# Thread: Lambda is eigenvalue of T if and only if inverse lambda is eigenvalue of inverse T

1. ## Lambda is eigenvalue of T if and only if inverse lambda is eigenvalue of inverse T

Question: Suppose that $T$ is an invertible linear operator. Prove that $\lambda$ is an eigenvalue of $T$ if and only if $\lambda^{-1}$ is an eigenvalue of $T^{-1}$.

Let $A$ be defined as an $n \times n$ matrix such that $T(x) = Ax$. Then I tried approaching this by demonstrating that
$det(A - \lambda I_n) = det(A^{-1} - \lambda^{-1} I_n) = 0$
But the only information that I have is $AA^{-1}=A^{-1} A = I_n$ - how does that get me to the desired result?

I suppose that if we know $det(A - \lambda I_n) = 0$, when the rank of $A - \lambda I_n$ is not full (i.e. - less than $n$). And then we can try to prove that because of this, the rank of $A^{-1} - \lambda^{-1} I_n$ is also less than $n$...?

2. If $\lambda$ is an eigenvalue of $T$ then there exists $v\in V$ such that $T(v)=\lambda v$ which happens iff $v= \lambda T ^{-1}(v)$ (since both $T$ and $T^{-1}$ are linear) and since $T$ is invertible it implies that $0$ can't be an eigenvalue so $\lambda ^{-1} v = T^{-1} (v)$