I have the following assignment question, but I can't seem to be able to do it. Can somebody guide me through?
Let $A_{1} \in \mathbb{R}^{n \times n}$ be an unreduced Hessenberg matrix. Given $\mu\in \mathbb{R}$, which is an eigenvalue of $A_{1}$.
We compute the QR factorization $A_{1} - \mu I = Q_1R_1$ by Givens rotations. Then we compute $A_{2} = R_1Q_1 +\mu I $.
a) Show that $R_{1}(n,n) = 0$.
b) Show that the last row of $A_2$ is $[0, ..., 0, \mu ]$. Thus from this we see $\mu$ is an eigenvalue of $A_{2}$ too.
c) Show that $A_{2}(i+1, i)) \neq 0$ for $i = 1 : n-2$.\
d) Suppose there is another orthogonal $\hat{Q_{1}} \in \mathbb{R}^{n \times n}$ such that $\hat{Q_{1}}e_{1} = \pm Q_{1}e_{1}$ and $\hat{A_{2}} = \hat{Q_{1}}^{T}A_{1}\hat{Q_{1}}$ is also Hessenberg. Show that $\hat{Q_{1}}e_{i} = \pm Q_{1}e_{i}$ for $i = 2 : n$ and $\hat{A_{2}} = DA_{2}D$, where $D = diag(\pm 1, ..., \pm 1)$.
If $ \mu $ is an eigenvalue of $ A_1 $, then $ 0 $ is an eigenvalue of $ A_1 - \mu I $. But that means $ A_1 - \mu I $ is singular. What does that mean about the QR factorization of $ A_1 $? What does that mean about the QR factorization of $ A_1 $ given that $ A_1 $ is unreduced? (Meaning it has no zeroes on the subdiagonal)? The answer to this should get you past a).
Parts b) and c) are then a matter of thinking of what happens when you apply the Givens rotations in the appropriate order to an upper triangular matrix that has a zero in the n,n entry.
For part d), get your inspiration from the "implicit Q" theorem.