# Why does linearly independent spanning set imply minimal spanning set for a vector space?

Suppose β is a linearly independent spanning set of some vector space V. Why must it be the minimal spanning set?

In other words, why can there not be two linearly independent spanning sets of a vector space V with different sizes?

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There are usually many minimal spanning sets. "Minimal" means you cannot remove any element without losing the spanning property. "Minimal" is different from "smallest"; smallest things are unique, minimal things usually not. – Daniel Fischer Jun 12 '14 at 13:01
OK, I was a bit confused on that point, thanks. But what about my second question? Why can there not be two linearly independent spanning sets of a vector space V with different sizes? – user156724 Jun 12 '14 at 13:06

## 3 Answers

Suppose $v_1,\ldots,v_n$ and $w_1,\ldots,w_m$ are bases for $V$ where $n<m$.

By writing the $w_i$ in terms of the $v_j$, you create a system $w_i=\sum_j \alpha_{ij}v_j$.

In terms of coordinates, this means that the vector $w_i$ has coordinates $[\alpha_{i1},\alpha_{i2},\ldots,\alpha_{in}]$ in the basis of $v_i$'s. In other words, you have $m$ length $n$ vectors that you believe are linearly independent (since the $w_i$ are linearly independent.) You can put these in a matrix, and keep in mind you believe the rows to be linearly independent.

By using elementary row operations, though, you can put this matrix into row-echelon form, where it must have all zeros in the last row. This says that the last row is a linear combination of the rows above it, and that means that the $w_i$ weren't linearly independent after all. This contradiction shows why the assumption in the first line can never occur.

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Sorry, can you clarify why in row-echelon form, it must have all zeroes in the last row? – user156724 Jun 12 '14 at 14:57
@user156724 Row echelon form makes everything below the diagonal $a_{ii}$ entries zero. If the matrix is taller than it is wide, then an entire row lies underneath the diagonal. – rschwieb Jun 12 '14 at 15:10

Suppose that $S$ is a proper subset of $\beta$ such that $\text{span}(S)=\text{span}(\beta)$. Choose $v\in \beta$ such that $v \not \in S$. So, there exist $a_i\in F$, $v_i\in \beta$, $1\leq i\leq n$ such that $\displaystyle v=\sum_{i=1}^n a_iv_i$ i.e. $$0=\sum_{i=1}^n a_iv_i+(-1)v$$ which implies that $\beta$ is linearly dependent, a contradiction!So, $\beta$ is a minimal spanning set.

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Just a quick explanation from dimension theorem of vector space.

http://en.wikipedia.org/wiki/Dimension_theorem_for_vector_spaces

Suppose you have $v_1$ and $v_2$; they are linearly independent and span $V$. Then they form the basis for $V$. As such the dimension of $V$ is $2$ and according to the dimension theorem, all set of linearly independent vectors that span $V$must be of size $2$ as well.

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This is not very useful since the question is essentially "why is the dimension theorem true?" We can't invoke the dimension theorem to answer the reader's question. – rschwieb Jun 12 '14 at 13:45
Sorry I'm student as well. I thought refering to the standard theorem might be useful. – vTx Jun 12 '14 at 13:53
In many cases it is useful, but in this case it is circular reasoning. The dimension theorem can't be used to justify itself ;) – rschwieb Jun 12 '14 at 13:58