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 - **λ**I 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.

Re: Finding eigenvectors for a matrix

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

Now to find the eigen vectors

$\displaystyle \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 $\displaystyle \begin{bmatrix} -1 & -1 \\ -1 & -1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} $. Which yields infinetly many solutions of the form $\displaystyle \begin{bmatrix} -p \\ p \end{bmatrix}$ So one eigen vector for the eigen value $\displaystyle \lambda = 3 $ would be p = 1, $\displaystyle \begin{bmatrix} -1 \\ 1 \end{bmatrix} $

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

Re: Finding eigenvectors for a matrix

Quote:

Originally Posted by

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

Now to find the eigen vectors

$\displaystyle \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 $\displaystyle \begin{bmatrix} -1 & -1 \\ -1 & -1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} $. Which yields infinetly many solutions of the form $\displaystyle \begin{bmatrix} -p \\ p \end{bmatrix}$ So one eigen vector for the eigen value $\displaystyle \lambda = 3 $ would be p = 1, $\displaystyle \begin{bmatrix} -1 \\ 1 \end{bmatrix} $

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

I appologize, the matrix A is actually [2,1 ; 1,2]

could you check it once again with the right matrix

Re: Finding eigenvectors for a matrix

Quote:

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

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

Re: Finding eigenvectors for a matrix

Quote:

Originally Posted by

**jakncoke** with this matrix the eigen vectors are merely exchanged.

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

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

Re: Finding eigenvectors for a matrix

Yea, even books make mistakes