How do you calculate this third eigenvector in this 3x3 matrix?

Question

Scroll down to the bottom if you don't want to read how I arrived at my original two answers. My question is how are all the online calculators I check coming up with this third eigenvector (1, 1, 0), when I come up with only two of them (1, 0, 1), and (0, 0, 1).

I have the matrix

-1     2    0
 0     1    0
-2     2    1

I use the Rule of Sarrus to find the determinant, which ends up being λ = 1 and λ = -1.

I then attempt to calculate the eigenvectors, first with λ = 1:

2     -2     0
0     -2     0
2     -2     0

which reduces to:

1     0     0
0     1     0
0     0     0

So with this, I come up with the eigenvector (0, 0, 1), since v3 is arbitrary (I assign it a 1).

Then I calculate the eigenvector for λ = -1:

0     -2     0
0     -2     0
2     -2    -2

which reduces to

1      0    -1
0      1     0
0      0     0

And so I assign v3 = 1 since it's arbitrary, and come up with the eigenvector (1, 0, 1)

The thing is, when I calculate this with the eigenvector calculators online, it comes up with three eigenvectors:

(1, 0, 1)

(0, 0, 1)

(1, 1, 0)

My question is, where is the third one coming from? I do not see how (1, 1, 0) is a possible eigenvector for this matrix. Can someone explain this to me?

Answer 1

$v=(1,1,0)$ is not an eigenvector, as we can verify by multiplying $Mv$ (where $M$ is your matrix) to get $(1,0,0)$, which is not a multiple of $v$.

Also, notice that $0$ is clearly an eigenvalue, because one of the rows is identically zero, which means the rows are not linearly independent, which means the matrix is not invertible, which means the null space is nontrivial, which means there is a nonzero vector $v$ such that $Mv = 0 = 0v$.

Answer 2

Since there's a row of $0$'s, the matrix is singular, and $0$ is an eigenvalue. An eigenvector for this eigenvalue is $$\pmatrix{1\cr 1/2\cr 1\cr}$$ I don't know where $\pmatrix{1\cr 1\cr 0\cr}$ came from, unless you made a mistake in entering the matrix.