886: Trace of Vectors Space Endomorphism

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definition of trace of vectors space endomorphism

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Topics

About: vectors space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Note

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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.

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Target Context

• The reader will have a definition of trace of vectors space endomorphism.

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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.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$F$: $\in \{\text{ the fields }\}$
$V$: $\in \{\text{ the finite-dimensional }F\text{ vectors spaces }\}$
$f$: $:V\to V$, $\in \{\text{ the }F\text{ vectors space endomorphisms }\}$
$B$: $\in \{\text{ the bases of }V\}$
$M_{f,B}$: $=\text{ the matrix for }f\text{ with respect to }B$
$\ast tr(f)$: $=tr(M_{f,B})$
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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 $\tilde{v}$, the components representation with respect to another basis, $B^{\prime }$, of $v$, is $U\tilde{v}$ where $U$ is an invertible matrix; the components representation with respect to $B^{\prime }$ of $f(v)$ is $U{M}_{f,B}\tilde{v}=U{M}_{f,B}{U}^{-1}U\tilde{v}$, which means that the matrix for $f$ with respect to $B^{\prime }$ is $M_{f,B^{\prime }}:=U{M}_{f,B}{U}^{-1}$; then, $tr(M_{f,B^{\prime }})=tr(U{M}_{f,B}{U}^{-1})=tr(U^{-1}U{M}_{f,B})$, 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.

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References

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