Eigenvalues are Vectors in Kernels

Theorem

$T(v)=\lambda v⟺v\in ker(T-\lambda I)$

Proof

Proving $⟹$

1. Suppose $T(v)=\lambda v$
2. $⟹T(v)=\lambda I(v)$ by identity transform
3. $⟹T(v)-\lambda I(v)=0$
4. $⟹(T-\lambda I)(v)$ as $T,\lambda I$ are both linear
5. Thus, $v\in ker(T-\lambda I)$

Proving $⟸$

1. Suppose $v\in ker(T-\lambda I)$
2. $⟹(T-\lambda I)(v)=0$ by defn of kernel
3. $⟹T(v)-\lambda I(v)=0$ as $T,\lambda I$ are both linear
4. $⟹T(v)=\lambda I(v)$
5. $⟹T(v)=\lambda v$ by identity transform