# Yet another eigenvalue proof

1. Apr 9, 2009

### sana2476

1. The problem statement, all variables and given/known data

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).

3. The attempt at a solution

I have no idea where to start this proof.

2. Apr 9, 2009

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 $$x$$.

Last edited: Apr 9, 2009
3. Apr 9, 2009

### sana2476

Ok...so heres 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 dont know if im begging the question by doing it this way though

4. Apr 9, 2009

No. Look at

\begin{align*} (A - bx')x & = (A - \lambda I + \lambda I - bx') x \\ & = (A - \lambda I) x + (\lambda I - bx')x \end{align*}

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

use that, together with what you get when you expand

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

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