Show that the trace $c_{ii}$ of a quadratic form Q does not depend on the choice of basis (invariant)

Question

So, I want to Show that the trace $c_{ii}$ of a quadratic form Q does not depend on the choice of basis (that is trace is invariant). I have the following information and ideas : I would like to try to show this in the language of tensors. I know that the coeffisients $c_{ij}$ of a quadratic form Q = $c_{ij}x^ix^j $ transforms as a covariant tensor: $$ c_{k'm'} = c_{ij}A^{i}_{k'}A_{m'}^{j} $$ where the A's are the transformation coeffisients of the vector components of the x's : $x^i = A^{i}_{k'}x^{k'}$ and $x^j = A_{m'}^{j}x^{m'} $ Also, I guess that we can assume that the x-vector is written initially in the standard orthonormal euclidean basis, so that the vector $x = x^ie_i$ , these basisvectors as I understand transforms covariantly as $e_i = A^{k'}_{i}e_{k'}$ and $e_{k'} = A^{i}_{k'}e_{i}$ Note that the coefficient matrix ($A^{k'}_{i}$) is the inverse of the coeffisientmatrix ($A^{i}_{k'}$) found in the transformation of the vector coordinate components. question is, how can I use this information to show the invariance of the trace ?

Answer 1

If the trace of quadratic form is the trace of a representative matrix, then the result you want to prove is false, unless you restrict to orthogonal transformations (aka isometries of your euclidean space).

In this case, if $e$ and $f$ are two orthonormal bases , then $Mat(Q,f)=P^t Mat(Q, e)P$, where $P$ is the appropriate base change matrix. Now since $e$ and $f$ are both orthonormal, $P$ is orthogonal, that is $P^t= P^{-1}$. Now the trace is invariant by conjugation and we are done.