A, B, C are three real-square matrices. A is an upper triangular matrix with all of its diagonal entries equal to zero. B is a matrix such that $b_{ij}=-b_{ji}$, and C is a matrix such that $\sum_j c_{ij}c_{jk} = \delta_{ik}$ in which $\delta_{ik}$ is a Kronecker delta. It's obvious that since A has determinant zero, at least one of its eigenvalue is zero, is there any more we could deduce about eigenvalues of A just from its properties? Also, for B and C, I am not quite sure what to say about their eigenvalues... Any hints or references would be greatly appreciated.

  • $\begingroup$ $B$ is skew-symmetric and has det=0 if the dimension is odd. $\endgroup$ – vadim123 Mar 17 '14 at 19:59
  • $\begingroup$ The matrix $\;A\;$ is in fact nilpotent and thus zero is its only eigenvalue... $\endgroup$ – DonAntonio Mar 17 '14 at 20:04
  • The matrix $A$ is (strictly) upper triangle matrix so its eigenvalues are the entries on its diagonal so all the eigenvalues are $0$.
  • The matrix $B$ is skew-symmetric and if $\lambda$ is an eigenvalue of $B$ i.e. $B^T=-B$ and $x$ is an eigenvector associated to it then

$$\lambda ||x||^2=\langle Bx, x\rangle=\langle x, B^Tx\rangle=-\overline\lambda||x||^2$$ so we conclude that $\lambda=-\overline\lambda$ and then $\lambda$ is $0$ or pure imaginary.

  • The matrix $C$ verify: $C^2=I_n$ so the polynomial $x^2-1$ annihilates $C$ and then $$\operatorname{sp}(C)\subset\{-1,1\}$$ or we can found this previous result by saying that $C$ is a matrix of a symmetry.
  • $\begingroup$ How about $ C=-I_n $ with all eigenvalues $-1 $? $\endgroup$ – DKal Mar 17 '14 at 20:14
  • $\begingroup$ Sorry I fixed the mistake: the roots of $x^2-1$ are obviously $\pm1$. $\endgroup$ – user63181 Mar 17 '14 at 20:16
  • $\begingroup$ I just found the answer (not the proof), and it says for B, all non-zero eigenvalues are pure imaginary, and for C, each real eigenvalue must be -1, not -1 or 1. For B, I probably could apply spectral theorem to find it out, but I'm not sure why for C, the real eigenvalue is -1 instead of 1 or -1 (you're right about A). $\endgroup$ – user98235 Mar 17 '14 at 20:20
  • $\begingroup$ I think that the answer that you have is wrong!@user98235 $\endgroup$ – user63181 Mar 17 '14 at 20:23
  • $\begingroup$ @SamiBenRomdhane well, it was from UPenn prelim hans.math.upenn.edu/amcs/AMCS/prelims/prelim_review.pdf in linear algebra section, problem 2. For c), (which corresponds to matrix A) the matching condition is (B), and for d), (which corresponds to matrix C) the matching condition is (F), and for f), (which corresponds to matrix B) the matching condition is A). $\endgroup$ – user98235 Mar 18 '14 at 8:31

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