Proving a property of eigenvalues and their eigenvectors.

[pondzo]

Homework Statement

I am asked to prove that if λ is an eigenvalue of A then λ + k is an eigenvalue of
A + kI.

The Attempt at a Solution

$A\vec{v}=\lambda\vec{v}$

$(A+kI)\vec{v}=\lambda\vec{v}$
$A\vec{v}+k\vec{v} = \lambda\vec{v}$ → $A\vec{v} = \lambda\vec{v} - k\vec{v}$
$A\vec{v} = (\lambda-k)\vec{v}$ however, this is not λ+k
I must be missing something pretty obvious, but i can't see what it is...

 

[Matterwave]

Your second step is the error. You have assumed in that step that lambda is an eigenvalue of A+kI in that step.

 

[pondzo]

Ahh thank you! I didn't even realize that, I am so use to denoting the eigenvalues lambda..

 

[STEMucator]

Homework Statement

I am asked to prove that if λ is an eigenvalue of A then λ + k is an eigenvalue of
A + kI.

The Attempt at a Solution

$A\vec{v}=\lambda\vec{v}$

$(A+kI)\vec{v}=\lambda\vec{v}$
$A\vec{v}+k\vec{v} = \lambda\vec{v}$ → $A\vec{v} = \lambda\vec{v} - k\vec{v}$
$A\vec{v} = (\lambda-k)\vec{v}$ however, this is not λ+k
I must be missing something pretty obvious, but i can't see what it is...

If $\lambda + k$ is an eigenvalue of $A + kI$, then:

$Av = \lambda v$
$Av + (kI)v = \lambda v + (kI)v$
$(A + kI)v = (\lambda + kI)v$