- Let $A$ be an invertible matrix. If $\lambda$ is an eigenvalue of $A$, show that $\lambda\neq 0$ and that $\lambda ^{-1}$ is an eigenvalue of $A^{-1}$.
My proof trying. Assume $\lambda$ is an eigenvalue of $A$. Since $A$ is an invertible matrix, $Det(A)\neq 0$. Since $\lambda$ is an eigenvalue of $A$,
$Det(A-\lambda I_2)=0$. Let $A=\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$.Thus, $Det(A)\neq 0$. Then, $Det(A-\lambda I_2)=\lambda ^2-5\lambda -2=0$. So, solution of this equation, $x_1,x_2= \dfrac {\pm5+4\sqrt {2}}{2}$. Now, what should I do? Can you help, can you check my proof-trying?
Best Answer
Your proof is wrong, $A$ has to be any square matrix. Let $\lambda \neq 0$ be an eigenvalue of $A$, by definition $$Av=\lambda v,$$ where $v \neq \mathbf{0}$ is a vector. Multiplying by $A^{-1}$ both sides of the equation yields $$A^{-1}Av=A^{-1}\lambda v \iff v=A^{-1}\lambda v \iff \lambda^{-1}v=A^{-1}v.$$ Hence $\lambda^{-1}$ is a eigenvalue of $A^{-1}$.