Finding the Jordan canonical form when the characteristic polynomial does not split?

A problem on the 2009 qualifying exam for Harvard is the following:

Suppose $$\phi$$ is an endomorphism of a 10-dimensional vector space over $$\mathbb{Q}$$ with the following properties:

1. The characteristic polynomial is $$(x-2)^4(x^2-3)^3$$.
2. The minimal polynomial is $$(x-2)^2(x^2-3)^2$$.
3. The endomorphism $$\phi-2I$$, where $$I$$ is the identity map, is of rank 8.

Find the Jordan canonical form for $$\phi$$.

Unless I'm missing something about this problem, $$\phi$$ should not have a Jordan canonical form, right? Not all of its eigenvalues lie in the field $$\mathbb{Q}$$, so are we done? I guess it feels hard to believe that a qualifying exam problem would be solved this way.

Am I missing something here? Or would you assume that the exam writers wanted us to extend the field to $$\mathbb{R}$$ and find the Jordan form there?

• From the way I understand the question, the exam writers meant that you have a $10$-by-$10$ matrix with rational coefficients. The Jordan canonical form usually assumes that you are working over $\mathbb{C}$ (though there is a real version which is more complicated). I am guessing they want you to extend the field to $\mathbb{C}$ (or at least to $\mathbb{R}$). – Malkoun Sep 6 '19 at 21:42
• Basically $2$ has algebraic multiplicity $4$, and since the minimal polynomial has degree $2$ in $(x-2)$, then the maximal size of a Jordan block for $\lambda = 2$ is $2$. But we could have either two $2$-by-$2$ blocks, or one $2$-by-$2$ block and two $1$-by-$1$ blocks. Use the information in property $3$ to rule out one of these two cases. Do the same kind of analysis for the other eigenvalues. – Malkoun Sep 6 '19 at 21:53
• Op, are you familiar with the construction of the real Jordan canonical form? – Luca Morstabilini Sep 6 '19 at 22:20
• @Malkoun I see, thank you for clearing that up for me. – Bundle_Time Sep 6 '19 at 22:34
• @LucaMorstabilini Yes, thanks. I had no problem coming up with the Jordan form following the same logic that Malkoun mentioned. I was just confused on the wording of the problem since there should be no Jordan form over $\mathbb{Q}$. – Bundle_Time Sep 6 '19 at 22:35

What we will do is construct a generalization of the real Jordan canonical form. I will mostly omit proofs, since they are simple computations. This will work for any endomorphism $$\phi$$ over a finite dimensional $$\mathbb{K}$$-vector space $$V$$ such that the splitting field of the polynomial characteristic of $$\phi$$ is of degree $$2$$. Let $$L$$ be said extension of $$\mathbb{K}$$, i.e. $$L = \mathbb{K}(\alpha)$$. Every element of $$L$$ can be written as $$z = x + \alpha y$$, with $$x$$ and $$y \in \mathbb{K}$$. In lack of a better notation, since i do not know if there is one, i will use $$\text{Re}(z)$$ to mean $$x$$ and $$\text{Im}(z)$$ for y, and call them respectively real and imaginary part of $$z$$, taking inspiration from the real and complex numbers. If you know a better one, or a more used one, please let me know. We consider now the endomorphism $$\phi \otimes Id_L$$ of $$V \otimes_\mathbb{K} L$$, and by abuse of notation denote it by $$\phi$$ again. In the real case, this corresponds to complexification. The important thing is that the matrix representing $$\phi$$ and $$\phi \otimes Id_L$$ are the same, and have the same characteristic polynomial (since the matrices are the same) and the same minimal polynomial (see, for example Minimal polynomial is invariant under field extensions).

We first note that the spectrum of $$\phi$$ is the union of the eigenvalues belonging to $$\mathbb{K}$$ and those belonging to $$L \setminus \mathbb{K}$$. Those belonging to $$L \setminus \mathbb{K}$$ come in "conjugate" pairs, in the sense that if $$\mu = a + \alpha b$$ is an eigenvalue, $$\bar{\mu} = a - \alpha b$$ is another eigenvalue and both have the same algebraic multiplicity. Denoting $$V(\lambda)'$$ the generalized eigenspace relative to the eigenvalue $$\lambda$$, we can decompose $$V \otimes L$$ as $$V \otimes L = \bigoplus_{\lambda_j} V(\lambda_j)'$$. This is possible since the characteristic polynomial of $$\phi$$ is split in $$L[t]$$, so it admits a Jordan canonical form as $$L$$-matrix. For the eigenvalues belonging to $$\mathbb{K}$$, all $$\ker(\phi - \lambda I)^j$$ have the same dimensions both when seeing them as $$\mathbb{K}$$-vector spaces and as $$L$$-vector spaces, so we can find a real Jordan basis for $$V(\lambda)$$. Regarding the eigenvalues in $$L \setminus \mathbb{K}$$, it can be easily proven that if $$\{z_1, \dots, z_k\}$$ is a Jordan basis for $$V(\lambda)'$$, then $$\{\bar{z_1}, \dots, \bar{z_k}\}$$ is a Jordan basis for $$V(\bar{\lambda})'$$. Moreover, there is a basis made of elements of $$V$$ for $$V(\lambda)' \oplus V(\bar{\lambda})'$$ by noting that $$\text{Re}(z) = \frac{z + \bar{z}}{2}$$ and $$\text{Im}(z) = \frac{z - \bar{z}}{2 \alpha}$$. From this, it is easy to see the real and imaginary parts of the elements of the basis $$\{z_1, \dots..., z_k, \bar{z_1}, \dots, \bar{z_k}\}$$ form a basis.

We can now find, supposing $$\mu \in L \setminus \mathbb{K}$$, what is the matrix associated to $$\phi$$ restricted to $$V(\mu)' \oplus V(\bar{\mu})'$$ with respect to a real basis of a Jordan basis. I will denote the real parts of $$z$$ by $$x$$ and the imaginary ones by $$y$$, and let $$\{z_1, \dots, z_k, \bar{z_1}, \dots, \bar{z_k}\}$$ a Jordan basis for $$V(\mu)' \oplus V(\bar{\mu})$$. If $$z_j$$ is such that $$\phi (z_j) = \mu z_j$$, then $$\phi(x_j) = \phi(\frac{z_j + \bar{z_j}}{2}) = \text{Re}(\mu z_j) = \frac{\mu z_j + \bar{\mu} \bar{z_j}}{2} = \text{Re}(\mu)x_j + \alpha^2 \text{Im}(\mu)y_j$$ and $$\phi (y_j) = \text{Im}(\mu z_j) = \text{Im}(\mu)x_j + \text{Re}(\mu)y_j$$ If, on the other hand, $$z_j$$ is such that $$\phi(z_j) = \mu z_j + z_{j-1}$$, we get $$\phi(x_j) = \phi(\frac{z_j + \bar{z_j}}{2}) = \frac{\mu z_j +z_{j-1} + \bar{\mu} \bar{z_j} + \bar{z_{j_1}}}{2} = \text{Re}(\mu)x_j + \alpha^2 \text{Im}(\mu)y_j + x_{j-1}$$ and $$\phi(y_j) = \text{Im}(\mu)x_j + \text{Re}(\mu)y_j + y_{j-1}$$ Then it is immediate to find the matrix representing $$\phi$$ restricted to $$\text{Span} \{x_i, y_i \}$$ wrt the basis $$\{x_i, y_i \}$$: $$\begin{bmatrix} \text{Re}(\mu) & \alpha^2 \text{Im}(\mu) \\ \text{Im}(\mu) & \text{Re}(\mu) \end{bmatrix}$$

In our case, the eigenvalue is $$\mu = \sqrt{3} = 0 + 1 \sqrt{3}$$ so the matrix will be

$$\begin{bmatrix} 0 & 3 \\ 1 & 0 \end{bmatrix}$$

and the Jordan generalized form is $$\begin{bmatrix} J(2, 2) & 0 & 0 & 0 & 0 \\ 0 & J(2, 2) & 0 & 0 & 0 \\ 0 & 0 & J(\sqrt{3}, 2) & 0 & 0 \\ 0 & 0 & 0 & J(\sqrt{3}, 2) & I_2 \\ 0 & 0 & 0 & 0 & J(\sqrt{3}, 2) \end{bmatrix}$$ where $$J(2, 2)$$ is a block of size two corresponding to the eigenvalue 2, and $$J(\sqrt{3}, 2)$$ is the $$2 x 2$$ matrix written above and $$I_2$$ is an identity block of size $$2$$.

• Wow, thank you for the very detailed response! I will work through the details of your work later tonight. After a quick read-through it seems to make sense: I can see the analogy with the complex numbers. In the final form of your matrix, would there be another three blocks corresponding to $-\sqrt{3}$? – Bundle_Time Sep 7 '19 at 1:09
• Oh nevermind, I missed the $\alpha^2$ bit. – Bundle_Time Sep 7 '19 at 1:11
• ok, I am not really sure how hard Harvard comprehensive exams are. You really think they wanted the students to find a rational canonical form? But then they should have made that more clear, in my opinion, and not call it a Jordan canonical form. In any case, @Bundle_Time, good luck with the comprehensive exams! I had a perfect $10$ many years back at Stony Brook. – Malkoun Sep 7 '19 at 6:43
• @Bundle_Time, do note though that sometimes some problems are easier than others. I kind of have a feeling the exam writers just wanted the Jordan canonical form, in the usual meaning of these terms. – Malkoun Sep 7 '19 at 6:47
• @Malkoun I agree, it would be very weird of Harvard to ask for JCF and mean Frobenius normal form instead. – A.Γ. Sep 7 '19 at 8:06