Linear Algebra, geometric multiplicity I have a matrix and the question says I that I have an eigenvalue of 0.
 The question asks me to find the geometric multiplicity of that eigenvalue. I know the answer      is 4. I just don't understand how it is 4 since this matrix can be reduced to just one row of 1 1 1 1 1and the rest of the rows are 0's. Thanks in advance for your comments/
  1 1 1 1 1 
  1 1 1 1 1
  2 2 2 2 2
  2 2 2 2 2
  2 2 2 2 2

 A: The geometric multiplicity of an eigenvalue $\lambda$ for the $n\times n$ matrix $A$ is, by definition, the dimension of the subspace
$$
E_A(\lambda)=\{v\in K^n:Av=\lambda v\}
$$
where $K$ is the base field, in your case probably $\mathbb{R}$ or $\mathbb{C}$, and the elements of $K^n$ are column vectors.
This subspace is just the null space (also known as kernel) of the matrix $A-\lambda I_n$, because
$$
Av=\lambda v
\quad\text{if and only if}\quad
Av=\lambda I_nv
\quad\text{if and only if}\quad
(A-\lambda I_n)v=0
$$
The dimension of this subspace is computed easily:
$$
\dim E_A(\lambda)=\dim N(A-\lambda I_n)=n-\operatorname{rank}(A-\lambda I_n)
$$
by the rank-nullity theorem.
The case of $\lambda=0$ is no different:
$$
\dim E_A(0)=\dim N(A-0I_n)=\dim N(A)=n-\operatorname{rank}(A)
$$
and in your case $n=5$ and $\operatorname{rank}(A)=1$.
A: Well, precisely because you can reduce your matrix to that is that the geometric multiplicity of zero is four...
Think it this way: you're looking for the dimension of the solution subspace of the homogeneous linear system whose coefficients' matrix is your matrix, so getting only one
 non-zero row after reducing the matrix means you have $\;4\;$ degrees of freedom to choose your general solution vector...voila!
