Theorem:
Let $T$ be a linear operator on a vector space $V$ and let $\lambda$ be an eigenvalue of $T$. A vector $v\in V$ is an eigenvector of $T$ corresponding to $\lambda$ if and only if $v\not=0$ and $v\in N(T-\lambda I)$.
Question Motivation:
By definition a nonzero vector $v\in V$ is eigenvector if there is a scalar $\lambda$ such that $T(v)=\lambda v$
In the proof they use $Tv=\lambda v\Rightarrow(T-\lambda I)v=0$ since $v$ is not zero, then $\det(T-\lambda I)=0$
Questions:
Why? Is it all because if $\det(T-\lambda I)\not=0$ we can take inverse of $T-\lambda I$ from left and $v$ would be equal to $0$?
Why2? All we have is $T(v)=\lambda v$. How can it imply that $Tv=\lambda v$. What does $Tv=\lambda v$ means? Matrix multiplication or composition or …?
Best Answer
When $T$ is a linear transformation sometimes people write just $Tv$ instead of $T(v)$. It makes sense in the finite dimensional case (though it is used when the dimension is infinite as well) because of the strong connection between linear transformations and multiplication of a matrix by vector. Anyway, $Tv$ and $T(v)$ is the same thing.
And yes, if $A$ is a matrix then $\det(A)\ne 0$ if and only if the system of linear equations $Ax=0$ has only the trivial solution. This is a basic theorem in linear algebra. The corresponding theorem for an operator $T:V\to V$ is that $\det(T)\ne 0$ if and only if $Ker(T)=\{0\}$.