1025: For Finite-Dimensional Vectors Space, Transition of Components of Vector w.r.t. Change of Bases Is This

<The previous article in this series | The table of contents of this series | The next article in this series>

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

---

Topics

About: vectors space

---

The table of contents of this article

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

---

Starting Context

• The reader knows a definition of %field name% vectors space.

---

Target Context

• The reader will have a description and a proof of 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.

---

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:
$F$: $\in \{\text{ the fields }\}$
$V$: $\in \{\text{ the }F\text{ vectors spaces }\}$
$B$: $\in \{\text{ the bases for }V\}=\{b_s|1\le s\le dimV\}$
$B^{\prime }$: $\in \{\text{ the bases for }V\}=\{b_s^{\prime }=b_j{M}_s^j|1\le s\le dimV\}$
//

Statements:
$\mathrm{\forall }v=v^j{b}_j=v^{\prime j}{b}_j^{\prime }\in V$
(
$v^{\prime j}={M^{-1}}_l^j{v}^l$
)
//

---

2: Proof

Whole Strategy: Step 1: for $b_j^{\prime }{v}^{\prime j}=b_j{v}^j$, expand $b_j^{\prime }$ with $b_l$ s, and compare the coefficients of $b_l$ s on the both hand sides.

Step 1:

From $b_j^{\prime }{v}^{\prime j}=b_j{v}^j$, $b_j^{\prime }{v}^{\prime j}=b_l^{\prime }{v}^{\prime l}=b_j{M}_l^j{v}^{\prime l}=b_j{v}^j$.

So, $M_l^j{v}^{\prime l}=v^j$.

So, $v^{\prime j}={M^{-1}}_l^j{v}^l$.

---

References

<The previous article in this series | The table of contents of this series | The next article in this series>