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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)

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    In that case, $\lambda$ is not an eigenvalue – 2017-02-27
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    @Exodd See my worked example in the question – 2017-02-27
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    Write the original matrix $A$ and the value of $\lambda$ – 2017-02-27
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    @Exodd I've made a mistake on the calculations, so It has actually rank $1$. I'll try to find a correct example – 2017-02-27
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    Why the downvote? – 2017-02-27
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    @Michael If I remember correctly, your example did indeed have rank $2$. I think your problem lies somewhere deeper. Give us the matrix whose eigenvalues and eigenvectors you want to compute. Then we can show you step by step how to find eigenvalues and eigenvectors – 2017-02-27
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    @VishalGupta Yes, I'm looking for examples. Any matrix, whose one eigenvalue gives a system $(A-λI)X=0$ with the zero solution, will do, if you'd like to help me out a bit ($2\times 2$ matrix is fine). – 2017-02-27
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    Do you understand what eigenvalue means? – 2017-02-27
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    @VishalGupta Haha, I think I saw somewhere such an example this is why I'm confused, yes I know that there needs to exist a non-zero vector such that $AX=λX$. Maybe I had something else in my mind, and I didn't see such an example. – 2017-02-27
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    You could surely find examples of matrix $A$ and a *number* $\lambda$ such that $(A - \lambda I)X = 0$ has only trivial solution. But in that case, $\lambda$ is not called an eigenvalue **by definition**. – 2017-02-27
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    @VishalGupta so no eigenvectors (and no eigenvalue) – 2017-02-27
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/54451/discussion-between-vishal-gupta-and-michael). – 2017-02-27

2 Answers 2

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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

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

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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.