a linear algebra proof

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 7 '20 at 21:00
• yes, I 'd assume it's real (the questions doesn't really say anything about that though...) – user716286 May 7 '20 at 21:05
• This is equivalent to showing that $1$ is an eigenvalue of $Q$. Are you familiar with eigenvalues. – Dave May 7 '20 at 21:05
• A relevant link: math.stackexchange.com/questions/653133/… – Theo C. May 7 '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 - $\lambda$ $I_n$? – user716286 May 7 '20 at 21:07

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.
• 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? – user716286 May 7 '20 at 21:18
• 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 7 '20 at 21:27