let $V$ be a finite dimension complex vector space and $T$ a normal operator. I'm looking for an example that on a $T$-invariant subspace ,$T$ is $not$ normal,for I really can't prove it must be.Any advise? Thank you

  • $\begingroup$ Sorry but to me, this is a bit confusing. can you explain to me the steps of how you want to disprove the given statement? to my understanding, you want to prove that restriction of normal operator need not be a normal operator. $\endgroup$
    – IamKnull
    Commented Sep 19, 2019 at 13:49
  • $\begingroup$ I think a normal matrix which is two-block upper triangular is hopefully to be a counterexample.it is necessarily not symmetric or the subspace be one dimension, for in these two cases the assertion is obviously right. $\endgroup$
    – ysTuan
    Commented Sep 19, 2019 at 13:57
  • $\begingroup$ just like the matrix you gave, but i want a normal one.. $\endgroup$
    – ysTuan
    Commented Sep 19, 2019 at 13:58
  • $\begingroup$ Real 2x2 nor9mal matrix are symmetric or has form $\begin{bmatrix} a & b\\ -b & a \end{bmatrix} $ 1 $\endgroup$
    – IamKnull
    Commented Sep 19, 2019 at 14:06
  • $\begingroup$ @KNilesh sorry i meant two-block uppertriangular not 2 dim..for example a 4×4 matrix with left-down 2×2 block be zero $\endgroup$
    – ysTuan
    Commented Sep 19, 2019 at 14:07


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