# Show property of eigenvectors on block triagular matrix

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.

• 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. – weee Nov 6 '18 at 23:02
• @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 – MTLaurentys Nov 7 '18 at 11:29

$$A = \begin{bmatrix} A_{11} & A_{12}\\ 0 & A_{22}\\ \end{bmatrix},\\ v = \begin{bmatrix} u\\ w\\ \end{bmatrix},\\ Av = \lambda v\\$$ 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 $$A$$.