# $(A−\lambda I) x=0$: eigenvalues, nontrivial solutions, and invertibility

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?

• 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. May 3, 2018 at 20:45
• @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/…
– user
May 31, 2018 at 19:51

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.

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.

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.