1029: For Finite-Dimensional Vectors Space and Vectors Space Endomorphism, Transition of Endomorphism Matrix w.r.t. Change of Bases Is This

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description/proof of that for finite-dimensional vectors space and vectors space endomorphism, transition of endomorphism matrix w.r.t. change of bases is this

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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
• 3: Proof

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Starting Context

• The reader knows a definition of %field name% vectors space.
• The reader admits the proposition that for any finite-dimensional vectors space, the transition of the components of any vector with respect to any change of bases is this.

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Target Context

• The reader will have a description and a proof of the proposition that for any finite-dimensional vectors space and any vectors space endomorphism, the transition of the endomorphism matrix with respect to any change of bases is this.

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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:
$F$: $\in \{\text{ the fields }\}$
$V$: $\in \{\text{ the }F\text{ vectors spaces }\}$
$B$: $\in \{\text{ the bases for }V\}=\{b_s|1\le s\le dimV\}$
$B^{\prime }$: $\in \{\text{ the bases for }V\}=\{b_s^{\prime }=b_j{M}_s^j|1\le s\le dimV\}$
$f$: $:V\to V$
$N$: $=\text{ the matrix of }f\text{ with respect to }B$
$N^{\prime }$: $=\text{ the matrix of }f\text{ with respect to }{B}^{\prime }$
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Statements:
$N=M{N}^{\prime }{M}^{-1}$
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2: Note

A motivation for this proposition is to get $N$ from a simple $N^{\prime }$.

For example, for $V={\mathbb{R}}^3$ with the Euclidean inner product, in order to get $N$ for the $\theta$ rotation around the axis, $(n^1,n^2,n^3)$, which is with respect to $B$, we can get any orthonormal $B^{\prime }$ with $(n^1,n^2,n^3)$ as $b_3^{\prime }$, then, $N^{\prime }$ is simple because it is the rotation around the $b_3^{\prime }$ axis, and we can get $N$ from $N^{\prime }$.

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3: Proof

Whole Strategy: Step 1: for each $v\in V$, let the components column vectors with respect to $B$ and $B^{\prime }$ be $\bar{v}$ and ${\bar{v}}^{\prime }$; Step 2: see that ${\overline{f(v)}}^{\prime }=N^{\prime }{\bar{v}}^{\prime }$; Step 3: see that ${\bar{v}}^{\prime }=M^{-1}\bar{v}$ and ${\overline{f(v)}}^{\prime }=M^{-1}\overline{f(v)}$; Step 4: see that $M^{-1}\overline{f(v)}=N^{\prime }{M}^{-1}\bar{v}$, and conclude the proposition.

Step 1:

For each $v\in V$, let the components column vectors with respect to $B$ and $B^{\prime }$ be $\bar{v}$ and ${\bar{v}}^{\prime }$.

Step 2:

${\overline{f(v)}}^{\prime }=N^{\prime }{\bar{v}}^{\prime }$.

Step 3:

${\bar{v}}^{\prime }=M^{-1}\bar{v}$ and ${\overline{f(v)}}^{\prime }=M^{-1}\overline{f(v)}$, by the proposition that for any finite-dimensional vectors space, the transition of the components of any vector with respect to any change of bases is this.

Step 4:

So, $M^{-1}\overline{f(v)}=N^{\prime }{M}^{-1}\bar{v}$.

So, $\overline{f(v)}=M{M}^{-1}\overline{f(v)}=M{N}^{\prime }{M}^{-1}\bar{v}$, which means that $N=M{N}^{\prime }{M}^{-1}$.

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References

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