1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Will Fisher Aug 23 '16 at 13:30
  • $\begingroup$ Hi @WillFisher so then I can say that $Span(v_1,v_2)$ is the eigenspace? $\endgroup$ – Tom Aug 23 '16 at 13:41
  • $\begingroup$ @Tom Yes, you can. $\endgroup$ – DonAntonio Aug 23 '16 at 14:56
0
$\begingroup$

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.

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?

It's impossible to find $4$ linearly independant vectors which has $0$ as eigenvalue, since it will imply that $A$ is the zero matrix.

$\endgroup$
  • $\begingroup$ So then I can say that $Span(v_1,v_2)$ is the eigenspace? $\endgroup$ – Tom Aug 23 '16 at 13:40
  • $\begingroup$ @Tom If only two linear independent eigenvectors can be obtained, then these two eigenvectors form an eigenspace. $\endgroup$ – imranfat Aug 23 '16 at 13:53
  • $\begingroup$ @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) $\endgroup$ – user171326 Aug 23 '16 at 14:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.