Eigenvalues are Roots of Characteristic Polynomial

Theorem

A scalar $\lambda$ is an eigenvalue of $T$ $⟺$ if it is a root of Characteristic Polynomial $X_T$

Proof

Proving $⟹$

• Suppose $\lambda$ is a Eigenvalue.
• Then, $T(v)=\lambda v⟹(T-\lambda I)(v)=\overrightarrow{0}$
• Thus, $v\in ker(T-\lambda I)$
• It follows that $T-\lambda I$ is non-invertible by Determinants and Invertiblity Theorem
• Since, $\det (T-\lambda I)=0$ it follows $X_T(\lambda )=0$

Proving $⟸$

• Suppose $X_T(\lambda )=0$
• Then, we get $\det (T-\lambda I)=0$
• Therefore, $ker(T-\lambda I)\ne \{0\}$
• Thus, there is $v\ne 0$ in $ker(T-\lambda I)$
• This gives:
  • $(T-\lambda I)(v)=0$
  • $⟺T(v)=\lambda v$
  • Thus, there is some $\lambda$ eigenvector $v$
  • With $\lambda$ as eigenvalue of $T$