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Suppose ##v_i## is an eigenvector of ##A## with eigenvalue ##\lambda_i## and multiplicity ##1##.
##AA^{-1}v_i=A^{-1}Av_i=A^{-1}\lambda_iv_i=\lambda_iA^{-1}v_i##
Thus ##A^{-1}v_i## is also an eigenvector of ##A## with the same eigenvalue ##\lambda_i##.
Since the multiplicity of ##\lambda_i## is ##1##,
##A^{-1}v_i=k_iv_i##, where ##k_i## is a constant.
Thus ##v_i## is also an eigenvector of ##A^{-1}## with the same eigenvalue ##\lambda_i##.
What's wrong with this proof?
Counterexample:
##A=\begin{pmatrix}\frac{\sqrt3}{2}&\frac{1}{2}&0\\-\frac{1}{2}&\frac{\sqrt3}{2}&0\\0&0&1\end{pmatrix}##
The eigenvector corresponding to eigenvalue ##\frac{\sqrt3}{2}+\frac{1}{2}i## is ##\begin{pmatrix}-\frac{\sqrt2}{2}i\\ \frac{\sqrt2}{2}\\0\end{pmatrix}##, but that for ##A^{-1}## is ##\begin{pmatrix}\frac{\sqrt2}{2}\\-\frac{\sqrt2}{2}i\\0\end{pmatrix}##.
##AA^{-1}v_i=A^{-1}Av_i=A^{-1}\lambda_iv_i=\lambda_iA^{-1}v_i##
Thus ##A^{-1}v_i## is also an eigenvector of ##A## with the same eigenvalue ##\lambda_i##.
Since the multiplicity of ##\lambda_i## is ##1##,
##A^{-1}v_i=k_iv_i##, where ##k_i## is a constant.
Thus ##v_i## is also an eigenvector of ##A^{-1}## with the same eigenvalue ##\lambda_i##.
What's wrong with this proof?
Counterexample:
##A=\begin{pmatrix}\frac{\sqrt3}{2}&\frac{1}{2}&0\\-\frac{1}{2}&\frac{\sqrt3}{2}&0\\0&0&1\end{pmatrix}##
The eigenvector corresponding to eigenvalue ##\frac{\sqrt3}{2}+\frac{1}{2}i## is ##\begin{pmatrix}-\frac{\sqrt2}{2}i\\ \frac{\sqrt2}{2}\\0\end{pmatrix}##, but that for ##A^{-1}## is ##\begin{pmatrix}\frac{\sqrt2}{2}\\-\frac{\sqrt2}{2}i\\0\end{pmatrix}##.
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