I need to confirm if the eigenvalue given is a complete eigenvalue and also need to determine the dimension of the associated eigenspace.
I know that an eigenvalue is complete if the geometric and algebraic multiplicity are equal, but I don't totally understand what this means. If the eigenvalue is repeated then its algebraic multiplicity is 2 and if it just appears once it is 1 I think. And I read that the geometric multiplicty is the number of lin. ind. eigenvectors you can find for an eigenvalue, si this correct?
$$3, \begin{bmatrix}
2 & 1
\\ 1 &2
\end{bmatrix}$$
I solved this for $\lambda=1,3$ so I know that the algebraic multiplicity of 3 is 1. But I was wondering if I am doing too much work calculating the eiegnvectors since 3 is already given to me, but this was the only way I could think of checking the algebraic multiplicity.
for $\lambda_1=3$:
$$v_1=\begin{bmatrix}
1
\\ 1\end{bmatrix}$$
so the geometric multiplicty is 1, and thus the eigenvalue is complete.
My book agrees that it is a complete eigenvalue and says that the basis$=[1 1]^T$ (so the basis is just the same as the eigenvector?) with $dim=1$
I am not sure how they calculated the dimension of the eigenspace but I assume it's related to the multiplicity somehow?
Thanks!
Best Answer
Since it doesn't ask you to find all eigenvalues, you don't need to find the eigenvalue $1$. To check whether $3$ is an eigenvalue, check $\det{(A-3I)}=0$.
Your steps to find eigenvector and eigenspace is correct. The eigenspace is the span of all eigenvectors corresponding to this eigenvalue. Since you only found one linearly independent eigenvector, which constitute the basis, the dimension of this space is $1$. The dimension is the number of the basis vectors.
Also, once you find out the algebraic multiplicity is $1$, the geometric multiplicity has to be $1$ since it is less than or equal to the algebraic multiplicity.