Let $Q \in \mathbb{M}_n(\mathbb{R})$ an orthogonal matrix. Show that $\mathbb{K}_2(Q)=1$
By the definition of condition number
$$\mathbb{K}_2(A)=\frac{\lambda_{max}}{\lambda_{min}}$$ where $\lambda$ are the eigenvalues of the matrix $A$.
$$\lambda_{max}=max(|\lambda_i|,i=1,…,n)$$
$$\lambda_{min}=min(|\lambda_i|,i=1,….n)$$
What is the trick to prove that the eigenvalues of the orthogonal matrix are 1 and -1.
Best Answer
Hint
Let $\lambda$ be an eigenvalue and $v$ be an eigenvector.
Estimate $$\bar{v}^T(Q^TQ)v=(\bar{v}^T Q^T)(Qv)$$