# Thread: Find the Eigenvalues and Eigenspaces

1. ## Find the Eigenvalues and Eigenspaces

Is my working correct?

Question: Find the eigenvalues and the corresponding eigenspaces for matrix A:

A=
6 -4
3 -1

|A - lambda * I| = 0

6-lambda -4
3 -1-lambda

(6-lambda)(-1-lambda) - (3)(-4)

-6 -6*lambda + lambda + lambda^2 + 12

6 -5*lambda + lambda^2

Factorize 6 -5*lambda + lambda^2

(lambda - 3)(lambda - 2)

Eigenvalues of A:

lambda = 3
lambda = 2

E(3) =

6x - 4y = 3x
3x - y = 3y

-4y = -3x
3x = 4y

Span = ?

The eigenspaces are ?

E(2) =

6x - 4y = 2x
3x - y = 2y

-4y = -4x
3x = 3y

Span = ?

The eigenspaces are ?

Where did I go wrong? How do I find the eigenvalues, what is the "span", and how do I find it?

2. Your work so far is just fine, I would say. You are a bit confused in finding the eigenvectors (they span the eigenspace). Solve the system $\displaystyle (A-\lambda I)x=0$ for $\displaystyle x.$ Since you have chosen $\displaystyle \lambda$ such that the matrix $\displaystyle A-\lambda I$ is singular, you're guaranteed that the eigenvectors you find are not unique. In fact, any scalar multiple of an eigenvector is an eigenvector. So what do you get when you solve the system $\displaystyle (A-\lambda I)x=0$ for $\displaystyle x?$

Here is what I got so far...

(A - lambda I)x = 0

6-lambda -4
3 -1-lambda

4. Originally Posted by sparky

Eigenvalues of A:

lambda = 3
lambda = 2

E(3) =

6x - 4y = 3x
3x - y = 3y

-4y = -3x
3x = 4y

Span = ?

The eigenspaces are ?

E(2) =

6x - 4y = 2x
3x - y = 2y

-4y = -4x
3x = 3y

Span = ?

The eigenspaces are ?

Where did I go wrong? How do I find the eigenvalues, what is the "span", and how do I find it?
I find it easier to do:

$\displaystyle \displaystyle\lambda=3 \ \ \ \displaystyle\begin{bmatrix}6-3&-4\\3&-1-3\end{bmatrix}\Rightarrow\begin{bmatrix}3&-4\\0&0\end{bmatrix}$

$\displaystyle 3x_1=4x_2$

$\displaystyle 3x_2$

$\displaystyle \displaystyle x_2\begin{bmatrix}4\\3\end{bmatrix}$

The eigenvector for lamba = 3 is

$\displaystyle \displaystyle \begin{bmatrix}4\\3\end{bmatrix}$

Do the same for lambda = 2. Then your eigenspace will be

$\displaystyle \displaystyle\left\{\begin{bmatrix}4\\3\end{bmatri x},\begin{bmatrix}\alpha\\ \beta\end{bmatrix}\right\} \ \ \ \alpha,\beta\in\mathbb{R}$