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I would like to ask you something regarding the trace of a matrix (the value of the diagonal after adding all its members, a value which is said to remain constant independently from base changes):

Do matrices with trace value zero constitute a (sub)group? If so, which properties tell them apart from matrices with non-zero trace values? Should a matrix with diagonal entries such as , say, [2, -3, 1], group together with that showing [0,0,0]?

I am particularly interested in those cases in which all the members of the diagonal equal zero. Do they have interesting algebraic properties as compared with the other zero trace matrices and the nonzero matrices altogether?

Thanks in advance.

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  • $\begingroup$ Useful Point: Skew Symmetric matrices have trace 0. (in fact, all diagonal elements zero.) $\endgroup$
    – Jesse P Francis
    Commented Apr 14, 2015 at 9:06
  • $\begingroup$ en.wikipedia.org/wiki/… has useful information towards it. $\endgroup$
    – Jesse P Francis
    Commented Apr 14, 2015 at 9:09
  • $\begingroup$ If you consider the $K$-vector space of square matrices $M_n(K)$, then the trace is a linear form $\operatorname{Tr}:M_n(K) \longrightarrow K$, hence the space of traceless matrices is the kernel of this form. This implies that it is a subspace of $M_n(K)$ of dimension $n^2-1$. $\endgroup$
    – Crostul
    Commented Apr 14, 2015 at 9:14

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For the group question, try computing the square of the $2\times2$ matrix describing a quarter turn.

The traceless matrices do form a Lie algebra (the bracket is the commutator $[A,B]=AB-BA$). In fact the commutator of any pair of square matrices is traceless.

There is nothing particularly interesting about matrices with all diagonal entries zero, since that condition is not invariant under change of basis (while the value of the trace is).

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  • $\begingroup$ How come the zero diaginal entries condition not be invariant? As you correctly state, the value fo the trace remains constant. All diagonal entries being zero means the trace value is zero, and that zero value will remain invariant across changes of basis. Do you mean that value would remain zero while the zero entries rearrange themselves with different values which nonetheless yield zero as their sum? Could you provide a specific example withchange of base? Thanks. $\endgroup$
    – Javier Arias
    Commented Apr 14, 2015 at 10:41
  • $\begingroup$ Is it fair to say that traceless matrices represent a scenario where the group does not undergo any movement whatsoever? $\endgroup$
    – Javier Arias
    Commented Apr 14, 2015 at 11:16
  • $\begingroup$ @JavierArias: For instance the permutation matrix $(\begin{smallmatrix}0&1\\1&0\end{smallmatrix})$ has all diagonal entries zero, but it is diagonalisable with diagonal form $(\begin{smallmatrix}1&0\\0&-1\end{smallmatrix})$, which does not have all diagonal entries zero (though it is still traceless). Diagonalisation is obtained by an appropriate change of basis. $\endgroup$
    – Marc van Leeuwen
    Commented Apr 14, 2015 at 11:26
  • $\begingroup$ @JavierArias: For your second comment, I don't know what you mean by "group does not undergo any movement whatsoever"? What group, which action? What one can say is that for the adjoint (conjugation) action of $GL(n)$ on itself, the derivatives (which is what the Lie bracket gives) are always in the direction of the codimension$~1$ subspace of the tangent space given by the traceless matrices. This does not mean the derivative is zero; certainly things do undergo change. $\endgroup$
    – Marc van Leeuwen
    Commented Apr 14, 2015 at 11:29
  • $\begingroup$ Well, a Lie Algebra represents a Lie Group, right? A continuous movement group, right? I am sorry, I am not a mathematician, just someone trying to understand certain structures in order to better understand linguistics, for instance; someone who relies on books such as the Encyclopedia of Mathematics by Kolmogorov or on Jules Vuillemin's chapter on Lie in his "La philosophie de l'algèbre". I make an effort to understand, but probably have many formal lacunae. Sorry if that is the case. $\endgroup$
    – Javier Arias
    Commented Apr 14, 2015 at 11:32

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