Relation between eigenvector/values of A and R in QR factorization

Question

It's intuitively clear that in QR factorization of given matrix, $A \in \mathbb{R}^{n\times n}$, i.e. $A = QR\,$ [where $Q \in \mathbb{R}^{n\times n}$ is orthogonal ($Q^T Q = I$) and $R \in \mathbb{R}^{n\times n}$ is upper triangular] because $R$ is undergoing an orthogonal transformation (via $Q$) the eigenvalues of A and R must be the same and that we would expect the eigenvectors of $A$ to be orthogonally transformed versions of those of $R$. However, I am not sure how to prove this. Any suggestions?

Answer 1

The relationship that you suspect does not hold. As an example, consider $$ A = \left(\begin{array}{cc} \frac{3}{5} & \frac{7}{5}\\ -\frac{4}{5} & -\frac{1}{5} \end{array}\right) \implies Q = \left(\begin{array}{cc} \frac{3}{5} & \frac{4}{5}\\ -\frac{4}{5} & \frac{3}{5} \end{array}\right), \quad R = \left(\begin{array}{cc} 1 & 1\\ 0 & 1 \end{array}\right). $$ $R$ has $1$ as its only eigenvalue, and the associated eigenspace is spanned by $(1,0)$. $A$ has eigenvalues $\frac{1}{5}\pm\frac{2\sqrt{6}\,{}}{5}\mathrm{i}$, each with a corresponding complex eigenvector.