2
$\begingroup$

Given the linear operator $T: \mathbb{R}^4 \rightarrow \mathbb{R}^4: T^3=-4T^2$ and $\dim \text{Im}\ T=\dim\ker T$.
Prove that exists Jordan Form of $T$ in $\mathbb{R}$ and find all of the possible Jordan Forms of $T$.
I'm sorry for asking without providing my own thoughts, but I don't have any ideas where to start from yet. Any help is very appreciated. Thank you in advance.

$\endgroup$
4
$\begingroup$

Below I use the fact that (1) the roots of the minimal polynomial are eigenvalues of the operator, (2) multiplicity of an eigenvalue as a root in the minimal polynomial is the size of the largest Jordan block corresponding to that eigenvalue, and (3) the number of Jordan blocks corresponding to an eigenvalue is the dimension of its eigenspace.

By the assumption on the dimension and the rank theorem, the kernel and image are both 2-dimensional. The fact that $T^3 = -4T^2$ means that the minimum polynomial $p(t)$ of this operator must divide $t^2(t+4)$. Checking the possibilities, we see that:

  1. $p(t) = t$ would imply $ T \equiv 0$, which is not true.
  2. $p(t) = t^2$ implies that there is a Jordan block of size 2 corresponding to the eigenvalue 0. Since the kernel has size 2, we need exactly 2 blocks corresponding to this eigenvalue.
  3. $p(t) = t+4$ would imply $T = -4I$ which is not true.
  4. $p(t) = t(t+4)$ means the operator is diagonalisable (since the minimum polynomial splits) with 0 or -4 along the diagonal, so it must be $\operatorname{diag}(0, 0, -4, -4)$ upto ordering.
  5. $p(t) = t^2(t+4)$: in this case there is a Jordan block of size 2 corresponding to the eigenvalue 0, and 2 Jordan blocks of size 1 corresponding to the eigenvalue -4.

Thus the possibilities are
$$\begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & -4 & 0\\ 0 & 0 & 0 & -4 \end{bmatrix}, \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & -4 & 0\\ 0 & 0 & 0 & -4 \end{bmatrix}$$ upto permutation of the blocks.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.