# Properties of a matrix and eigenvalues

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.

• $B$ is skew-symmetric and has det=0 if the dimension is odd. – vadim123 Mar 17 '14 at 19:59
• The matrix $\;A\;$ is in fact nilpotent and thus zero is its only eigenvalue... – 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.
• How about $C=-I_n$ with all eigenvalues $-1$? – DKal Mar 17 '14 at 20:14
• Sorry I fixed the mistake: the roots of $x^2-1$ are obviously $\pm1$. – user63181 Mar 17 '14 at 20:16