# Canonical form of normal operator in Euclidean space

Let $$V$$ be a finite-dimensional Euclidean space and $$f$$ be a normal operator on $$V$$. Then there is an orthonormal basis of $$V$$ such that the matrix of $$f$$ in this basis is block-diagonal consisting of blocks of the size $$1\times 1$$ and blocks of the size $$2\times 2$$ of the form $$\begin{pmatrix} \mu & -\nu \\ \nu & \mu \end{pmatrix}$$. This matrix is called canonical form of normal operator and corresponding basis is called canonical basis.

I was able to prove this theorem by myself using my previous post. So let's discuss the algorithm how to find such canonical form:

Consider complexification $$f_{\mathbb{C}}$$ of our operator $$f$$. Operator $$f_{\mathbb{C}}$$ has the same matrix as $$f$$. Let $$\lambda=\mu+i\nu$$ be a root of characteristic polynomial of this matrix. We'll find the basis of corresponding eigenspace $$V_{\lambda}\subset V$$ and orthogonalize it (with respect to Hermitian inner product): let $$z_1,\dots,z_l$$ - desired orthogonal basis, where $$x_j, y_j$$ are real and imaginary parts of $$z_j$$: $$z_j=x_j+iy_j$$, $$j=1,\dots,l$$.

Then the system of real vectors $$\{x_1,y_1,x_2,y_2,\dots,x_l,y_l\}$$ is orthogonal with $$|x_j|=|y_j|$$, $$j=1,\dots,l$$ (with respect to Euclidean inner product), and operator $$f$$ acts on them in the desired way: $$f(x_j)=\mu x_j+\nu y_j, f(y_j)=-\nu x_j+\mu y_j$$. Let's normalize vectors $$x'_j=\frac{x_j}{|x_j|}, \quad y'_j=\frac{y_j}{|y_j|}$$ and include them to the canonical basis: each such pair of vectors $$x'_j,y'_j$$ corresponds to the block $$\begin{pmatrix} \mu & -\nu \\ \nu & \mu \end{pmatrix}$$.

I was trying to understand some moments of the second excerpt but I failed, so let me clarify my questions:

Questions: 1) How to show that the system of real vectors $$\{x_1,y_1,x_2,y_2,\dots,x_l,y_l\}$$ is orthogonal with $$|x_j|=|y_j|$$, $$j=1,\dots,l$$ (with respect to Euclidean inner product)?

2) Since $$f_{\mathbb{C}}(z_k)=\lambda z_k$$ then $$f(x_k)+if(y_k)=(\mu+i \nu)(x_k+iy_k)=(\mu x_k-\nu y_k)+i(\nu x_k+\mu y_k)$$ it follows $$f(x_k)=\mu x_k-\nu y_k$$ and $$f(y_k)=\nu x_k+\mu y_k$$. And we see that my answer differs from the above one a bit. So I guess in this case it is more natural to consider the system $$\{y_1,x_1,\dots,y_l,x_l\}$$. Right?

3) Suppose that roots of characteristic polynomial are $$\lambda_1,\dots,\lambda_k$$ and $$V_{\lambda_1},\dots,V_{\lambda_k}$$ are corresponding eigenspaces. It could happen that $$V_{\lambda_1}\oplus \dots \oplus V_{\lambda_k}\subsetneq V$$. Using the above algorithm I can construct basis for each $$V_{\lambda_i}$$ but what I should do with the rest, since their direct sum is a proper subspace?

EDIT (Possible answer to question 1): I guess the algorithm above is applied for purely complex root, i.e. $$\nu \neq 0$$. We see that $$Az_k=\lambda z_k$$ then $$A\overline{z_k}=\overline{\lambda}\overline{z_k}$$ because $$A$$ is real matrix. Since $$\nu \neq 0$$ as I said then $$z_k \perp \overline{z_k}$$, i.e. $$0=\langle x_k+iy_k,x_k-iy_k\rangle=|x_k|^2-|y_k|^2+2i\langle y_k,x_k\rangle$$ which shows us that $$\langle x_k,y_k\rangle=0$$ and $$|x_k|=|y_k|$$.

And somehow I need to show that $$x_i\perp x_j, y_i\perp y_j$$ and $$x_i\perp y_j$$ for $$i\neq j$$.

Let $$k\neq l$$ and I am going to show that $$\langle x_k,x_l\rangle=\langle y_k,y_l\rangle=\langle x_k,x_l\rangle=0$$. I will show just the first one. We see that $$2x_k=z_k+\overline{z_k}$$ and $$2x_l=z_l+\overline{z_l}$$ then it follows that $$\langle x_k,x_l\rangle=\frac{1}{4}\langle z_k+\overline{z_k},z_l+\overline{z_l}\rangle=\dfrac{1}{4}\left[\langle z_k,z_l\rangle+\langle z_k,\overline{z_l}\rangle+\langle\overline{z_k}, z_l\rangle+\langle\overline{z_k},\overline{z_l}\rangle\right].$$

We already know that $$\langle z_k,z_l\rangle=0$$. Since $$Az_k=\lambda z_k$$ then $$A\overline{z_l}=\overline{\lambda}\overline{z_l}$$ and because $$\lambda \neq \overline{\lambda}$$ and since $$z_k$$ and $$\overline{z_l}$$ are eigenvectors corresponding to different eigenvalues then it follows that $$\langle z_k,\overline{z_l}\rangle=0$$. In the same way one can show $$\langle \overline{z_k},z_l\rangle=0.$$

It is easy to show that $$\langle\overline{z_k},\overline{z_l}\rangle=0$$ and it follows because $$\langle z_k,z_l\rangle=0$$. More precisely, $$\langle\overline{z_k},\overline{z_l}\rangle=\operatorname{Re}\langle z_k,z_l\rangle -i\operatorname{Im}\langle z_k,z_l\rangle=0.$$

The same reasoning can be applied for the rest pairs.

Regarding question 1: your proof is correct and seems to be the most reasonable approach.

Regarding question 2: I agree. You could also use blocks of the form $$\pmatrix{\mu & \nu\\-\nu & \mu}$$ instead.

Regarding question 3: The rest of the eigenvalues are real. So, we simply take the remaining blocks to be $$1 \times 1$$.

• Thanks a lot for your reply! I got your answers for questions 1 and 2. But the third was not so clear to me probably due to complexification but yesterday I spent almost all day trying to understand it and I think that I got it. Anyway it would be great if you can check my reasoning and details(I am wondering do I understand everything correctly) which I add soon as an answer. – ZFR Apr 10 at 17:49
• I added an answer. Please take a look. Thank you! – ZFR Apr 10 at 19:14
• @Zfr I took a look and your proof looks fine, but also I don’t understand what was supposed to be tricky here – Omnomnomnom Apr 11 at 13:31
• I had some issues with basis of complexification and original vector space since they are technically different structures. That's why I uploaded my answer in order to be sure. Thanks a lot for checking! – ZFR Apr 11 at 18:59
• BTW, I am a bit confused with this topic. Maybe you can give some clear answer, please? math.stackexchange.com/questions/3620888/… – ZFR Apr 11 at 19:01

Let $$V$$ be Euclidean space and $$f$$ be a normal operator on $$V$$. Consider complexification $$V_{\mathbb{C}}$$ and $$f_{\mathbb{C}}$$ of $$V$$ and $$f$$, respectively. One can check that $$f_{\mathbb{C}}$$ normal operator on Hermitian space $$V_{\mathbb{C}}$$. Let $$\text{Sp}f_{\mathbb{C}}=\{\lambda_1,\dots,\lambda_k,\dots,\lambda_m\}$$ be eigenvalues of $$f_{\mathbb{C}}$$ and suppose the first $$k$$ are complex and the last $$m-k$$ are real numbers. Since $$f_{\mathbb{C}}$$ is a normal operator on Hermitian space $$V_{\mathbb{C}}$$ then it is diagonalizable, i.e. $$V_{\mathbb{C}}=\bigoplus \limits_{j=1}^m V^{\mathbb{C}}_{\lambda_j},$$ where by $$V^{\mathbb{C}}_{\lambda_j}$$ I mean eigenspace corresponding to $$\lambda_j$$, i.e. $$V^{\mathbb{C}}_{\lambda_j}=\ker (f_{\mathbb{C}}-\lambda_j\cdot \text{id})$$.

Remark: Since $$V_{\mathbb{C}}$$ is $$V\oplus V$$ equipped with operator of complex structure $$J(u,v)=(-v,u)$$ then for my convenience I will denote elements of $$V_{\mathbb{C}}$$ as pair of elements from $$V$$.

1. If $$\lambda\in \mathbb{R}$$ then $$V^{\mathbb{C}}_{\lambda}$$ has "real" basis say $$\{(v_1,0),\dots,(v_n,0)\}$$.

Indeed, consider $$V_{\lambda}=\ker(f-\lambda\cdot \text{id})$$ and its easy to see that $$\dim V_{\lambda}=\dim V^{\mathbb{C}}_{\lambda}$$ because $$A_{f_{\mathbb{C}}}=A_f$$ in some basis. Let $$\{v_1,\dots,v_n\}$$ is a basis for $$V_{\lambda}$$ then one can show that $$\{(v_1,0),\dots,(v_n,0)\}$$ is a basis for $$V^{\mathbb{C}}_{\lambda}$$.

2. Let $$\lambda=\mu+i\nu$$ with $$\nu\neq 0$$ then $$\lambda\neq \overline{\lambda}$$. Also easy to see that $$\dim V^{\mathbb{C}}_{\lambda}=\dim V^{\mathbb{C}}_{\overline{\lambda}}=n$$ (this $$n$$ is not the same as the previous one). Then $$\dim V^{\mathbb{C}}_{\lambda}\oplus V^{\mathbb{C}}_{\overline{\lambda}}=2n$$. Let $$\{z_1,\dots,z_n\}$$ is orthogonal basis of $$V^{\mathbb{C}}_{\lambda}$$, where $$z_j=(x_j,y_j)$$. Then by reasoning which have demonstrated in my question one can show that $$(x_1,0),(0,y_1),\dots,(x_n,0),(0,y_n)$$ is orthogonal and $$|x_j|=|y_j|$$ and since $$z_j$$ is basis then it follows that those $$2n$$ vectors are linearly independent and forms the basis of $$V^{\mathbb{C}}_{\lambda}\oplus V^{\mathbb{C}}_{\overline{\lambda}}$$. It is easy exercise to show that $$(x_1,0),(y_1,0),\dots,(x_n,0),(y_n,0)$$ is still a basis (I just changed $$(0,y)$$ to $$(y,0))$$.

Since each complex eigenvalue of $$f_\mathbb{C}$$ appears in pair (with its conjugate) then we can do that for each such pair. In general we will get a basis for $$V_{\mathbb{C}}$$ of the form $$\{(e'_i,0)\}_{i=1}^n$$, i.e. "real" basis of $$V_{\mathbb{C}}$$. Then this is also an easy exercise to show that $$\{e'_i\}_{i=1}^n$$ is basis for $$V$$.

P.S. May be I was to picky in the proof but I just want to know that I am understanding all this correctly. Thanks a lot for your attention, Omnomnomnom!