# About definition and theorem of eigenvector/eigenvalue

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?

• 1. always, 2. yes.
– Jan
Feb 25, 2020 at 15:28
• @Jan Thanks! can you explain further why has to be always nonzero? Feb 25, 2020 at 15:29
• One explanation is maybe that $v = 0$ is $always$ a solution of $Tv = \lambda v$ for $every$ $\lambda$ and this is not really interesting.
– Jan
Feb 25, 2020 at 15:31
• 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 $Av=\lambda v$. Feb 25, 2020 at 15:38
• Thanks guys, now I can tell why. Feb 25, 2020 at 15:39

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.