# Question about eigenvector and identity matrix

1. Oct 21, 2013

### Umayer

1. The problem statement, all variables and given/known data

I was doing this practice exam and I had to calculate the eigenvalues en vectors. The matrix had two eigenvalues, I calculated one eigenvector. But when I was performing row operations for the second eigenvector, the matrix with the second eigenvalue substitued became an identity matrix, which kinda blew my mind.

So my question is what does this mean? Does it mean that the matrix doesn't have any eigenvectors? And is it possible that it can become an identity matrix? Also I'm pretty sure that I didn't make a mistake, I put the matrix on my calculator and used the funcion "Rref" on it and the result was the identity matrix. Any help would be appreciated!

2. Relevant equations

3. The attempt at a solution

2. Oct 21, 2013

### Bryson

Every vector is an Eigenvector of the identity matrix. Perhaps I do not understand what your saying. . .

3. Oct 21, 2013

### Staff: Mentor

When you solve for the eigenvalues and eigenvectors of some square matrix A, you are trying to find nonzero vectors x for which Ax = λx. That's the same as saying (A - λI)x = 0, for some nonzero vector x. This can happen only if det(A - λI) = 0. In your row reduction, you should have ended up with a matrix with one or more rows of zeros.

It would help if you showed us the matrix you started with, and what you ended with after row reduction.

4. Oct 21, 2013

### Ray Vickson

Please show us the actual matrix.

5. Oct 22, 2013

### Umayer

1. The problem statement, all variables and given/known data
Actually I figured it out, I forgot to put a minus when calculating the determinant. But I'll write down the matrix. This is the matrix given: $$\begin{pmatrix} 1 & 1 & 2\\ 4 & 0 & 2\\ -2 & 1 & 1 \end{pmatrix}$$

So when determining the eigenvalues the matrix will become: $$\begin{pmatrix} 1-λ & 1 & 2\\ 4 & -λ & 2\\ -2 & 1 & 1-λ \end{pmatrix}$$

2. Relevant equations
This is how I calculated the determinant:

$A11: (1-λ)([-λ(1-λ)]-2)=(1-λ)(λ^2-λ-2)$
$A12: -1([4(1-λ)]-[2*-2])=-8+4λ$ (This is where I forgot the minus in front of the 1.)
$A13: 2([1*4]-[-2*-λ])=8-4λ$

A12 and A13 cancels each other out so:

$detA = 0 → (1-λ)(λ^2-λ-2)=0$

So the eigenvalues are: $λ=1$, $λ=2$ , $λ=-1$

I then had to calculate two eigenvectors, I chose 1 and 2 which are respectively:$$\frac{√6}{3} \begin{pmatrix} 0,5\\ 1\\ -0,5 \end{pmatrix}$$

$$\frac{1}{√35} \begin{pmatrix} -3\\ -5\\ 1 \end{pmatrix}$$

I believe this is the correct answer.

3. The attempt at a solution
So now my question is when I was calculating the determinant, I forgot the minus so my answers became this: $λ=2$, $λ^2=9$

I inserted $λ=3$ into the eigenvalue matrix, it became this:$$\begin{pmatrix} -2 & 1 & 2\\ 4 & -3 & 2\\ -2 & 1 & -2 \end{pmatrix}$$

Performing row operations the matrix will become:$$\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix}$$

Can a situation like this ever occur? I hope this clears a little bit up.