0
$\begingroup$

I have been learning basic types of matrices, when I came across the following problem: Prove that the eigenvalues of orthogonal matrices are of unit modulus. And the solution mentioned that since Orthogonal matrices are unitary matrices, so the result held good as the same result had already been proven in case of Unitary Matrices. I had the following questions in my mind next:

  1. Are all elements of an orthogonal matrix real? Why can't they be complex?
  2. Can a complex matrix be orthogonal?

Any productive discussion is looked forward to. And I think what might better help the discussion is an example of a "COMPLEX ORTHOGONAL MATRIX".

$\endgroup$
6
  • 4
    $\begingroup$ By definition. ${}$ $\endgroup$
    – anon
    Commented Jun 17, 2021 at 7:40
  • $\begingroup$ So that would mean all orthogonal matrices are unitary $\endgroup$
    – Krishan
    Commented Jun 17, 2021 at 7:43
  • $\begingroup$ To expound further, the definition is due to the fact that in the complex case, it is more interesting to consider the conjugate transpose rather than just the transpose (so it yields the unitary matrices instead) $\endgroup$
    – Evariste
    Commented Jun 17, 2021 at 7:43
  • $\begingroup$ @Krishan Yes, orthogonal matrices are a subset of unitary matrices $\endgroup$
    – Evariste
    Commented Jun 17, 2021 at 7:44
  • $\begingroup$ Just so that it is interesting to consider the conjugate transpose we can't neglect the transpose, which also might be an inverse, and thus making the complex matrix orthogonal. $\endgroup$
    – Krishan
    Commented Jun 17, 2021 at 15:33

2 Answers 2

2
$\begingroup$

Here's a small orthogonal matrix with complex elements $$\eqalign{ Q &= \frac 15\left[ \begin{array}{r} -1-8\textit{i} & 2-4\textit{i} & -10+0\textit{i} \\ -10+0\textit{i} & -5+0\textit{i} & 0+10\textit{i} \\ -2+4\textit{i} & 4+2\textit{i} & 5+0\textit{i} \\ \end{array} \right] \\\\ Q^HQ &= \frac 15\left[ \begin{array}{r} 37+0\textit{i} & 16+0\textit{i} & 0-40\textit{i} \\ 16+0\textit{i} & 13+0\textit{i} & 0-20\textit{i} \\ 0+40\textit{i} & 0+20\textit{i} & 45+0\textit{i} \\ \end{array} \right], &\qquad Q^TQ=I \\\\ QQ^H &= \frac 15\left[ \begin{array}{r} 37+0\textit{i} & 0+40\textit{i} & -16+0\textit{i} \\ 0-40\textit{i} & 45+0\textit{i} & 0+20\textit{i} \\ -16+0\textit{i} & 0-20\textit{i} & 13+0\textit{i} \\ \end{array} \right], &\qquad QQ^T=I \\ }$$ which was generated with the following simple recipe $$\eqalign{ A &= {\rm rand}(3,3) + i\cdot{\rm rand}(3,3) \quad&\big({\rm random\,matrix}\big) \\ K &= A-A^T \quad&\big({\rm skew\,matrix}\big) \\ Q &= (I+K)^{-1}(I-K) \quad&\big({\rm Cayley\,transform}\big) \\ }$$ So it's possible to create such matrices, but they're not really useful for anything.

$\endgroup$
0
1
$\begingroup$

Orthogonal matrices can be complex.

A square matrix $A$ is called orthogonal if $A^TA=I$. This definition is valid over any field, such as $\mathbb R,\mathbb C$ or $GF(2)$. E.g. here is a complex orthogonal matrix: $$ A=\pmatrix{i&-\sqrt{2}\\ \sqrt{2}&i} \Rightarrow A^TA=\pmatrix{i&\sqrt{2}\\ -\sqrt{2}&i}\pmatrix{i&-\sqrt{2}\\ \sqrt{2}&i}=I. $$ However, most people only work with real or complex matrices, and real orthogonal matrices are more useful and appear far more frequently than their complex counterparts. When one speaks of an ‘orthogonal matrix’ without specifying the underlying field, one more often than not is talking about a real orthogonal matrix.

$\endgroup$
4
  • $\begingroup$ If the field is $\mathbb{C}$, isn't the "orthogonally" lost since the standard inner product on $\mathbb{C}^n$ is $\sum a_i \bar{b_i}$? $\endgroup$
    – E.E.
    Commented Jun 17, 2021 at 8:41
  • $\begingroup$ @E.E. When the field is not $\mathbb R$, we no longer consider orthogonal matrices in the context of inner product spaces. $\endgroup$
    – user1551
    Commented Jun 17, 2021 at 8:45
  • $\begingroup$ @user1551 But can complex matrices be orthogonal, though we talk only of the tranjugate of complex matrices. I thought the question was simple. $\endgroup$
    – Krishan
    Commented Jun 17, 2021 at 15:35
  • $\begingroup$ @Krishan Sorry, I don't understand your question. What is ‘tranjugate’? $\endgroup$
    – user1551
    Commented Jun 17, 2021 at 16:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .