729: For Linear Surjection from Finite-Dimensional Vectors Space, if Dimension of Codomain Is Equal to or Larger than That of Domain, Surjection Is Bijection

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

description/proof of that for linear surjection from finite-dimensional vectors space, if dimension of codomain is equal to or larger than that of domain, surjection is bijection

---

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 linear map.
• The reader knows a definition of dimension of vectors space.
• The reader knows a definition of bijection.
• The reader admits the proposition that for any vectors space, any finite generator can be reduced to be a basis.

---

Target Context

• The reader will have a description and a proof of the proposition that for any linear surjection from any finite-dimensional vectors space, if the dimension of the codomain is equal to or larger than that of the domain, the surjection is a bijection.

---

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:
$F$: $\in \{\text{ the fields }\}$
$V_1$: $\in \{\text{ the }{d}_1\text{ -dimensional }F\text{ vectors spaces }\}$
$V_2$: $\in \{\text{ the }F\text{ vectors spaces }\}$
$f$: $:V_1\to V_2$, $\in \{\text{ the linear surjections }\}$
//

Statements:
$dim{V}_1\le dim{V}_2$
$⟹$
$f\in \{\text{ the bijections }\}$
//

---

2: Natural Language Description

For any field, $F$, any $d_1$-dimensional $F$ vectors space, $V_1$, any $F$ vectors space, $V_2$, and any linear surjection, $f:V_1\to V_2$, if $dim{V}_1\le dim{V}_2$, $f$ is a bijection.

---

3: Note

$V_2$ is not presupposed to be finite-dimensional, although $V_2$ will be concluded to be finite-dimensional, in fact $d_1$-dimensional.

---

4: Proof

Whole Strategy: Step 1: choose any basis for $V_1$; Step 2: express the range of $f$ with the image of the basis and see that $V_2$ is spanned by the image of the basis; Step 3: suppose that there were some distinct elements of $V_1$ that were mapped to the same element, and find a contradiction.

Step 1:

Let us choose any basis for $V_1$, $\{e_1,...,e_{d_1}\}$.

Step 2:

For each $v\in V_1$, $v=\sum _{j\in \{1,...,d_1\}}{v}^j{e}_j$, where $v^j\in F$, $f(v)=\sum _{j\in \{1,...,d_1\}}{v}^{j}f(e_j)$, and $f(V_1)=V_2=\{\sum _{j\in \{1,...,d_1\}}{v}^{j}f(e_j)|v^j\in F\}$, which means that $V_2$ is spanned by $\{f(e_1),...,f(e_{d_1})\}$.

Step 3:

Let us suppose that there were some $v_1,v_2\in V_1$ such that $v_1\ne v_2$ and $f(v_1)=f(v_2)$.

$v_1=\sum _{j\in \{1,...,d_1\}}{v_1}^j{e}_j$ and $v_2=\sum _{j\in \{1,...,d_1\}}{v_2}^j{e}_j$. $f(v_1)=\sum _{j\in \{1,...,d_1\}}{v_1}^{j}f(e_j)=\sum _{j\in \{1,...,d_1\}}{v_2}^{j}f(e_j)=f(v_2)$. As $v_1\ne v_2$, ${v_1}^k\ne {v_2}^k$ for a $k$. Then, $\sum _{j\in \{1,...,d_1\}}({v_1}^j-{v_2}^j)f(e_j)=0$, but as $v_1-v_2\ne 0$, $\{f(e_1),...,f(e_n)\}$ would be linearly dependent.

So, $\{f(e_1),...,f(e_n)\}$ would not be any basis of $V_2$, but $\{f(e_1),...,f(e_n)\}$ could be reduced to be a basis of $V_2$, by the proposition that for any vectors space, any finite generator can be reduced to be a basis, which would mean that $dim{V}_2<dim{V}_1$, a contradiction against $dim{V}_1\le dim{V}_2$.

So, $f$ is injective, and $f$ is bijective.

In fact, $\{f(e_1),...,f(e_n)\}$ is linearly independent and is a basis, because otherwise, $dim{V}_2<dim{V}_1$, so, $dim{V}_2=dim{V}_1$.

---

References

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