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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.

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    $\begingroup$ Please clarify what you want to prove. Do you want to show, that for any eigenpair $(\lambda,v)$ of $A$ there exist $u,w,\lambda$ such that $v=(u,w)^T$ and $(\lambda,u)$ is an eigenpair of $A_{11}$? You won't be able to prove this, since its not true. $\endgroup$ – weee Nov 6 '18 at 23:02
  • $\begingroup$ @weee Sorry for the late reply. That is what I want to prove. The book says it is true. If you say $w = 0$, for example, you get the answer expected $\endgroup$ – MTLaurentys Nov 7 '18 at 11:29
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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

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