# Can a real matrix have arbitrary complex eigenvalues?

Given a set $$S$$ of complex numbers such that $$z\in S \implies \bar{z}\in S$$ can I find a real matrix $$M$$ (in a space of dimension $$|S|$$) whose eigenvalues are precisely those in $$S$$? More generally is there a way to know when a real matrix M does exist?

I'm struggling to come up with any counter examples. I thought about proving the stronger statement "Every complex matrix is similar to a real matrix" but I don't think thats easier to show. Something like jordan normal form could be useful but we don't really care about the jordan structure, just the eigenvalues. Another stronger (but less so than the last) that could work is "Every diagonalisable complex matrix is similar to a real matrix". There are also some nice ideas like knowing that the determinant and trace of a real matrix are real but the conjugate property probably makes this useless.

• Can you find a real matrix such that $M^2 = - I$? Hint: Think of rotations. Oct 23 at 7:08
• The characteristic polynomial of a real matrix has real coefficients. Complex conjugate eigenvalues are possible. However, the matrix cannot be symmetric because symmetric real matrices have real eigenvalues. Oct 23 at 7:10
• The determinant is the product of eigenvalues counting multiplicity and must be real. Thus if you want eigenvals be $1, 2+i$ you won't find a real matrix with these and only these eigenvals. Oct 23 at 7:16

For the case $$S=\{a+ib,a-ib\}$$ we have the matrix $$\begin{pmatrix}a & b\\-b &a\end{pmatrix}$$.

For larger sets $$S$$ just take a matrix with appropriate $$2\times 2$$ blocks down the diagonal.

For your subsidiary questions clearly $$\begin{pmatrix}i\end{pmatrix}$$ is not similar to a real matrix.

[Note: Some people even define complex numbers to be these $$2\times 2$$ matrices.]

• "Note: Some people even define complex numbers to be these matrices" It is well known that the real numbers are defined up to some isomorphism by the characterization as a complete ordered field. Can something similar be done with the complex numbers? Oct 23 at 8:01
• @Filippo It's doubtful that one can "approach" $\mathbb{C}$ as an ordered field. Which kind of natural order would be a candidate ? Oct 23 at 8:42
• @JeanMarie I think there has been a missunderstanding. I do not think that the complex numbers are an ordered field. I mentioned the real numbers to illustrate the kind of definition I am looking for: A definition that does not tell me in what set complex numbers live - i.e. $\mathbb{R}\times\mathbb{R}$ or some set of matrices - but that rather defines the complex numbers up to some isomorphism. Oct 23 at 9:55
• In fact, "up to isomorphism" necessitates to inspect different means of construction of $\mathbb{C}$ that can be found for example here : math.stackexchange.com/q/1777120 Oct 23 at 10:00
• @JeanMarie Thank you for the link! Do you think this is an open task? Oct 23 at 10:22

If a matrix $$A$$ is real then $$P = det(A- \lambda E) = 0$$ is a set of zeros of polynomial with real coefficients. If $$z_0$$ is a zero of $$P$$, then $$\bar{z}$$ is also a zero. Hence, e.g., for $$S = \{1, 1+i \}$$ the answer is negative.

Now suppose that $$S$$ consists of real numbers $$c_1, c_2, \ldots, c_k$$ and pairs of complex numbers: $$z_1, \bar{z}_1, ..., z_m, \bar{z}_m$$ where $$c_j = a_j + i b_j$$. Consider $$P(u)= (u-c_1)*...*(u-c_k)((u-a_1)^2+b_1^2)*...*((u-a_m)^2+b_m^2)$$. It's possible to make a block matrix A with $$det(A-uE)= P$$: indeed, consider $$1*1$$ matrices $$(c_i)$$ and $$2*2$$ matrices $$\begin{pmatrix}a_j & b_j\\-b_j &a_j\end{pmatrix}$$ and make zero matrix A, which is $$N*N$$ matrix (with $$N=k+2m$$) and then place our $$1*1$$ and $$2*2$$ matrices onto diagonal of A.

• This doesn't answer the question. PP assumes conjugate eigenvalues and asks for a method to find a convenient matrix. Oct 23 at 7:21
• @user376343 PP also asks, if such a method exists, and the hint was that it's a good idea to try to find it, because it exists:) But I agree with you and made an addition. Oct 23 at 7:55

The product $$P=\prod_{a\in S}(X-a)$$ is by hypothesis equal to its complex conjugate, so it has real entries (and it has of course $$S$$ as spectrum). Like any monic polynomial $$P$$ is the minimal (and characteristic) polynomial of the companion matrix of $$P$$, which therefore has real entries and has $$S$$ as spectrum.