Show that orthogonal matrices have eigenvalues with magnitude $1$ without a sesquilinear inner product

Question

Is it possible to consider complex eigenvalues without a Hermitian (i.e. sesquilinear) inner product over a complex vector space?

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For instance: let $A$ be a real orthogonal matrix (so $A^TA = I$). Without referencing a Hermitian inner product, is it possible to show that the complex eigenvalues of $A$ have magnitude $1$?

The usual proof of this fact is as follows: if $x,\lambda$ is an eigenpair of $A$, then we have $$ \|x\|^2 = x^*x = x^*(A^*A)x = (Ax)^*(Ax) = \lambda\overline{\lambda} (x^*x) = |\lambda|^2 \|x\|^2 $$ from which it follows that $|\lambda| = 1$. Note: this proof required the use of the sesquilinear inner product $\langle x,y \rangle = y^*x$.

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A rephrasing of the original question: consider $\Bbb C^n$ with the bilinear from $$ \langle x,y \rangle = y^Tx $$ note that this bilinear form is not an inner product. The complex-orthogonal matrices are those matrices $A$ that satisfy $A^TA = I$, where $T$ is the entrywise transpose. Notably, the complex-orthogonal matrices preserve the above bilinear form. How can we show that if $A$ is complex-orthogonal with real entries, then the eigenvalues of $A$ have magnitude $1$?

Answer 1

Well, in a way, yes. The real Jordan normal form theorem yields that every real matrix has an invariant subspace $W$ (i.e. $AW\subseteq W$) such that $1\le\dim W\le 2$. Since $\dim W^\perp +\dim W=n$ and, if $A$ is orthogonal, the orthogonal complement of an $A$-invariant subspace is $A$-invariant, there is an orthonormal basis $B=(b^1,\cdots, b^n)$ such that $B^{-1}AB=B^TAB=\begin{pmatrix}U &0\\ 0&U'\end{pmatrix}$, for some $U\in O(2)$ and $U'\in O(n-2)$ - or, repsectively, $O(1)$ and $O(n-1)$. Now, the eigenvalues of $A$ are either eigenvalues of $U$ or eigenvalues of $U'$. The form and eigenvalues of a $2\times 2$ orthogonal matrix can be calculated explicitly, and the rest can be done by induction.

That being said, as far as I know the real Jordan normal form theorem is proved by complexifying the endomorphism of $A$, using the machinery of Jordan normal form in $\Bbb C^n$, and then bringing it back to $\Bbb R^n$. So one could argue that this is just hiding the issue under the carpet.

Answer 2

Here is an argument without using the sesquilinear inner product on $\mathbb C^n$, but the usual bilinear inner product $b$ on $\mathbb R^n$ is still used. Essentially:

1. By using the normality of $Q$ with respect to the real inner product $b$, it can be shown that, up to a change of orthonormal basis on $\mathbb R^n$, we may assume that $Q=(-I_s)\oplus R$ for some real orthogonal matrix $R$ that doesn't possess $-1$ in its spectrum.
2. By using Rayleigh quotients on $\mathbb R^n$ (and so we are still using the real inner product $b$), it can be shown, without stepping into the complex field, that every real symmetric matrix has a real orthonormal eigenbasis. It follows that every real positive semidefinite matrix has a complete and nonnegative spectrum.
3. By Cayley transform, $R=(I-K)(I+K)^{-1}$ for some real skew-symmetric matrix $K$. As $-K^2=K^TK$ is positive semidefinite, all complex eigenvalues of $K$ are purely imaginary (this is straightforward if we can use the sesquilinear inner product on $\mathbb C^n$; since we cannot use it here, we need item 2). Since $|(1-z)/(1+z)|=1$ for every $z\in i\mathbb R$, we conclude that all eigenvalues of $R$ lie on the unit circle.

Answer 3

As long as you have a notion of complex scaling (which you have by definition of "vector space over $\Bbb C$"), you can define eigenvalues and eigenvectors. You cannot, without a norm or inner product, say anything about normalised eigenvectors, or whether the eigenvectors are orthogonal, or anything like that, but the eigenvalues are just complex numbers. As such they behave as complex numbers always do, which specifically means that they have an absolute value / norm of their own.

Answer 4

You can show this in a way which seems to me more elementary than the answers above suggest. The key is simply that if $A \in \text{Mat}_n(\mathbb R)$ satisfies $A^tA$ then $A$ is an isometry of $\mathbb R^n$ (with respect to the usual Euclidean distance). Since $A$ has to preserve the length of a vector, and in particular any eigenvector, its real eigenvalues must lie in $\{\pm 1\}$.

For the complex eigenvalues of $A$, we have to consider $A$ as a linear map on $\mathbb C^n$, but as a real vector space this is just $\mathbb R^n \oplus i\mathbb R^n$, with $A$ acting "diagonally'', that is, $A(v_1+iv_2) = Av_1+iAv_2$. By imposing the condition that the two copies of $\mathbb R^n$ are orthogonal to each other, and using the usual dot product on each copy, $\mathbb C^n = \mathbb R^n \oplus i\mathbb R^n$ naturally inherits a (real) inner product from $\mathbb R^n$, and the diagonal action of $A$ still preserves distances.

Now if $\lambda=\lambda_1+i\lambda_2\in \mathbb C$ is an eigenvalue of $A$ thought of as an operator on $\mathbb C^n$, then we may find an eigenvector $v=v_1 +iv_2$ for $A$ with this eigenvalue. Now $$ \begin{split} A(v_1+iv_2) = (\lambda_1 + i\lambda_2)(v_1+iv_2) = (\lambda_1 v_1 -\lambda_2v_2)+ i(\lambda_2v_1 +\lambda_1v_2) \end{split} $$ and thus we must have $A(v_1) = \lambda_1 v_1-\lambda_2v_2$ and $A(v_2) = \lambda_2v_1 + \lambda_1 v_2$, and since $A$ is an isometry $\|A(v_1+iv_2)\| =1$, hence $$ \begin{split} \|v\|^2 &=\|v_1\|^2+\|v_2\|^2 = \|A(v_1)\|^2+\|A(v_2\|^2 \\ &= \lambda_1^2\|v_1\|^2 +\lambda_2^2\|v_2\|^2 -2\lambda_1\lambda_2\langle v_1,v_2 \rangle \|v_1\|^2 \\ & \quad + \lambda_2^2\|v_1\|^2 +\lambda_1^2\|v_2\|^2 +2\lambda_1\lambda_2\langle v_1,v_2 \rangle \|v_1\|^2 \\ &= (\lambda_1^2+\lambda_2^2)(\|v_1\|^2+\|v_2\|^2) = (\lambda_1^2+\lambda_2^2)\|v\|^2 \end{split} $$ Thus $|\lambda|^2=1$ as desired.