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For an odd n $\in$ $\mathbb{N}$, let Q be an n$\times$n matrix with orthonormal columns and detQ=1. Prove that T(x) = Q x admits nontrivial fixed points $x_0$ $\in$ $\mathbb{R}^{n}$, i.e. T($x_0$) = $x_0$.

I was given a hint: to consider det[$(Q - \lambda I_n)^T$].

I've gotten as far as saying det[$(Q - \lambda I_n)^T$] = det[($Q^T$ - $(\lambda I_n)^T$] but I'm not sure if this is even the right way to go about it or not... Because I don't think you can separate this into two different determinants, everywhere I've looked online only mention separating determinants of products.

Any help is appreciated, thanks in advance!

  • Is $Q$ a real matrix? – Amit Hochman May 07 '20 at 21:00
  • yes, I 'd assume it's real (the questions doesn't really say anything about that though...) –  May 07 '20 at 21:05
  • This is equivalent to showing that $1$ is an eigenvalue of $Q$. Are you familiar with eigenvalues. – Dave May 07 '20 at 21:05
  • A relevant link: https://math.stackexchange.com/questions/653133/eigenvalues-in-orthogonal-matrices – Theo C. May 07 '20 at 21:06
  • yeah, so we'd just have to show that? but is there a simpler way of showing this besides finding the determinant of Q - $\lambda$ $I_n$? –  May 07 '20 at 21:07
  • Well eigenvalue considerations can be done. Since the columns of $Q$ are orthonormal, the eigenvalues are necessarily on the unit circle in $\mathbb C$. Then one can play with how $n$ odd implies $Q$ has a real eigenvalue, and then the determinant being positive implies it should have a positive eigenvalue. – Dave May 07 '20 at 21:11

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Assuming $Q \in \mathbb{R}^{n\times n}$, its eigenvalues are either real or complex conjugate pairs (the characteristic polynomial has real coefficients). As $n$ is odd, there is an odd number of real eigenvalues, and as the matrix is orthogonal, all eigenvalues have unit modulus. Hence, there’s and odd number of real eigenvalues equal to either 1 or -1. They cannot all be -1 because the determinant is positive and it is equal to the product of eigenvalues (the product of all complex conjugate pairs is positive). So there must be an eigenvalue equal to 1, which is what you are trying to prove.

Amit Hochman
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  • when you say unit modulus what do you mean? Also how would I show that this transformation admits "nontrivial" points in $\mathbb{R}^n$ Edit: wait does unit modulus mean it's absolute value is 1? –  May 07 '20 at 21:18
  • Unit modulus means the modulus, or absolute value is 1. A nontrivial point means it is not a vector of zeros, which is part of the definition of an eigenvalue/eigenvector pair, so it fits your question. – Amit Hochman May 07 '20 at 21:21
  • For $\lambda$ an eigenvalue, the determinant in your hint vanishes. This means the matrix $A = Q-\lambda I$ is singular and the system $Ax=0$ has a nontrivial solution. In your case $\lambda=1$ is such a value, so Qx=x has a nontrivial solution. – Amit Hochman May 07 '20 at 21:27
  • how do you mean the determinant vanishes? –  May 07 '20 at 22:03
  • Eigenvalues are roots of the characteristic polynomial, which is the determinant in your hint, viewed as a polynomial in \lambda – Amit Hochman May 07 '20 at 22:08
  • ah okay, thank you! –  May 07 '20 at 22:14