How would you prove the trace of a transformation from V to V (where V is finite dimensional) is independent of the basis chosen?
$\begingroup$
$\endgroup$
2
-
6$\begingroup$ Give a basis-independent definition (for example as a coefficient of the characteristic polynomial). $\endgroup$– Qiaochu YuanOct 13, 2011 at 14:16
-
$\begingroup$ 1. If you have two different matrices representing the same linear operator, how are they related? 2. Use elementary properties of the trace. $\endgroup$– user2055Oct 13, 2011 at 14:18
Add a comment
|
4 Answers
$\begingroup$
$\endgroup$
The simplest way it to note that a basis transformation of a transformation $T$ is done via $ATA^{-1}$ where $A$ is an invertible matrix, and that the trace has the property $\operatorname{tr}(AB)=\operatorname{tr}(BA)$. Putting this together, you get $$\operatorname{tr}(ATA^{-1}) = \operatorname{tr}(A^{-1}AT) = \operatorname{tr}T$$
$\begingroup$
$\endgroup$
2
An elementary proof could be the following. First, let $A$ be the matrix of your linear transformation in any basis of $V$. The characteristic polynomial of $A$ is
$$ Q_A(t) = \mathrm{det}\ (A - tI) = \begin{vmatrix} a^1_1 - t & a^1_2 & \dots & a^1_n \\ a^2_1 & a^2_2 - t & \dots & a^2_n \\ \vdots & \vdots & \ddots & \vdots \\ a^n_1 & a^2_2 & \dots & a^n_n - t \end{vmatrix} $$
You can compute easily at least the first terms of this polynomial taking into account that, by the definition of the determinant:
$$ Q_A(t) = (a^1_1 -t)\cdot \dots \cdot (a^n_n - t) + \quad \text{sums of products with at most $n-2$ terms in the diagonal} \quad \ . $$ Hence $$ Q_A(t) = (-1)^n t^n +(-1)^{n-1} (a^1_1 + \dots + a^n_n) t^{n-1} + \quad ( \text{terms of degree}\ \leq n-2 ) \ . $$
That is, up to a sign, the trace of $A$ is the coefficient of $t^{n-1}$ in the characteristic polynomial of $A$, $Q_A(t)$.
Now you can prove that $Q_A(t)$ does not depend on the basis you have chosen. Indeed, if $B$ is the matrix of the same linear transformation in another basis, then $A$ and $B$ are related through an equality like $B= S^{-1}A S$, where $S$ is the change of basis matrix. So
$$ Q_B(t) = \mathrm{det}\ (S^{-1}A S - tI ) = \mathrm{det}\ (S^{-1}A S - S^{-1}tIS ) = \mathrm{det}\ (S^{-1}(A - tI)S ) $$
Thus
$$ Q_B(t) = \mathrm{det}\ (S^{-1}) \ \mathrm{det}\ (A - tI )\ \mathrm{det}\ (S) = \mathrm{det}\ (A - tI ) = Q_A(t) \ . $$
Hence, the characteristic polynomial is invariant through basis change. In particular, so are its coefficients. More particularly, the trace.
answered Oct 13, 2011 at 15:55
-
1
-
$\begingroup$ NB: This uses the very-often-used-and-very-seldom-proved fact that two real-valued polynomials are (numerically) equal for all values of t if and only if their individual coefficients are equal. (E.g., this is not true over all fields, such as t^2 = t over the field of size 2.) $\endgroup$– LabradorJul 28, 2022 at 4:44
$\begingroup$
$\endgroup$
There are a few ways to view this. The simplest is that if you have two $n \times n$ matrices $A, B$ then $$ \operatorname{tr}(AB) = \operatorname{tr}(BA). $$ This is an easy calculation.
Alternatively (some of the words may be unfamiliar to you, in which case ignore this bit), note that we have a canonical isomorphism $\operatorname{End}(V) \approx V \otimes V^\vee$, and that the trace on $\operatorname{End}(V)$ corresponds to the functional on $V \otimes V^\vee$ induced by the natural bilinear pairing of evaluation between $V$ and $V^\vee$. Verifying this is probably just as much work, but I think it's comforting to know that matrices are unnecessary for the definition.
answered Oct 13, 2011 at 14:25
$\begingroup$
$\endgroup$
Let $A,B$ be $n \times n$ matrices, it applies $tr(AB) = tr(BA)$.
proof $\: tr(AB) = \sum_{i} (AB)_{ii} = \sum_{i}(\sum_{k}A_{ik}B_{ki})= \sum_{k}\sum_{i} (B_{ki} A_{ik}) = \sum_{k} (BA)_{kk} = tr(BA)$ q.e.d.
Let $A$ be a $n \times n$ matrix. The change of basis is given by $U^{-1}AU$, where $U$ is an invertible $n \times n$ matrix.
We finally can see that: $\:$ $tr(U^{-1}AU) = tr(AUU^{-1}) = tr(A)$.
