# Eigenspace for $4 \times 4$ matrix

The matrix $A=\begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix}$ has eigenvalue $\lambda=0$ of multiplicity $4$. Solving $(A-\lambda I)v=0$ for $\lambda=0$ only get two eigenvectors:

$v_1=\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}$ and $v_2=\begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}$. So can I say the eigenspace corresponding to $\lambda=0$ is equal to $Span(v_1,v_2)$? Or do I need four vectors to form the eigenspace, and if I do need 4 then how do I find the other two to form the eigenspace?

• The eigenspace for a given lambda is the union of all eigenvectors corresponding to that eigenvalue and the zero vector. Here if you just choose two independent eigenvectors that are a basis for all eigenvectors corresponding to $\lambda =0$, then their span also creates the eigenspace. – Will Fisher Aug 23 '16 at 13:30
• Hi @WillFisher so then I can say that $Span(v_1,v_2)$ is the eigenspace? – Tom Aug 23 '16 at 13:41
• @Tom Yes, you can. – DonAntonio Aug 23 '16 at 14:56

That means simply the algebraic multiplicity of $\lambda = 0$ is $4$ and the geometric multiplicity of $\lambda$ is $2$. As you noticed there is no other eigenvalue.
It's impossible to find $4$ linearly independant vectors which has $0$ as eigenvalue, since it will imply that $A$ is the zero matrix.
• So then I can say that $Span(v_1,v_2)$ is the eigenspace? – Tom Aug 23 '16 at 13:40
• @Tom : yes, sure. The eigenspace of an eigenvalue is by definition a subspace of $V$. When the eigenspaces generate the whole space, this is a special case and the matrix/endomorphism is called diagonalisable. You just have a non-diagonalisable matrix !! (Notice for example that every non-zero nilpotent matrix is non-diagonalisable) – user171326 Aug 23 '16 at 14:12