Answer to Question #147783 in Linear Algebra for Saranya

Question

if 2,-1,-3 are the eigen values of the matrix A then find the eigen values of the matrix A Inverse

Answer 1

We claim that the eigenvalues of $A^{-1}$ are $\frac{1}{2},-1,-\frac{1}{3}$ . We will prove that in 2 steps :

• Suppose that $v_\lambda\neq 0$ is an eigenvector of $A$ associated to an eigenvalue $\lambda$ (where $\lambda \in \{2, -1, 3\}$). Therefore we have :

$\frac{1}{\lambda} v_\lambda = \frac{1}{\lambda} id \cdot v_\lambda = \frac{1}{\lambda} A^{-1}Av_\lambda = \lambda \cdot \frac{1}{\lambda}(A^{-1}v_\lambda) = A^{-1} v_\lambda$

So $v_\lambda$ is an eigenvector of $A^{-1}$ associated to an eigenvalue $\frac{1}{\lambda}$ .

• Now suppose that $v_\alpha\neq 0$ is an eigenvector $A^{-1}$ associated to an eigenvalue $\alpha$ . We have :

$A(v_\alpha)=A (\frac{1}{\alpha}A^{-1}v_\alpha) = \frac{1}{\alpha} (A A^{-1}v_\alpha) = \frac{1}{\alpha} v_\alpha$

So we have that $\frac{1}{\alpha}$ is an eigenvalue of $A$ .

Thus we have proven that eigenvalues of $A^{-1}$ are $\frac{1}{2}, -1, -\frac{1}{3}$ and there is no other eigenvalues.