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I A real matrix and its inverse share the same eigenvectors?

  1. Apr 6, 2016 #1
    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}##.
     
    Last edited: Apr 6, 2016
  2. jcsd
  3. Apr 6, 2016 #2

    Samy_A

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    If ##A v= \lambda v##, then ##A^{-1} v= \frac{1}{\lambda} v##.

    (Of course assuming here that ##A^{-1}## exists.)
     
  4. Apr 6, 2016 #3
    Thanks! I've found the mistake.
     
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