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
Topics
About: vectors space
The table of contents of this article
Starting Context
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.
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:
\(F\): \(\in \{\text{ the fields }\}\)
\(V\): \(\in \{\text{ the } F \text{ vectors spaces }\}\)
\(B\): \(\in \{\text{ the bases for } V\} = \{b_s \vert 1 \le s \le dim V\}\)
\(B'\): \(\in \{\text{ the bases for } V\} = \{b'_s = b_j M_s^j \vert 1 \le s \le dim V\}\)
\(f\): \(: V \to V\)
\(N\): \(= \text{ the matrix of } f \text{ with respect to } B\)
\(N'\): \(= \text{ the matrix of } f \text{ with respect to } B'\)
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Statements:
\(N = M N' M^{-1}\)
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2: Note
A motivation for this proposition is to get \(N\) from a simple \(N'\).
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 \(B'\) with \((n^1, n^2, n^3)\) as \(b'_3\), then, \(N'\) is simple because it is the rotation around the \(b'_3\) axis, and we can get \(N\) from \(N'\).
3: Proof
Whole Strategy: Step 1: for each \(v \in V\), let the components column vectors with respect to \(B\) and \(B'\) be \(\overline{v}\) and \(\overline{v}'\); Step 2: see that \(\overline{f (v)}' = N' \overline{v}'\); Step 3: see that \(\overline{v}' = M^{-1} \overline{v}\) and \(\overline{f (v)}' = M^{-1} \overline{f (v)}\); Step 4: see that \(M^{-1} \overline{f (v)} = N' M^{-1} \overline{v}\), and conclude the proposition.
Step 1:
For each \(v \in V\), let the components column vectors with respect to \(B\) and \(B'\) be \(\overline{v}\) and \(\overline{v}'\).
Step 2:
\(\overline{f (v)}' = N' \overline{v}'\).
Step 3:
\(\overline{v}' = M^{-1} \overline{v}\) and \(\overline{f (v)}' = M^{-1} \overline{f (v)}\), 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, \(M^{-1} \overline{f (v)} = N' M^{-1} \overline{v}\).
So, \(\overline{f (v)} = M M^{-1} \overline{f (v)} = M N' M^{-1} \overline{v}\), which means that \(N = M N' M^{-1}\).
