This is part 'a' of exercise 4.2.5 of the book Fundamentals Of matrix Computations 1st. ed.
$ A \in C^{nxn},\\ A = \begin{bmatrix} A_{11} & A_{12}\\ 0 & A_{22}\\ \end{bmatrix}, $
$A$ is triangular
$ A_{11} \in C^{jxj} and A_{22} \in C^{kk}, j + k = n $
a) If ($\lambda$, $u$) eigenpair of $A_{11}$, show that $\exists$ $w$ $\in$ $C^{k}$ such that ($\lambda, \begin{bmatrix} u\\ w\\ \end{bmatrix} $) autopair of A
It is previously proven that $\lambda$ eigenvalue of A if it is an eigenvalue of $A_{11}$. I understand that if $w$ = 0, the exercise is there, but I cannot get a general case: I know ($\lambda, v$) is eigenpair of A, but I cannot show v = $ \begin{bmatrix} u\\ w\\ \end{bmatrix} $.
Any help appreciated!
EDIT:
My idea for the solution was:
$$ A = \begin{bmatrix} A_{11} & A_{12}\\ 0 & A_{22}\\ \end{bmatrix},\\ v = \begin{bmatrix} u\\ w\\ \end{bmatrix},\\ Av = \lambda v\\ $$ For the first $j$ rows: $$ Av = \lambda b,\\ b = \begin{bmatrix} A_{11}[i] \cdot u + A_{12}[i] \cdot w\\ \end{bmatrix}, $$ For the following rows: $$ Av = \lambda c,\\ c = \begin{bmatrix} 0 \cdot u + A_{22}[k] \cdot w\\ \end{bmatrix}, $$ Therefore: $$ v = \begin{bmatrix} A_{11}[i] \cdot u + A_{12}[i] \cdot w\\ 0 \cdot u + A_{22}[k] \cdot w\\ \end{bmatrix}, $$ Since we know $A_{11}v = \lambda v$, we can just simplify by saying w = 0.
That does show $\exists w$, but I am not 100% confident on this. Again, I would love to know if this is correct and a possible better path.
Today I had the chance to talk to a math professor in my university and showed him this question. He said my solution not accurately written but was almost correct, except for the final conclusion
$$ A = \begin{bmatrix} A_{11} & A_{12}\\ 0 & A_{22}\\ \end{bmatrix},\\ v = \begin{bmatrix} u\\ w\\ \end{bmatrix},\\ Av = \lambda v\\ $$ For the first $j$ rows: $$ Av = \lambda b,\\ b = \begin{bmatrix} A_{11}[i] \cdot u + A_{12}[i] \cdot w\\ \end{bmatrix}, $$ For the following rows: $$ Av = \lambda c,\\ c = \begin{bmatrix} 0 \cdot u + A_{22}[k] \cdot w\\ \end{bmatrix}, $$ Therefore: $$ v = \begin{bmatrix} A_{11}[i] \cdot u + A_{12}[i] \cdot w\\ 0 \cdot u + A_{22}[k] \cdot w\\ \end{bmatrix}, $$
Since we know $A_{11}v = \lambda v$, we can just simplify by saying w = 0.Instead, the correct was: $$ A_{11} u = \lambda u\\ A_{11} u + A_{12} w = \lambda u\\ 0 + A_{22} w = \lambda w $$ which gives: $$ A_{12}w = 0\\ A_{22}w = \lambda w $$ Either $w$ = 0 or (($\lambda, w$) eigenpair of $A_{22}$ AND $A_{12}w = 0$[miracle]). Since $w$ = 0 is always valid, we showed there is always a $w$ such that ($\lambda, \begin{bmatrix} u\\ w\\ \end{bmatrix},\\$) eigenpair of $A$.
We were both in a hurry so we might have still missed something. However, I think this looks better now