# If $\dim V = n$, why does any set of $n$ linearly independent vectors span $V$?

I am currently learning about vector spaces, spaning and dimensions for the first time and am reading old posts on here. I came across this one: If $\dim(V) = n$, is every spaning set $\{v_1,v_2,\ldots,v_n\}$ a basis for $V$?

In it, it is mentioned that "If $$\dim V = n$$, then any set of $$n$$ linearly independent vectors will span $$V$$". I understand that any spanning set of $$n$$ vectors must consist of linearly independent vectors due to $$\dim V = n$$ but I don't quite get the other direction.

Could someone help me understand this better?

• How many subspaces of maximal (finite) dimension can exist? Commented Jan 30, 2021 at 20:55
• Can you show (i) every basis of a finite-dimensional vector space has the same size and (ii) $n+1$ vectors in an $n$-dimensional space are linearly dependent?
– KCd
Commented Jan 30, 2021 at 20:58
• Note that a spanning set can be redundant (have more vectors than necessary). A basis is a minimal spanning set. A linearly independent set can be insufficient (cover a subspace rather than the whole space). A basis is also a maximal linearly independent set. We can rightly call this set with the dual properties of being "minimal spanning" and "maximal linearly independent" a basis, because we can prove that (in a vector space) the two properties are equivalent. You perhaps need to review what you already know and test these two aspects. Commented Jan 30, 2021 at 21:03

Every linear independent family can be extended to a basis. By definition of $$\dim$$, this basis still has $$n$$ members, i.e., we did not really extend our family and already had a basis.

The dimension is defined as the cardinality of a maximal linearly independent set, or the minimal generating set. These two are shown to be the same in any textbook. We say basis to refer to these sets; being both maximal linearly independent and minimal generating sets.

So a set of $$\dim V$$ linearly independent vectors is a basis basically by definition

• The last assertion is true only in finite dimension. Commented Jan 30, 2021 at 23:03

Let $$\{v_1,\ldots,v_r\}\subset V$$ be a set of $$r$$ linearly independent vectors, and let $$\{w_1,\ldots,w_n\}$$ be a basis for $$V$$. Then there exist $$\alpha_{ij}$$ with $$i\in\{1,\ldots,r\}$$ and $$j\in\{1,\ldots,n\}$$ such that $$v_i=\sum_{j=1}^n \alpha_{ij}w_j$$. Consider the system of equations: $$\sum_{i=1}^r \alpha_{ij} x_i=0, \ j\in\{1,\ldots,n\}$$ Let $$(x_1,\ldots,x_n)$$ be a solution. Then $$\sum_{i=1}^r x_i v_i=\sum_{i=1}^r\sum_{j=1}^n\alpha_{ij}x_i w_j=\sum_{j=1}^n\left(\sum_{i=1}^r\alpha_{ij}x_i\right)w_j=\sum_{j=1}^n 0\cdot w_j=0$$ $$\{v_1,\ldots,v_r\}$$ is linearly independent, so it must be $$(x_1,\ldots,x_n)=0$$. Hence, the previous system of equations either has unique solution or hasn't solution at all. Anyways, this means that it must have at least as many equations as variables, i.e. $$r\leq n$$.

Then, any linearly independent set must have, at most, $$n$$ vectors. Let now $$\{v_1,\ldots,v_n\}$$ be a linearly independent set and let's prove it must span $$V$$. Assume it doesn't, and take $$v\in V\setminus\operatorname{Span}(v_1,\ldots,v_n)$$. Then $$\{v,v_1,\ldots,v_n\}$$ is a linearly independent set with $$n+1$$ elements; a contradiction. Hence $$\{v_1,\ldots,v_n\}$$ spans $$V$$ so it's a basis.

• i don't get the last part "so it must be $(x_1,...,x_n) = 0$". why does each $x_i$ have to equal $0$? I know that $\{v_1,...,v_r\}$ are linearly independent, I just can't seem to connect the two statements together. Commented Oct 6, 2021 at 21:15
• is each $x_i$ a coefficient? or a vector? Commented Oct 7, 2021 at 15:25

Assume that V is a vector space and that you have n linearly independent vectors from it. The dimension of the sub space spanned by the n vectors is n. However since dim V =n, this must be V itself.