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$M$ is a square matrix $M$ ( matrix representation of a linear operator $L$ acting on a hilbert space $H$ , $L: H \to H$ ) with eigen values $\lambda_i$ and corresponding eigen spaces $V_i$. I know if two eigen spaces are orthogonal eg. $V_i$ and $V_j$ then it means every vector of first eigen space is orthogonal to every vector of second eigen space. But I have two doubts I am having trouble proving ( I know that eigen spaces are linearly independent )

  • If two eigen spaces $V_i$ and $V_j$ are not orthogonal can we still find a pair of vectors $u \in V_i , w \in V_j$ such that $u$ and $w$ are orthogonal ?
  • I know that for a normal operator all eigen spaces are orthogonal to each other. Is it possible to have a linear operator where some pairs of eigen spaces are orthogonal but not all ?
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  • Yes, certainly. $0\in V_i$ and $0\in V_j$ and $\langle 0,0\rangle = 0$
  • Yes, certainly. If $L\colon H\to H$ has nonrthogonal eigenspaces and wlog. $1$ is not an eigenvalue, then consider $L\oplus\operatorname{id}\colon H\oplus \mathbb C\to H\oplus \mathbb C$.
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    $\begingroup$ Using 0 is kind of obtuse (no pun intended). The real answer to 1 is no, considering only the interesting case of two nonzero vectors. $\endgroup$ May 1 '15 at 17:33
  • $\begingroup$ Apart from null vector there won't be a vector , for the first case. Am I correct ? $\endgroup$
    – sashas
    May 1 '15 at 17:34
  • $\begingroup$ @MattSamuel thanks . I get it now. $\endgroup$
    – sashas
    May 1 '15 at 17:34
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    $\begingroup$ @sasha The answer is no in general. The answer is very often yes. $\endgroup$ May 1 '15 at 17:37
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  1. Yes, it is possible.

  2. Yes, in fact you can build a matrix with any eigenvectors/eigenvalues you like.

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