# Any two bases of a finite dimensional vector space must have the same number of elements.

Prove that any two bases of a finite dimensional vector space must have the same number of elements.

By considering the following two bases

$$S_1 = \{ \alpha_1, \alpha_2 , \ldots, \alpha_n \},$$

$$S_2 = \{ \beta_1, \beta_2, \ldots, \beta_m \},$$

how do I show that $$m = n$$?

Hints to get started. Thanks very much.

Quote from Wikipedia quoted from Linear Algebra Done Right of a bit more general result:

Theorem: If A = $$(a_1,\dots,a_n) \subseteq V$$ is a linearly independent tuple in a vector space $$V$$, and $$B_0 = (b_1,...,b_r)$$ is a tuple that spans $$V$$, then $$n\le r$$.

Proof: Since $$B_0$$ spans $$V$$, the tuple $$(a_1,b_1,\dots,b_r)$$ also spans $$V$$. Since $$a_1\ne 0$$ (because A is linearly independent), there is at least one $$t\in\{1,\ldots,r\}$$ such that $$b_t$$ can be written as a linear combination of $$B_1 = (a_1,b_1,\dots,b_{t-1}, b_{t+1}, ... b_r)$$. Thus, $$B_1$$ is a spanning tuple, and its length is the same as $$B_0$$'s.

Repeat this process. Because $$A$$ is linearly independent, we can always remove an element from the list $$B_i$$ which is not one of the $$a_j$$'s that we prepended to the list in a prior step (because $$A$$ is linearly independent, and so there must be some nonzero coefficient in front of one of the $$b_i$$'s). Thus, after $$n$$ iterations, the result will be a tuple $$B_n = (a_1, \ldots, a_n, b_{m_1}, \ldots, b_{m_k})$$ (possibly with $$k=0$$) of length $$r$$. In particular, $$A\subseteq B_n$$, so $$|A|\le |B_n|$$, i.e., $$n\leq r$$.

Assume that $$(\alpha_1,\cdots, \alpha_n)\supseteq (\beta_1,\cdots, \beta_m) \Rightarrow n \geq m$$

where $\{ \alpha_i\}$, $\{\beta_i\}$ are linearly independent.

Proof : Assume that $(\alpha_1) \supseteq (\beta_1,\cdots, \beta_m),\ m>1$ So $\beta_1=c_1\alpha_1,\ \beta_2=c_2\alpha_1$ so that $\{ \beta_i\}_{i=1}^2$ is dependent. Contradiction.

We will use induction : $n=k$ is fine. Then let $$(\alpha_1,\cdots, \alpha_{k+1})\supseteq (\beta_1,\cdots, \beta_m),\ m>k+1$$

Clearly we have $c_{ij}$ : $$\beta_j:= \sum_i c_{ji}\alpha_i$$ Note that for some $j$, $c_{j1}\neq 0$. If not $$(\alpha_2,\cdots, \alpha_{k+1})\supseteq (\beta_1,\cdots, \beta_m)$$ So we have $k\geq m$.

Let $c_{11}\neq 0$. Then $$(\beta_2- \frac{c_{21}}{c_{11}} \beta_1,\cdots, \beta_m-\frac{c_{m1}}{c_{11}} \beta_1 )\subset (\alpha_2,\cdots , \alpha_{k+1})$$

Note that $\{ \beta_2- \frac{c_{21}}{c_{11}} \beta_1,\cdots, \beta_m-\frac{c_{m1}}{c_{11}} \beta_1 \}$ is linearly independent. So $k\geq m-1$. Thus we have a contradiction.

Assuming you're proving solely by the following theorem: "If basis of B of V = {b1, ..., bn}, then any set in V containing more than n vectors must be linearly dependent" (1)

Note the inner contrapositive of (1), that is, if B contains n vectors, then, any set in V is linearly independent if it contains not more than n vectors (2). Since S1 is a basis with n vectors and S2 is linearly independent, then S2 contains not more than n vectors.

Note the contrapositive of (2), that is, if it is not the case that if any set in V is linearly independent then it contains not more than n vectors, then B does not contain n vectors. Now, we substitute n with 'not more than n', equivalently as two cases, 'less than n' or 'n':

'less than n' case: We see that it is not the case that S1 is linearly independent and also contains not more than 'less than n' vectors. (might be more helpful to sub 'less than n' with n-1). So, S2 does not contain less than n vectors, equivalently, S2 has at least n vectors.