# Proof

(related to Theorem: Finite Basis Theorem)

#### Proof for $$(1)$$: $$V$$ has a finitebasis.

Since $$V$$ is a vector space over a field $$F$$ with a finite dimension $$dim(V)=n < \infty$$, the number $$n$$ is by definition of the dimension the maximum number of linearly independent vectors contained in $$V$$. Let $$B:=\{v_1,\ldots,v_n\}\subseteq V$$ be any such linearly independent vectors. We note that, for any subset $$A\subseteq B$$, the vectors in $$A$$ have also to be independent. In order to show that $$B$$ is a basis, it remains to show that $$B$$ is a generating system of $$V$$.

If $$V=\{0\}$$, then $$B=\{0\}$$, and trivially the vector $$v_1=0\in V$$ can be represented as a linear combination of the basis1 $$v_1=0=\alpha v_1$$ for all $$\alpha\in F$$, so $$V=Span(B)$$.

If $$V\neq\{0\}$$, $$B:=\{v_1,\ldots,v_n\}$$ and $$b\in V$$ with $$b\neq 0$$, then, since $$n$$ is maximal by assumption, the set of vectors $$B\cup \{b\}$$ must be linearly dependent, i.e. the equation $\begin{array}{ccl} 0&=&\alpha_1v_1+\ldots+\alpha_nv_n+\beta b\\ &\Updownarrow\\ -\beta b&=&\alpha_1v_1+\ldots+\alpha_nv_n~~~~~~~~~~~~~~~~( * ) \end{array}$ can be solved either for $$\beta\neq 0$$ or $$\alpha_i\neq 0$$ for at least one $$i=1,\ldots,n$$. Note that $$\beta\neq 0$$, for if $$\beta=0$$, we would have the case $0=\alpha_1v_1+\ldots+\alpha_nv_n$ which can only be solved trivially by $$\alpha_1=\ldots=\alpha_n=0\in F$$ because of the linear independence of the $$v_i$$. Therefore, we can divide both sides of the equation $$( * )$$ by $$-\beta$$, resulting in $b=\frac{\alpha_1}{- \beta} v_1+\ldots+\frac{\alpha_n}{- \beta} v_n.$ Since $$\alpha_i,\beta\in F$$ were all arbitrary, we have shown that any non-zero vector $$b\in V$$ can be represented as a linear combination of the $$v_i\in B$$. Thus, we have again $$V=Span(B)$$.

#### Proof for $$(2)$$: Two different bases of $$V$$ have the same number of elements, and this number equals $$n=dim(V)$$.

In $$(1)$$ we have shown already the existence of a basis $$B=\{v_1,\ldots,v_n\}$$ with $$n$$ elements, $$n=dim(V)$$. It remains to show that no basis of $$V$$ can have a different number of elements. Suppose the existence of such a basis, say $$A:=\{w_1,\ldots,w_m\}$$ with $$m\neq n$$.

$$m > n$$ would mean a contradiction to the property of $$n$$ as the maximum number of linearly independent vectors in $$V$$ and the definition of $$A$$ as a basis. $$m < n$$ would mean that we can generate all vectors of $$V$$ using a smaller number of independent vectors then its dimension $$n$$. Since there are (!) some $$n$$ linearly independent vectors $$v_1,\ldots, v_n$$, after all, we shall be able to generate them with a smaller number the independent vectors $$w_1,\ldots, w_m$$. We will show that this is impossible.

Suppose in contrary that it is possible. In other words, we have

$\begin{array}{ccc} \alpha_{11}w_1+\ldots + \alpha_{1m}w_m&=&v_1\\ \alpha_{21}w_1+\ldots + \alpha_{2m}w_m&=&v_2\\ \vdots&\vdots&\vdots\\ \alpha_{n1}w_1+\ldots + \alpha_{nm}w_m&=&v_n\\ \end{array}$

In matrix form we can write $\pmatrix{ \uparrow & \uparrow & \ldots & \uparrow \cr w_1 & w_2 & \ldots & w_m \cr \downarrow & \downarrow & \ldots & \downarrow \cr }\cdot\pmatrix{ \alpha_{11} & \alpha_{12} & \ldots & \alpha_{1n} \cr \alpha_{21} & \alpha_{22} & \ldots & \alpha_{2n} \cr \vdots & \vdots & \ddots & \vdots \cr \alpha_{m1} & \alpha_{m2} & \ldots & \alpha_{mn} \cr }=\pmatrix{ \uparrow & \uparrow & \ldots & \uparrow \cr v_1 & v_2 & \ldots & v_n \cr \downarrow & \downarrow & \ldots & \downarrow \cr }$ Since $$m < n$$, the matrix with the coefficients $$\alpha_{ij}$$ has more columns then rows. Following the fundamental lemma, the homogeneous equation system

$\pmatrix{ \alpha_{11} & \alpha_{12} & \ldots & \alpha_{1n} \cr \alpha_{21} & \alpha_{22} & \ldots & \alpha_{2n} \cr \vdots & \vdots & \ddots & \vdots \cr \alpha_{m1} & \alpha_{m2} & \ldots & \alpha_{mn} \cr }\cdot\pmatrix{ X_{1} \cr X_{2} \cr \vdots \cr X_{n} \cr }=\pmatrix{ 0 \cr 0 \cr \vdots \cr 0 \cr }$ has more unknowns as equations, and as such, a nontrivial solution $$X_1 = \beta_1 , X_2=\beta_2,\ldots X_n = \beta_n$$: $\pmatrix{ \alpha_{11} & \alpha_{12} & \ldots & \alpha_{1n} \cr \alpha_{21} & \alpha_{22} & \ldots & \alpha_{2n} \cr \vdots & \vdots & \ddots & \vdots \cr \alpha_{m1} & \alpha_{m2} & \ldots & \alpha_{mn} \cr }\cdot\pmatrix{ \beta_{1} \cr \beta_{2} \cr \vdots \cr \beta_{n} \cr }=\pmatrix{ 0 \cr 0 \cr \vdots \cr 0 \cr }$ In other words $\pmatrix{ \uparrow & \uparrow & \ldots & \uparrow \cr w_1 & w_2 & \ldots & w_m \cr \downarrow & \downarrow & \ldots & \downarrow \cr }\cdot\pmatrix{ \alpha_{11} & \alpha_{12} & \ldots & \alpha_{1n} \cr \alpha_{21} & \alpha_{22} & \ldots & \alpha_{2n} \cr \vdots & \vdots & \ddots & \vdots \cr \alpha_{m1} & \alpha_{m2} & \ldots & \alpha_{mn} \cr }\cdot\pmatrix{ \beta_{1} \cr \beta_{2} \cr \vdots \cr \beta_{n} \cr }=\pmatrix{ \uparrow & \uparrow & \ldots & \uparrow \cr v_1 & v_2 & \ldots & v_n \cr \downarrow & \downarrow & \ldots & \downarrow \cr }\cdot\pmatrix{ \beta_{1} \cr \beta_{2} \cr \vdots \cr \beta_{n} \cr }$

$0 w_1 + \ldots + 0w_m = 0 = \beta_1v_1+\ldots+\beta_nv_n$ would be a non-trivial representation of $$0$$, which again is a contradiction to $$v_i$$ being independent.

Therefore $$m=n$$.

#### Proof for $$(3)$$: Linearly independent vectors $$v_1,\ldots,v_r$$ with $$r < n$$ can be completed by vectors $$v_{r+1},\ldots,v_{n}$$ to create a basis $$v_1,\ldots,v_n$$ of $$V$$.

Let $$A_r:=\{v_1,\ldots,v_r\}$$. Because $$A_r$$ is a set of linearly independent vectors and because $$r < n$$, the vector space $$Span(A_r)$$ a proper subspace of $$V$$, i.e. there is a vector $$v_{r+1}\in V$$ with $$v_{r+1}\not\in Span(A_r)$$. Therefore, $$v_{r+1}$$ is linearly independent from $$A_r$$. We can repeat the argument for $$A_{r+1}=A_r\cup\{v_{r+1}\}$$. This will end, once we have reached the dimension of $$V$$, which by definition is the maximum number of independent vectors in $$V$$.

Github: ### References

#### Bibliography

1. Koecher Max: "Lineare Algebra und analytische Geometrie", Springer-Verlag Berlin Heidelberg New York, 1992, 3rd Volume
2. Wille, D; Holz, M: "Repetitorium der Linearen Algebra", Binomi Verlag, 1994

#### Footnotes

1. $$v_1$$ is linearly independent, since $$0=\alpha\cdot v_1$$ can be solved only trivially by $$\alpha=0$$, since $$v_1\neq 0$$.