Confusion with eigenspaces

Question

Suppose we have a square matrix $A\in \Bbb F^{n\times n}$ and we want to find its eigenspaces.

We continue to find its eigenvalues and for each eigenvalue $λ$ we consider $(A-λI)X=0$

If $λ$ is such that this system has only the solution $x_i=0, \forall i\in \{1,..,n\}$. How do I proceed to findinging an eigenvector which corresponds to that eigenvalue?

This is something that causes confusion to a lot of students beginning linear algebra and I'd like to see an answer not only for me but also for those who might experience this too.

Any easy to use examples are welcome.

(trying to find a suitable example)

Answer 1

Observe that $\;\lambda\;$ is an eigenvalue of $\;A\;$ iff $\;\det(\lambda I-A)=0\;$ iff $\;\lambda I-A\;$ is singular iff the homogeneous system $\;(\lambda I-A)\vec x=\vec0\;$ has more than the trivial solution.

Answer 2

If $\lambda$ is such that the equation has only zerp vector as a solution, it is not an eigenvalue, since by definition $\lambda$ is an eigenvalue if and only if there is a non-zero vector which satisfies the equation

$$ (A - \lambda I)X = 0. $$

Also, to find eigenvalues, you should solve the polynomial equation

$$ \det(A - \lambda I) = 0. $$