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Definition: Let T be a linear operator on a vector space V. A nonzero vector v ∈ V is called an eigenvector of T if there exists a scalar λ such that T(v) = λv. The scalar λ is called the eigenvalue corresponding to the eigenvector v.

Question 1): When does eigenvector $v \in V$ have to be nonzero? And why?

I also come upon this: Theorem: Let $A \in M_{n \times n}(F)$, then a scalar $\lambda$ is an eigenvalue of A if and only if $det(A-\lambda I_n)=0$

Question 2: Does that mean the linear map $(A-\lambda I_n)$ is always noninvertible?

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    $\begingroup$ 1. always, 2. yes. $\endgroup$
    – Jan
    Feb 25, 2020 at 15:28
  • $\begingroup$ @Jan Thanks! can you explain further why has to be always nonzero? $\endgroup$
    – neveryield
    Feb 25, 2020 at 15:29
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    $\begingroup$ One explanation is maybe that $v = 0$ is $always$ a solution of $Tv = \lambda v$ for $every$ $\lambda$ and this is not really interesting. $\endgroup$
    – Jan
    Feb 25, 2020 at 15:31
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    $\begingroup$ For question 2: Since $A-\lambda I_n$ isn't invertible, $\ker(A-\lambda I_n)\neq\lbrace 0\rbrace $, thus there's a nonzero $v\in V$ such that $(A-\lambda I_n )v=0$ which is $Av=\lambda v$. $\endgroup$
    – Fakemistake
    Feb 25, 2020 at 15:38
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    $\begingroup$ Thanks guys, now I can tell why. $\endgroup$
    – neveryield
    Feb 25, 2020 at 15:39

1 Answer 1

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If you allow an eigenvector to be $\mathbf{0}$, then suddenly every scalar $\lambda$ is an eigenvalue. This is because $A\mathbf{0} = \lambda \mathbf{0} = \mathbf{0}$ now holds for every $\lambda$. This is not useful.

Now $A-\lambda I$ is no longer one-to-one if there is non-zero $\mathbf{v}$ with $A\mathbf{v} = \lambda \mathbf{v}$, because both $\mathbf{0}$ and $\mathbf{v}$ are in the kernel. Hence $A-\lambda I$ must be singular/non-invertible.

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