Proving Eigenvalues: A Unit Vector Approach for (A - λI)x = b

[sana2476]

Homework Statement

Let x be a unit vector. Namely x(Transpose)*x = 1. If (A − Let x be a unit vector. If (A − λI)x = b, then λ is an eigenvalue of A − bx(transpose).

The Attempt at a Solution

I have no idea where to start this proof.

 

[statdad]

Start like this.

$$(A - bx')x = (A - \lambda I + \lambda I - bx')x$$

Expand the right hand side, and use facts about $$A, \lambda, b$$ and $$x$$.

 

[sana2476]

Ok...so here's what I have so far..I'm not sure if I'm on the right track..

Since x(transpose)*x=1
Therefore, (A-b*x(transpose))*x=λx
So Ax - bx(transpose)x=λx
Ax-b=λx
Then Ax-λx=b
Hence you have (A-λI)x=b

I don't know if I am begging the question by doing it this way though

 

[statdad]

No. Look at

$$\begin{align*}<br /> (A - bx')x & = (A - \lambda I + \lambda I - bx') x \\<br /> & = (A - \lambda I) x + (\lambda I - bx')x<br /> \end{align*}$$

What do you know about

$$(A - \lambda I)x$$ ?

use that, together with what you get when you expand

$$(\lambda I - bx') x$$

(don't forget that $$x' x = 1$$)