definition of trace of vectors space endomorphism
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of %field name% vectors space.
- The reader knows a definition of %structure kind name% endomorphism.
- The reader knows a definition of trace of square matrix over ring.
Target Context
- The reader will have a definition of trace of vectors space endomorphism.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\( F\): \(\in \{\text{ the fields }\}\)
\( V\): \(\in \{\text{ the finite-dimensional } F \text{ vectors spaces }\}\)
\( f\): \(: V \to V\), \(\in \{\text{ the } F \text{ vectors space endomorphisms }\}\)
\( B\): \(\in \{\text{ the bases of } V\}\)
\( M_{f, B}\): \(= \text{ the matrix for } f \text{ with respect to } B\)
\(*tr (f)\): \(= tr (M_{f, B})\)
//
Conditions:
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2: Note
The point is that \(tr (f)\) does not depend on the choice of \(B\), which is the reason why this definition is well-defined: for any vector, \(v \in V\), letting the components representation with respect to \(B\) of \(v\) be \(\widetilde{v}\), the components representation with respect to another basis, \(B'\), of \(v\), is \(U \widetilde{v}\) where \(U\) is an invertible matrix; the components representation with respect to \(B'\) of \(f (v)\) is \(U M_{f, B} \widetilde{v} = U M_{f, B} U^{-1} U \widetilde{v}\), which means that the matrix for \(f\) with respect to \(B'\) is \(M_{f, B'} := U M_{f, B} U^{-1}\); then, \(tr (M_{f, B'}) = tr (U M_{f, B} U^{-1}) = tr (U^{-1} U M_{f, B})\), by the proposition that for any 2 same-dimensional square matrices over any commutative ring, the trace of each product of the matrices does not depend on the order of the product, \(= tr (M_{f, B})\).
So, talking about the trace of any vectors space endomorphism makes sense, without specifying any basis.
This definition makes sense because \(f\) is an endomorphism, instead of just a linear map between 2 different vectors spaces: for the between-2-different-vectors-spaces case, the basis of the codomain is inevitably different from the basis of the domain (in fact, there is no canonical sameness between 2 elements in 2 different vectors spaces), and by changing the basis of the codomain, the trace of the corresponding matrix can change: for example, make all the elements of the codomain basis half-lengthed, then, the matrix will be the double, and the trace will be the double.
