Let $A \in \mathrm{Mat_{3}}(\mathbb{C})$ with A invertible such that $A$ is conjugate to $A^{2}$. Find the Jordan form of $A$.
Suppose $B \in \mathrm{Mat_{3}}(\mathbb{C})$ such that $B$ is conjugate to $-B$. Find the Jordan form of $B$.
These problems have appeared in a certain college's Linear Algebra qual consecutively. First I thought the computations would be tedious, but I find nothing interesting at last perhaps much cleaner solutions could be beneficial as the ones here relying on brute force calculations do not extend to higher dimensions.
For problem 2, let $E_{B} =\{\lambda_{1}, \lambda_{2}, \lambda_{3} \}$ be the set of eigenvalues of $B$ and $E_{-B} =\{-\lambda_{1}, -\lambda_{2}, -\lambda_{3} \}$ be the set of eigenvalues of $-B$. Then note that B being similar to -B $\implies \mathrm{Trace}(B) = \mathrm{Trace}(-B) = - \mathrm{Trace}(B) \implies \mathrm{Trace}(B) = 0$ and $\mathrm{det}(B) = \mathrm{det}(-B) = (-1)^{3} \mathrm{det}(B) \implies 0 = \mathrm{det}(B)$.
Thus, without loss of generality, $0 = \mathrm{det}(B) \implies \lambda_{1} = 0$. But then $\mathrm{Trace}(B) = 0 \implies \lambda_{2} = -\lambda_{3}$.
Case 1: If any two or all three eigenvalues are the same, then they are zeros, which gives us the characteristic polynomial to be $C_{A}(x) = x^{3}$. From which we get three sets of invariant factors: $\{x, x, x \}, \{ x, x^{2}\}, \{x^{3} \}$ corresponding to the zero matrix, Jordan matrix with two blocks one with size 1 and the other with size $2 \times 2$, and then finally the one block Jordan form with $a_{21} = 1, a_{32} = 1$ and zeros everywhere else.
Case 2: If the eigenvalues are all distinct, then we get the characteristic polynomial to be $C_{A}(x) = x(x-\lambda_{3})(x+ \lambda_{3})$ and the Jordan form is just the diagonal matrix with diagonal entries being $0, -\lambda_{3}, \lambda_{3}$.
For problem 1, let $E_{A} =\{\lambda_{1}, \lambda_{2}, \lambda_{3} \}$ be the set of eigenvalues of $A$ and $E_{A^{2}} =\{\lambda_{1}^{2}, \lambda_{2}^{2}, \lambda_{3}^{2} \}$ be the set of eigenvalues of $A^{2}$. Then note that A being similar to $ A^{2}$ $ \implies \mathrm{Trace}(A) = \mathrm{Trace}(A^{2}) \implies \lambda_{1}+ \lambda_{2} +\lambda_{3} = \lambda_{1}^{2}+ \lambda_{2}^{2} + \lambda_{3}^{2}$ and $\mathrm{det}(A) = \mathrm{det}(A^{2}) \implies \lambda_{1} \lambda_{2}\lambda_{3} = \lambda_{1}^{2} \lambda_{2}^{2} \lambda_{3}^{2} \implies (\lambda_{1} \lambda_{2}\lambda_{3}) = 1$ since A is invertible.
Case 1: If all three eigenvalues are the same, say $\lambda_{1} = \lambda_{2} =\lambda_{3} = \lambda$, then by the similarity of $A$ and $A^{2}$, we have $E_{A} =\{\lambda_{1}, \lambda_{2}, \lambda_{3} \} = E_{A^{2}} =\{\lambda_{1}^{2}, \lambda_{2}^{2}, \lambda_{3}^{2} \} \implies \lambda^{2} = \lambda \implies \lambda = 1$. Therefore, the characteristic polynomial is $C_{A}(x) = (x-1)^{3}$.
Case 2: If any two of the eigenvalues are the same, say, $\lambda_{1} = \lambda_{2} = \lambda$, then the third one must be the same as well. From the equality of traces, we have $ 2\lambda + \lambda_{3} = 2\lambda^{2} + \lambda_{3}^{2}$. Note that if $\lambda = \lambda^{2}$, then $\lambda = 1$ and $(\lambda_{1} \lambda_{2}\lambda_{3}) = 1 \implies \lambda_{3} = 1$. If $\lambda = \lambda_{3}^{2}$ and $\lambda_{3} = \lambda^{2}$, then plugging these into the equality of the traces, we get: $2\lambda + \lambda_{3} = 2\lambda^{2} + \lambda_{3}^{2} \implies \lambda = \lambda^{2} \implies \lambda = 1$ and then we get the result again. Thus, the characteristic polynomial in this case is also: $C_{A}(x) = (x-1)^{3}$.
Case 3: For the third case here in which all the eigenvalues are distinct, I can show that none of the eigenvalues can be its own square as that leads to a contradiction. Hence we can have a unique Jordan form, which is diagonal with diagonal entries being the distinct squares of the eigenvalues [could this case be eliminated, I am not sure].
Can we generalize to $n \times n$ and still are able to compute Jordan forms?