682: For Finite-Dimensional Vectors Space and Basis, Linearly Independent Set of Elements Can Be Augmented with Some Elements of Basis to Be Basis

<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 and basis, linearly independent set of elements can be augmented with some elements of basis to be basis

---

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 basis of module.
• The reader admits the proposition that for any finite-dimensional vectors space, any linearly independent subset with dimension number of elements is a basis.

---

Target Context

• The reader will have a description and a proof of the proposition that for any finite-dimensional vectors space and its any basis, any linearly independent set of elements can be augmented with some elements of the basis to become a basis.

---

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:
$V$: $\in \{\text{ the }d\text{ -dimensional vectors spaces }\}$
$B$: $=\{b_1,...,b_d\}$, $\in \{\text{ the bases of }V\}$
$S$: $=\{v_1,...,v_k\}$, $\in \{\text{ the linearly independent subsets of }V\}$, where $0\le k\le d$
//

Statements:
$\mathrm{\exists }\{b_{j_1},...,b_{j_{d-k}}\}(\{v_1,...,v_k,b_{j_1},...,b_{j_{d-k}}\}\in \{\text{ the bases of }V\})$
//

---

2: Natural Language Description

For any finite $d$-dimensional vectors space, $V$, any basis of $V$, $B:=\{b_1,...,b_d\}$, and any linearly independent subset of $k$ elements, $S:=\{v_1,...,v_k\}\subseteq V$, where $0\le k\le d$, there are some elements of the basis, $b_{j_1},...,b_{j_{d-k}}$, such that $\{v_1,...,v_k,b_{j_1},...,b_{j_{d-k}}\}$ is a basis of $V$.

---

3: Note

This proposition is not about just augmenting the linearly independent set of $k$ elements to be a basis, but about augmenting the linearly independent set of $k$ elements to become a basis with some elements of any fixed basis.

---

4: Proof

Whole Strategy: Step 1: if $k<d$, add into $S$ an element of $B$ to form a linearly independent set, called $S^{\prime }$; Step 2: if $k+1<d$, add into $S^{\prime }$ an element of $B$ to form a linearly independent set, called again $S^{\prime }$; Step 3: and so on until $k+l=d$; Step 4: conclude that $S^{\prime }$ is a basis.

Let be $I:=\{1,...,d\}$ and $I^{\prime }:=\{1,...,k\}$.

Step 1:

If $k=d$, let us go to the last step.

If $k<d$, let us prove that there is an element of the basis, $b_{j_1}$, such that $S\cup \{b_{j_1}\}$ is linearly independent. Let us suppose otherwise. There would be a set of not-all-$0$ coefficients, $c_j^1,...,c_j^{k+1}$, such that $\sum _{l\in I^{\prime }}{c}_j^l{v}_l+c_j^{k+1}{b}_j=0$ for each $j\in I$. $c_j^{k+1}\ne 0$, because otherwise, $\sum _{l\in I^{\prime }}{c}_j^l{v}_l=0$, which would imply that all the coefficients were $0$, because $S$ was linearly independent, a contradiction. So, each $b_j$ would be a linear combination of $S$, which would imply that $S$ was a basis, a contradiction against that $V$ is $d$-dimensional. So, Let us choose such a $b_{j_1}$ and call $S\cup \{b_{j_1}\}$ "$S^{\prime }$".

Step 2:

If $k+1=d$, let us go to the last step.

If $k+1<d$, let us prove that there is an element of the basis, $b_{j_2}$, such that $S^{\prime }\cup \{b_{j_2}\}$ is linearly independent. Let us suppose otherwise. There would be a set of not-all-$0$ coefficients, $c_j^1,...,c_j^{k+2}$, such that $\sum _{l\in I^{\prime }}{c}_j^l{v}_l+c_j^{k+1}{b}_{j_1}+c_j^{k+2}{b}_j=0$ for each $j\in I$. $c_j^{k+2}\ne 0$, because otherwise, $\sum _{l\in I^{\prime }}{c}_j^l{v}_l+c_j^{k+1}{b}_{j_1}=0$, which would imply that all the coefficients were $0$, because $S^{\prime }$ was linearly independent, a contradiction. So, each $b_j$ would be a linear combination of $S^{\prime }$, which would imply that $(v_1,...,v_k,b_{j_1})$ was a basis, a contradiction against that $V$ is $d$-dimensional. Obviously, $b_{j_1}\ne b_{j_2}$, because if $b_{j_1}=b_{j_2}$, $S^{\prime }\cup b_{j_2}$ would not be linearly independent. So, Let us choose such a $b_{j_2}$ and call $S^{\prime }\cup \{b_{j_1}\}$ again "$S^{\prime }$".

Step 3:

And so on, which means that as far as $k+l<d$, there is a linearly independent $S^{\prime }\cup \{b_{j_{l+1}}\}$, called again $S^{\prime }$, and after all, it becomes that $k+l=d$.

Step 4:

Then, $S^{\prime }$ is a basis of $V$, by the proposition that for any finite-dimensional vectors space, any linearly independent subset with dimension number of elements is a basis.

---

References

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