# Finding eigenvalues and eigenvectors of a matrix A

## Homework Statement

Find the eigenvalues and eigenvectors of the matrix A = [2, 1; 8, 4]

## Homework Equations

det(A - lambda I) = 0

## The Attempt at a Solution

After expanding using the formula I have the equation (2 - $\lambda$)(4 - $\lambda$) - 8

Which gives $\lambda$ = 0, 6 (Should I be worried that I got a zero in my answer?)

Then substituting the values for lambda back into the problem and using (A - $\lambda$I)v = 0

For $\lambda$ = 6
[-4, 1; 8, -2] * [a; b] = [0; 0] and I get the eigenvector a[1; 4] from (-4a + b = 0)

For $\lambda$ = 0
[2, 1; 8, 4] * [a; b] = [0; 0] I get a second eigenvector of a[1; -2] from (2a + b = 0)

Did I get the right eigenvalues and corresponding eigenvectors?

## The Attempt at a Solution

Dick
Homework Helper
Yes, you did. And it's easy for you to check as well. A*[1;4] should give you 6*[1;4] and A*[1;-2] should give you 0*[1;-2], right? Does it work?

HallsofIvy
Homework Helper

## Homework Statement

Find the eigenvalues and eigenvectors of the matrix A = [2, 1; 8, 4]

## Homework Equations

det(A - lambda I) = 0

## The Attempt at a Solution

After expanding using the formula I have the equation (2 - $\lambda$)(4 - $\lambda$) - 8

Which gives $\lambda$ = 0, 6 (Should I be worried that I got a zero in my answer?)
Notice that the determinant of the matrix is 2(4)- 1(8)= 0. Since the determinant of a matrix is the product of its eigenvalues, 0 must be an eigenvalue.

Then substituting the values for lambda back into the problem and using (A - $\lambda$I)v = 0

For $\lambda$ = 6
[-4, 1; 8, -2] * [a; b] = [0; 0] and I get the eigenvector a[1; 4] from (-4a + b = 0)

For $\lambda$ = 0
[2, 1; 8, 4] * [a; b] = [0; 0] I get a second eigenvector of a[1; -2] from (2a + b = 0)

Did I get the right eigenvalues and corresponding eigenvectors?