807: For 'Vectors Spaces - Linear Morphisms' Isomorphism, Image of Linearly Independent Subset or Basis of Domain Is Linearly Independent or Basis on Codomain

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

description/proof of that for 'vectors spaces - linear morphisms' isomorphism, image of linearly independent subset or basis of domain is linearly independent or basis on codomain

---

Topics

About: vectors space

---

The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Natural Language Description
• 3: Note
• 4: Proof

---

Starting Context

• The reader knows a definition of %field name% vectors space.
• The reader knows a definition of %category name% isomorphism.
• The reader knows a definition of linearly independent subset of module.
• The reader knows a definition of basis of module.

---

Target Context

• The reader will have a description and a proof of the proposition that for any 'vectors spaces - linear morphisms' isomorphism, the image of any linearly independent subset or any basis of the domain is linearly independent or a basis on the codomain.

---

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:
$F$: $\in \{\text{ the fields }\}$
$V_1$: $\in \{\text{ the }F\text{ vectors spaces }\}$
$V_2$: $\in \{\text{ the }F\text{ vectors spaces }\}$
$f$: $:V_1\to V_2$, $\in \{\text{ the 'vectors spaces - linear morphisms' isomorphisms }\}$
$S$: $\subseteq V_1$, $\in \{\text{ the linearly independent subsets of }{V}_1\}$
$B$: $\subseteq V_1$, $\in \{\text{ the bases of }{V}_1\}$
//

Statements:
$f(S)\in \{\text{ the linearly independent subsets of }{V}_2\}$
$\wedge$
$f(B)\in \{\text{ the bases of }{V}_2\}$
//

---

2: Natural Language Description

For any field, $F$, any $F$ vectors spaces, $V_1,V_2$, any 'vectors spaces - linear morphisms' isomorphism, $f$, any linearly independent subset, $S\subseteq V_1$, and any basis, $B\subseteq V_1$, $f(S)$ is a linearly independent subset of $V_2$ and $f(B)$ is a basis of $V_2$.

---

3: Note

The vectors spaces do not need to be finite-dimensional; $S$ does not need to be finite.

This proposition is usually regarded to be obvious with $V_2$ treated as "the same" with $V_1$, but let us be at ease with conscience that we have indeed proved it once and for all.

---

4: Proof

Whole Strategy: Step 1: take any finite subset, $S^{\prime }\subseteq f(S)$, suppose that $\sum _{p_j\in S^{\prime }}{r}^j{p}_j=0,r^j\in F$, and see that $r^j=0$, by taking $f^{-1}(\sum _{p_j\in S^{\prime }}{r}^j{p}_j)=f^{-1}(0)$; Step 2: take any point, $p\in V_2$, and see that $p$ is a linear combination of a finite subset of $f(B)$, by taking $f^{-1}(p)$ and taking a finite subset, $S^{\prime }\subseteq B$, and some $r^j$ s such that $f^{-1}(p)=\sum _{p_j\in S^{\prime }}{r}^j{p}_j$.

Step 1:

Let $S^{\prime }\subseteq f(S)$ be any finite subset.

Let $\sum _{p_j\in S^{\prime }}{r}^j{p}_j=0$ for some $r^j\in F$ s.

What we need to see is that for each $j$, $r^j=0$.

$f^{-1}(\sum _{p_j\in S^{\prime }}{r}^j{p}_j)=f^{-1}(0)$. As $f^{-1}$ is linear, $f^{-1}(\sum _{p_j\in S^{\prime }}{r}^j{p}_j)=\sum _{p_j\in S^{\prime }}{r}^j{f}^{-1}(p_j)=f^{-1}(0)=0$. As $\{f^{-1}(p_j)\}$ is a finite subset of $S$, $r^j=0$, because $S$ is linearly independent.

Step 2:

By Step 1, we already know that $f(B)$ is linearly independent on $V_2$.

What we need to see is that for each point, $p\in V_2$, $p$ is a linear combination of a finite subset of $f(B)$.

Let us take $f^{-1}(p)\in V_1$. As $B$ is a basis, there is a finite subset, $S^{\prime }\subseteq B$, and some $r^j$ s, such that $f^{-1}(p)=\sum _{p_j\in S^{\prime }}{r}^j{p}_j$. Then, $p=f(f^{-1}(p))=f(\sum _{p_j\in S^{\prime }}{r}^j{p}_j)=\sum _{p_j\in S^{\prime }}{r}^{j}f(p_j)$. As $\{f(p_j)\}$ is a finite subset of $f(B)$, that is what we needed to see.

---

References

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