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I saw there is a proof for invertible matrices, but I don't know how to put this mathematically for a transformation. How do I prove an invertible linear transformation has the same eigenvectors as its inverse?

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  • $\begingroup$ Can't you just fix bases and consider your linear transformation as a matrix? $\endgroup$ – dcolazin Apr 13 at 7:22
  • $\begingroup$ yup...I suppose that's what should have been done in the first place. I have posted an algebraic solution. $\endgroup$ – DGR Apr 13 at 7:26
  • $\begingroup$ Wouldn’t the proof be exactly the same as the one for matrices? $\endgroup$ – amd Apr 13 at 18:42
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Observe that for an invertible matrix $A$ with eigenvector $\mathbf v$ and corresponding eigenvalue $\lambda\neq 0$, you have that \begin{align*} \mathbf v &= I \mathbf v\\&=A^{-1}A \mathbf v\\&=\lambda A^{-1} \mathbf v \end{align*} Hence $A^{-1} \mathbf v = \lambda^{-1}\mathbf v$

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OK, so Iv'e found a solution!

\begin{align*}\\ Tv &= \lambda v \ \ \vert *T^{-1} \\ T^{-1}Tv &= T^{-1}\lambda v \\ v &= \lambda T^{-1}v \ \ \vert *\lambda^{-1} & \text{(Invertible transformation, $\lambda\neq 0$)} \\ \lambda^{-1}v &= T^{-1}v \end{align*}

For $T^{-1}$, eigenvalue $\lambda^{-1}$, eigenvector is $v$ (same as eigenvector of $T$).

Hope this helps!

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