are linearly independent, then shouldn't the minimal polynomial of $T$ have degree $5$, so it is the characteristic polynomial? – user100101212 Dec 27 '19 at 16:27
  • Exactly. In other words, $T$ must be non-derogatory – Ben Grossmann Dec 27 '19 at 16:28
  • Can you lead me to reference which proves that if the characteristic polynomial is the same as the minimal polynomial then there is 1 Jordan Block for each eigenvalue. Also, in this situation would there be 4 distinct Jordan Canonical Forms: – user100101212 Dec 27 '19 at 16:45
  • $\begin{bmatrix} 1 & 0 & 0 & 0 & 0 \ 0 & 2 & 1 & 0 & 0 \ 0 & 0 & 2 & 1 & 0 \ 0 & 0 & 0 & 2 & 1 \ 0 & 0 & 0 & 0 & 2 \ \end{bmatrix} , \begin{bmatrix} 1 & 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 & 0 \ 0 & 0 & 2 & 1 & 0 \ 0 & 0 & 0 & 2 & 1 \ 0 & 0 & 0 & 0 & 2 \ \end{bmatrix}, \begin{bmatrix} 1 & 1 & 0 & 0 & 0 \ 0 & 1 & 1 & 0 & 0 \ 0 & 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 2 & 1 \ 0 & 0 & 0 & 0 & 2 \ \end{bmatrix}, \begin{bmatrix} 1 & 1 & 0 & 0 & 0 \ 0 & 1 & 1 & 0 & 0 \ 0 & 0 & 1 & 1 & 0 \ 0 & 0 & 0 & 1 & 0 \ 0 & 0 & 0 & 0 & 2 \ \end{bmatrix}$ – user100101212 Dec 27 '19 at 16:49
  • That's correct. That there is 1 Jordan block for each eigenvalue is a consequence of the fact that the exponent in the minimal polynomial is the size of the largest Jordan block associated with a given eigenvalue. One suitable reference is Horn and Johnson's Matrix Analysis. – Ben Grossmann Dec 27 '19 at 16:50
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The matrix could be $\begin{bmatrix}1 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0 & 2\end{bmatrix}$ if the eigen-space corresponding to eigenvalue 1 has dimension 3 and the eigen-space corresponding to eigenvalue 2 has dimension 2.

The matrix could be $\begin{bmatrix}1 & 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0 & 2\end{bmatrix}$ if the eigen-space corresponding to eigenvalue 1 has dimension 2 and the eigen-space corresponding to eigenvalue 2 has dimension 2.

The matrix could be $\begin{bmatrix}1 & 1 & 0 & 0 & 0\\ 0 & 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 0 & 2\end{bmatrix}$ if the eigen-space corresponding to eigenvalue 1 has dimension 1 and the eigen-space corresponding to eigenvalue 2 has dimension 2.

The matrix could be $\begin{bmatrix}1 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 0 & 2\end{bmatrix}$ if the eigen-space corresponding to eigenvalue 1 has dimension 3 and the eigen-space corresponding to eigenvalue 2 has dimension 1.

The matrix could be $\begin{bmatrix}1 & 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 0 & 2\end{bmatrix}$ if the eigen-space corresponding to eigenvalue 1 has dimension 2 and the eigen-space corresponding to eigenvalue 2 has dimension 1.

The matrix could be $\begin{bmatrix}1 & 1 & 0 & 0 & 0\\ 0 & 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 0 & 2\end{bmatrix}$ if the eigen-space corresponding to eigenvalue 1 has dimension 2 and the eigen-space corresponding to eigenvalue 2 has dimension 1.

user247327
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  • The reference by @Omnomnomnom says that the JCF has one Jordan Block for each eigenvalue, why doesn't this then give only 4 distinct JCF's. Some of the JCF's you have given have multiple blocks for the same eigenvalue – user100101212 Dec 27 '19 at 16:52
  • This answer does nothing to address the fact that $\exists v \in V$ with ${T^{i}(v)}_{i=0}^{4}$ a spanning set of $V$ – Ben Grossmann Dec 27 '19 at 16:52