Does a set of $n$ independent vectors in $R^{n}$ always span $R^{n}$?
Does a set of $n$ independent vectors in $R^{n}$ always span $R^{n}$?
If so, what's the reasoning behind this?
Note that any set of independent vectors can always be extended to be a basis (i.e. independent spanning set) of the vector space.
Now, $\mathbb{R^n}$ has dimension $n$- which means that any basis can have no more than $n$ elements (in fact a basis has exactly $n$ elements). Now if your linearly ind. set of $n$ vectors didn't span $\mathbb{R^n}$ then we would have a basis consisting of ore than $n$ elements by extending the set of l.i. vectors.
So, any set of $n$ linearly ind. vectors in $\mathbb{R^n}$ spans $\mathbb{R^n}$.
In Linear Algebra, I learned a theorem which our professors called "Third One Free" (free translation). If I remember it correctly (and I'm not sure I do), it states that given a set of vectors $B$ in a vector space $V$ - we say that if two of these three are correct, then so is the third:
- $span(B)=V$ (i.e. $B$ spans $V$).
- $B$ is linearly independent.
- $|B|=dim(V)$
So, in that case, we have the 2nd and the 3rd statements are correct (size and linear independence), hence, according to the "Third One Free" theorem, the 1st statement (the vector set spans the space) is correct.
Hope I helped, and that my English wasn't too horrible :)