# Thread: Finding eigenvectors for a matrix

1. ## Finding eigenvectors for a matrix

Need some help with this, the book is written in a very confusing way. I'm following the examples exactly and not getting the correct result.

The matrix A = [2,-1 ; -1,2]

I found the two eigenvalues, to be λ = 1 and λ = 3

Now when I do the A - λI calculation and reduce the matrix I get

for λ = 1

x1 + x2 = 0 --> x1 = -x2

however, the solution in the book says the answer is

[1,1]

for λ = 3

-x1 + x2 = 0 --> x1 = x2

but again, the solution says

[-1, 1]

Can someone please clarify or point out what I am doing wrong?

Thanks.

2. ## Re: Finding eigenvectors for a matrix

So you solved the characteristic equation to get $\lambda = 3$ and $\lambda = 1$. I got the same eigen values.

Now to find the eigen vectors

$\begin{bmatrix}2 & -1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \lambda \begin{bmatrix} x \\ y \end{bmatrix}$ in which you get to solve a homogenous equation $\begin{bmatrix} -1 & -1 \\ -1 & -1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$. Which yields infinetly many solutions of the form $\begin{bmatrix} -p \\ p \end{bmatrix}$ So one eigen vector for the eigen value $\lambda = 3$ would be p = 1, $\begin{bmatrix} -1 \\ 1 \end{bmatrix}$

For $\lambda = 1$ i got eigen vectors using the same processing above, infinitely many solutions of the form $\begin{bmatrix} p \\ p \end{bmatrix}$. So p = 1, $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$ is a eigen vector for eigen value $\lamba = 1$

3. ## Re: Finding eigenvectors for a matrix

Originally Posted by jakncoke
So you solved the characteristic equation to get $\lambda = 3$ and $\lambda = 1$. I got the same eigen values.

Now to find the eigen vectors

$\begin{bmatrix}2 & -1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \lambda \begin{bmatrix} x \\ y \end{bmatrix}$ in which you get to solve a homogenous equation $\begin{bmatrix} -1 & -1 \\ -1 & -1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$. Which yields infinetly many solutions of the form $\begin{bmatrix} -p \\ p \end{bmatrix}$ So one eigen vector for the eigen value $\lambda = 3$ would be p = 1, $\begin{bmatrix} -1 \\ 1 \end{bmatrix}$

For $\lambda = 1$ i got eigen vectors using the same processing above, infinitely many solutions of the form $\begin{bmatrix} p \\ p \end{bmatrix}$. So p = 1, $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$ is a eigen vector for eigen value $\lamba = 1$
I appologize, the matrix A is actually [2,1 ; 1,2]

could you check it once again with the right matrix

4. ## Re: Finding eigenvectors for a matrix

Originally Posted by zdravkoBG
I appologize, the matrix A is actually [2,1 ; 1,2]

could you check it once again with the right matrix
with this matrix the eigen vectors are merely exchanged.

for $\lambda = 3$ i got eigen vectors of the form $\begin{bmatrix} p \\ p \end{bmatrix}$ so for p = 1 $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$ is the one of the eigen vector for $\lambda = 3$

for $\lambda = 1$ i got eigen vectors of the form $\begin{bmatrix} -p \\ p \end{bmatrix}$ so for p = 1 $\begin{bmatrix} -1 \\ 1 \end{bmatrix}$ is the one of the eigen vector for $\lambda = 1$

5. ## Re: Finding eigenvectors for a matrix

Originally Posted by jakncoke
with this matrix the eigen vectors are merely exchanged.

for $\lambda = 3$ i got eigen vectors of the form $\begin{bmatrix} p \\ p \end{bmatrix}$ so for p = 1 $\begin{bmatrix} 1 \\ 1 \end{bmatrix}$ is the one of the eigen vector for $\lambda = 3$

for $\lambda = 1$ i got eigen vectors of the form $\begin{bmatrix} -p \\ p \end{bmatrix}$ so for p = 1 $\begin{bmatrix} -1 \\ 1 \end{bmatrix}$ is the one of the eigen vector for $\lambda = 1$
That's exactly what I thought the correct answer was. But in the back of the book with the answers, it states that for eigenvalue 3, the vector is [-1,1] and that for eigenvalue 1, the vector is [1,1].
Could they have just switched them by accident? Because I am getting the same results as you are.

6. ## Re: Finding eigenvectors for a matrix

Yea, even books make mistakes