Jordan Canonical Form for characteristic polynomial $x^4+1$

I encountered the following problem on my algebra qualifying exam yesterday: If $T: \mathbb{R}^4 \to \mathbb{R}^4$ is a linear transformation which satisfies $T^4 = -I$, where $I$ is the identity map, find all possible Jordan canonical forms for $T$ over $\mathbb{C}$.

I started by writing the characteristic polynomial $x^4+1$ and factoring it over $\mathbb{C}$. I did this by making the substitution $y=x^2$, which yields $y=i$ or $y=-i$, which in turn yields $x=i^{1/2},-i^{1/2},i^{3/2},-i^{3/2}$. So, I concluded that the characteristic polynomial splits over $\mathbb{C}$ and is in fact separable. Then by Cayley-Hamilton, the minimal polynomial must be the same as the characteristic polynomial in order to share the same roots, hence our elementary divisors are $x-i^{1/2},x+i^{1/2},x-i^{3/2},x+i^{3/2}$, which yields a matrix in Jordan form as follows: $$\begin{pmatrix} i^{1/2} & 0 & 0 & 0 \\ 0 & -i^{1/2} & 0 & 0 \\ 0 & 0 & i^{3/2} & 0 \\ 0 & 0 & 0 & -i^{3/2} \end{pmatrix}.$$

Most of the problems we did in class similar to this one had many possibilities for the Jordan form, so I was surprised to find only one in this problem. I'm not sure where I might have made a mistake -- could anyone correct my work?

• Well, what about the scalar matrix with just $i^{\frac{1}{2}}$ on the diagonals? I think the mistake you are making is that $T^{4} + I = 0$ does not imply that $x^{4} + 1$ is the characteristic polynomial of $T$. The way to approach this problem is to note that $T^{4} + I$ does give you a list of 4 possible eigenvalues and then to write out the different Jordan Decompositions you get by considering different combinations of these four elements on the diagonal. – Siddharth Venkatesh May 8 '14 at 11:48
• No, that's the characteristic polynomial of the matrix I mentioned. The minimal polynomial is linear. All $T^{4} + I = 0$ tells you is that the minimal polynomial of $T$ divides $x^{4} + I.$ – Siddharth Venkatesh May 8 '14 at 11:52
• Oohhh I see what you are saying. I assumed that 4 is the smallest power of $T$ which makes the equation hold, which doesn't have to be the case...? – phaiakia May 8 '14 at 11:53
• No, that might still be true. It is in the example I gave. But it is still not the smallest degree polynomial that $T$ satisfies. – Siddharth Venkatesh May 8 '14 at 11:53
• I think where I got confused was that we were usually given problems like "find all possible Jordan forms for a linear transformation $T$ which has precise order 4." And I didn't translate this well into the $T^4+1$ case. – phaiakia May 8 '14 at 11:55

Just because $T$ satisfies $T^4 +I = 0$, $T$ doesn't need to have an eigenvalue for all of the zeros of $x^4 +1$. In other words, $x^4+1$ isn't necessarily the characteristic polynomial of $T$. What you do know, however, is that the minimal polynomial of $T$ divides $x^4 +1$.
So, for example, $T$ could have the characteristic polynomial $(x - e^{i\frac{\pi}{4}})^4$, because it then has the minimal polynomial $x - e^{i\frac{\pi}{4}}$, and that divides $(x^4 + 1)$.
Another possibility is that $T$ has the characteristic polynomial $(x - e^{i\frac{\pi}{4}})^2(x - e^{-i\frac{\pi}{4}})^2=$ $(x^2 - x\sqrt{2}+ 1)^2$, in which case the minimal polynomial is $(x^2 - x\sqrt{2}+ 1)$ which also divides $(x^4 + 1)$.
Your $T$ is annihilated by the polynomial $X^4+1=(X^2+\sqrt2X+1)(X^2-\sqrt2X+1)$, and those two factors are irreducible over $\Bbb R$. As $X^4+1$ does not have multiple roots, one already concludes that the matrix of $T$ is diagonalisable over$~\Bbb C$.
The minimal polynomial, which must be a monic divisor of $X^4+1$ distinct from$~1$, can be either $X^4+1$ or one of those factors $X^2\pm\sqrt2X+1$. In the former case the minimal polynomial is also the characteristic polynomial and the complexification of $T$ is diagonalisable with $4$ distinct eigenvalues (as noted in the question). In the latter case the characteristic polynomial is the square of the minimal polynomial, one one has two conjugate complex eigenvalues, each with multiplicity$~2$.