I need to construct a Hemintian Matrix with some constraint of its eigenvectors.

Suppose the eigenvectors of the matrix are $u_1=[u_{1,1},u_{1,2},u_{1,3}]$, $u_2=[u_{2,1},u_{2,2},u_{2,3}]$, and $u_3=[u_{3,1},u_{3,2},u_{3,3}]$, and I want $|u_{n,m}|=1$ for all $n=1,2,3$ and $m=1,2,3$. How can I construct these kind of matrices?


  • $\begingroup$ Did you really mean to have two $u_1's$? $\endgroup$
    – Amzoti
    Jul 6, 2013 at 2:06
  • $\begingroup$ Sorry for the typo, I've corrected it, the last one should be $u_3$. $\endgroup$ Jul 6, 2013 at 2:54
  • $\begingroup$ The eigenvectors of a Hermitian matrix are orthogonal, no? So that's another constraint on your eigenvectors --- not easy to find three of them with unit entries and mutually perpendicular. Maybe they have to be (essentially) $(1,1,1)$, $(1,\omega,\omega^2)$, and $(\omega^2,1,\omega)$, where $\omega$ is a complex cube root of 1. $\endgroup$ Jul 6, 2013 at 3:00
  • $\begingroup$ @GerryMyerson Thanks! $\endgroup$ Jul 6, 2013 at 17:43
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    $\begingroup$ @awl, is it obvious that there is an orthogonal basis for, say, ${\bf R}^{36}$ in which the all 36 components of all 36 vectors have modulus 1? Also, I think OP is after examples beyond the identity matrix. $\endgroup$ Jul 7, 2013 at 9:53

1 Answer 1


Let $U=\begin{pmatrix}1&e^{-i\pi/3}&e^{i\pi/3}\\1&e^{i\pi}&-1\\e^{2\pi i/3}&e^{\pi i/3}&-1\end{pmatrix}$. Then, since $1+e^{2i\pi/3}=e^{i\pi/3}$, we find that $V\overline V^T=I_3$, where $V=U/\sqrt3$. Now let $A=V\begin{pmatrix}\lambda_1&0&0\\0&\lambda_2&0\\0&0&\lambda_3\end{pmatrix}\overline V^T$, where $\lambda_1, \lambda_2, \lambda_3\in \mathbb C$.

It is then immediate that $A$ is a Hermitian matrix, with eigen-vecotrs of the required type, since the eigen-spaces still contain the vectors in $U$.
Essentially it is to find a set of three orthonormal vectors with unit entries. And we can assign any eigen-value that we want to the matrix. And I just find three orthogonal such vectors.

The following is about the verification of the properties of $V$. Skip it if needed.
Let me show that $U$ is orthogonal:
$(1,e^{-i\pi/3},e^{i\pi/3})\cdot(1,e^{i\pi},-1)=1+e^{2i\pi/3}-e^{i\pi/3}=0$, $(1,e^{-i\pi/3},e^{i\pi/3})\cdot(e^{2\pi/3},e^{i\pi/3},-1)=e^{2i\pi/3}-e^{i\pi/3}+1=0$,
$(1,e^{i\pi},-1)\cdot(e^{2i\pi/3},e^{i\pi/3},-1)=e^{2i\pi/3}-e^{i\pi/3}+1=0$. (Notice that $e^{i\pi}+1=0$)
In the end, to see the claimed equality $e^{2i\pi/3}+1=e^{i\pi/3}$, see the following (rough) picture:



In fact, the space of eigen-vectors of the required Hermitian matrix is unique, and we shall show it in the following:
Suppose $u_1\cdot u_2=0$ where $u_1, u_2$ have unit entries. Then we obtain an identity $x_1+x_2+x_3=0$, where each $x_i$ is of the form $e^{i\theta}$. By factoring out a constant, which affects not the eigen-space, we might assume that $x_1=1$. Now it is easily verified that such an equality $1+e^{i\theta_1}+e^{i\theta_2}=0$ is of the above type. It is now a simple but tedious matter to verify that the choice of three vectors which satisfy the identities right above the picture is unique, in the sense that the eigen-space is unchanged. Thus the requirement defines a unique eigen-space in $\mathbb C^3$.

Hope the above is right, and that this helps.

  • $\begingroup$ Thanks a lot! It is exactly I need to construct. $\endgroup$ Jul 6, 2013 at 17:50
  • $\begingroup$ I was think if there is a general method to construct such matrix if the matrix is arbitrary dimension? By the way, even for the $3 \times 3$ matrix, what is the intuition to find the vectors to be orthogonal? $\endgroup$ Jul 6, 2013 at 18:01
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    $\begingroup$ user, did you not check my remark about the eigenvectors of Hermitian matrices being orthogonal? $\endgroup$ Jul 7, 2013 at 3:23
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    $\begingroup$ As for $n\times n$ matrices: if $n$ is prime, then there will be a similar construction with $n$th roots of unity. If there is a Hadamard matrix of size $n$, its columns have entries $\pm1$ and are mutually orthogonal, so will serve as eigenvectors. E.g., $n=4$: $(1,1,1,1)$, $(1,1,-1,-1)$, $(1,-1,1,-1)$, $(1,-1,-1,1)$. I suspect there's a roots-of-unity construction for every $n$. $\endgroup$ Jul 7, 2013 at 3:27
  • $\begingroup$ @GerryMyerson Thanks a lot. Yes I checked it, sorry for the confusion, I understand that the vectors must be orthogonal. What I didn't understand is that how did you find such construction that will lead to orthogonality. To me seems that it involves some fundamental theory about Algebra, now I kind understand maybe it is an open question and looking at roots-of-unity seems to be a good way. $\endgroup$ Jul 7, 2013 at 19:35

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