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If you solve $(A−\lambda I) \mathbf{x}=0$ to check if $\lambda$ is an eigenvalue, and get nontrivial solutions how does it prove that $\lambda$ is indeed an eigenvalue?

In other words, why can't $\mathbf{x} = 0$ vector be a solution? And what's the reasoning between a $0$ eigenvector and invertibility?

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  • $\begingroup$ Can you show us your thoughts on the problem? For example, state definition of eigenvalue and eigenvectors so that we know your understanding on these concepts. $\endgroup$ May 3, 2018 at 20:45
  • $\begingroup$ @Melanie Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/… $\endgroup$
    – user
    May 31, 2018 at 19:51

3 Answers 3

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  1. If $\mathbf{x} \neq 0$, we have $Ax = \lambda x$, which is the definition of eigenvectors.

  2. Eigenvectors are defined to be not the $0$-vector. In an eigenproblem formulation that would allow the $0$-vector to be an eigenvector, the $0$-vector would be an eigenvector for every matrix and and every value in $\mathbb{C}$ is a corresponding eigenvalue. This makes the whole concept pretty pointless.

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Of course that the null vector is a solution. The point is: if $(a-\lambda\operatorname{Id})v=0$ for some npn-null vector $v$ (that's what non-trivial solution means here), then $v$ is an eigenvector, since $v\neq0$ and $A.v=\lambda v$.

And if $0$ is an eigenvalue, then there is a non-null vector $v$ such that $A.v=0\times v=0$. Therefore, $A$ is not invertible.

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By definition an eigenvalue $\lambda$ is that value such that for some $\mathbf{x}\neq 0$

$$A\mathbf{x}=\lambda \mathbf{x} \iff (A−\lambda I) \mathbf{x}=0$$

indeed we are interested to non trivial solution since wheter $\mathbf{x}=0$ we have that $A\mathbf{x}=\lambda \mathbf{x}=0$ $\forall \lambda$.

If exist $\lambda=0$ it means that $A\mathbf{x}=0$ for some $\mathbf{x}\neq 0$ then $A$ is not invertible.

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