Quick Eigenvalues Question I'm asked to find the eigenvalues and bases of the eigenbases for this 3x3 matrix.
$$\begin{pmatrix}
1    &0   &  -2\\   
0     &0   & 0  \\  
-2     &0  &  4\end{pmatrix}$$
I got eigenvalues of $0 , 0$ and $5$. Now assuming I choose the free variable $Z = 1$, I should get:
Eigenvalue: 0     ----> $V = (2, 0 , 1)$
Eigenvalue: 5     ----> $V = (-1, 0, 2)$
But apparently their is one more eigenbase and thats:
Eigenvalue: 0     -----> $V= (0 ,1, 0)$
Where did this one come from, why are their 2 eigenbases for the eigenvalue of 0. Aren't they meant to me the same answer?
Any help would be greatly appreciated.
 A: Don't do it by "choosing free variables", at least not if you are just learning the subject - it's an almost certain way to miss out on parts of the solution.  (Though it is good and quick once you definitely know what you are doing.)
Instead solve $A-\lambda I=0$ by row-reduction.  In this case $A$ is your given matrix, $\lambda$ is zero and the echelon form is
$$\pmatrix{1&0&-2\cr0&0&0\cr0&0&0\cr}\ .$$
You have to substitute parameters for the non-leading (non-pivot) columns, which are columns 2 and 3.  So
$$x_3=t\ ,\quad x_2=s\ ,\quad x_1=2t\ .$$
The eigenvectors are
$$\pmatrix{2t\cr s\cr t\cr}=t\pmatrix{2\cr0\cr1\cr}+s\pmatrix{0\cr1\cr0\cr}\ ,$$
and you can see two independent eigenvectors.
A: It doesn't mean that you have two bases, but it means that for the eigenbasis, you need more than one vector, i.e. the eigenspace has more than dimension one.
In which cases can this happen? When you calculate the characteristic polynomial and determine the zeros, you might have noticed that $\lambda = 0$ is a root of multiplicity $2$, i.e. the characteristic polynomial is equal to $\lambda^2(\lambda - 5)$.
We say the the eigenvalue $0$ has an algebraic multiplicity of $2$. What you just noticed is that the eigenspace $\ker(A-0)$ is (at least) of dimension $2$. The dimension of the eigenspace is also called geometric multiplicity of the eigenvalue.
In fact, we always have that the geometric multiplicity is always equal to or less than the algebraic multiplicity. So if an eigenvalue is a double root, it can happen that the eigenspace is two-dimensional.
As simple examples consider the zero matrix and the identity ($3 \times 3$) matrix. Both have just one eigenvalue (namely $0$ and $1$, respectively), the algebraic multiplicity is $3$ and since the whole $\mathbb R^3$ is the eigenspace, the geometric multiplicity is $3$ as well.
