# Why is this always the case A=$\begin{array}{cc}1 & 6 1. Nov 4, 2011 ### KingBigness Why is this always the case... A=[itex]\begin{array}{cc} 1 & 6 \\ 4 & 3 \\ \end{array}$

$λ_{1}$=7

$λ_{2}$=-3

A=$\begin{array}{cc} 1-7 & 6 \\ 4 & 3-7 \\ \end{array}$

A=$\begin{array}{cc} -6 & 6 \\ 4 & -4 \\ \end{array}$

Why is $A_{11}$ and $A_{12}$
always a multiple of
$A_{21}$ and $A_{22}$?

Is this a feature of Eigenvalues or is this done on purpose to make solving the eigenvectors easier?

2. Nov 4, 2011

### Fredrik

Staff Emeritus
Re: Eigenvalues/Eigenvectors

You seem to mean $(A-\lambda I)_{11}$, $(A-\lambda I)_{12}$ and so on.

$Ax=\lambda x$ implies $(A-\lambda I)x=0$. So if $A-\lambda I$ is invertible, x=0. An eigenvector of A with eigenvalue $\lambda$ is by definition a non-zero x that satisfies $Ax=\lambda x$. So A has an eigenvector with eigenvalue $\lambda$ if and only if $A-\lambda I$ is not invertible. For a matrix to be not invertible, its rows must be linearly dependent.

3. Nov 4, 2011

### KingBigness

Re: Eigenvalues/Eigenvectors

Clear as day now, thank you