Background: I have some background in abstract and linear algebra.

In my undergrad complex calculus class, I have to write a $5$ page paper about "complex matrices".

I don't know exactly what the teacher wants me to talk about, so then I went to ask him and he said

"Well, first of all, a complex matrix (which means a matrix with complex entries) can be written as $A+iB$, where $A$ and $B$ are real matrices. We have very special such complex matrices, which are those whose complex conjugate is equal to itself. What can we say about the determinant of such matrices? How about the eigenvalues and eigenvectors? These kind of things."

Firstly, I get that complex matrices can be written in such a way, but I don't know why it is any useful to know that. Maybe just to say that $M_{m\times n}(\mathbb{C})\cong M(\mathbb{R})_{m\times n}\oplus M_{m\times n}(\mathbb{R})$ as vector spaces? Is there a nice interpretion of this in terms of the linear maps that those matrices represent? I don't think there is much to see here, since it seems to me that it is just saying that a complex-vector space linear map can be decomposed in what is happening in the real and imaginary parts.

About the second part, I know he is talking about Hermitian operators and I'd like to do something more profound than just these seemily easy computations, so I ask you: what results about hermitian operators (and maybe some other kind of special operator) are interesting?

I was reading in Wikipedia and I feel like the spectral theorem could be one result to put in my paper. However, I haven't seen the proof yet and I'd like to know if putting it on the paper would be an exageration.

Also, notice that things I'm talking about are very algebraic, and the class is about calculus, so this is weird. Am I missing something analytical about these complex matrices? Also, if you have any ideas or suggestions, feel free to say it! I appreciate your help. Thanks.

  • $\begingroup$ $M_{m\times n}(\Bbb{C})$ is not isomorphic to $M_{m \times n}(\Bbb{R}) \times M_{m \times n}(\Bbb{R}) $ but $M_{m \times n}(\Bbb{R}) \oplus M_{m \times n}(\Bbb{R}) $. $\endgroup$ – walkar May 4 '16 at 14:46
  • $\begingroup$ There is a lot to say about complex matrices regarding eigenvalues and eigenvectors. Escpecially since complex polynomials always have as much complex roots as their degree (taking into account the root's degree), which is not the case for real polynomials $\endgroup$ – Vincent May 4 '16 at 14:48
  • 1
    $\begingroup$ @walkar Aren't those the same when we're talking about vector space structure? $\endgroup$ – Shoutre May 4 '16 at 14:49
  • $\begingroup$ Not necessarily, the Cartesian product could have more than one vector space structure on it. The direct sum is a specific one. $\endgroup$ – walkar May 4 '16 at 14:58
  • $\begingroup$ The space of invertible complex matrices is connected but the space of invertible real matrices is not. $\endgroup$ – copper.hat May 4 '16 at 15:09

Here is a surprising application. First of all, I claim that $\forall n\in \mathbb{N}:\exists x,y\in \mathbb{Z}$ such that $5^n=x^2+y^2$. We can easily prove this by induction, however the induction proof I will show uses a nice trick in the indcution step.

Obviously, $5^0=1^2+0^2$, now suppose that $5^n=x_{n}^2+y_n^2$ for some $n>0$. Then $5^{n+1}=5(x_n^2+y_n^2)= (x_n-2y_n)^2 + (2x_n+y_n)^2$. This shows the claim.

Now notice that $$\begin{pmatrix}x_{n+1}\\y_{n+1}\end{pmatrix}=\begin{pmatrix}1&-2\\2&1 \end{pmatrix}\begin{pmatrix}x_n\\y_n\end{pmatrix}.$$ It follows that $$\begin{pmatrix}x_{n}\\y_{n}\end{pmatrix}=\begin{pmatrix}1&-2\\2&1 \end{pmatrix}^{n}\begin{pmatrix}x_{0}\\y_{0}\end{pmatrix}=\begin{pmatrix}1&-2\\2&1 \end{pmatrix}^{n}\begin{pmatrix}1\\0\end{pmatrix}.$$

So if we want a closed formula for $(x_n,y_n)$, we should compute high powers of the matrix $A=\begin{pmatrix}1&-2\\2&1 \end{pmatrix}$. We can do this by diagonalizing $A$. We first compute the characteristic polynomial of $A$. We find that $\text{char}_A(X)=(X-1)^2+4$. It follows that the eigenvalues are complex numbers and the eigenvectors belong to $\mathbb{C}^2$. You can then diagonalize $A$ over $\mathbb{C}$ to obtain a closed form expression of $(x_n,y_n)$ and hence of $5^n$ as a sum of two squares.

Even in this simple problem that deals with integers, complex matrices can be used to get some closed form expression. One of the most important differences between real matrices and complex matrices is that $\mathbb{C}$ is algebraically closed and hence we can always talk about (generalized) eigenvalues and put any matrix in the Jordan canonical form (generalizes diagonalization).

  • $\begingroup$ This is a very cool example, thanks! Isn't the characteristic polynomial $(x-1)^{2}+4$, though? $\endgroup$ – Shoutre May 4 '16 at 15:28
  • $\begingroup$ @Shoutre: It certainly is, thanks for pointing that out! $\endgroup$ – Mathematician 42 May 4 '16 at 15:30
  • $\begingroup$ Is it a coincidence that the determinant of the matrix is $5$? (I explain more precisely below) $\endgroup$ – Shoutre May 4 '16 at 15:36
  • $\begingroup$ And that this is exactly the matrix of the ring endomorphism of $\mathbb{Z}[i]$ given by multiplication by $1+2i$, a generator of one of the prime ideals of $\mathbb{Z}[i]$ above $(5)$? (Assuming you are familiar with some algebraic number theory) $\endgroup$ – Shoutre May 4 '16 at 15:40
  • $\begingroup$ Hmm, I'd say yes, but only because I do not see a reason for these things to be connected. Interesting though, I'll have a look at it. $\endgroup$ – Mathematician 42 May 4 '16 at 15:46

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.