# Same vector space for arbitrary independent vectors?

If we use $n$ linearly independent vectors $x_1,x_2,\dots,x_n$ to form a vector space $V$ and use another set of $n$ linearly independent vectors $y_1,y_2,\dots,y_n$ to form a vector space $S$, is it necessary that $V$ and $S$ are the same? Why?

If we have a vector space $Q$ that the dimension is $n$, can we say that any set of $n$ linearly independent vectors $k_1,k_2,\dots,k_n$ can form a basis of $Q$? Why?

Suppose only real numbers are involved.

• There is a theorem which says that on a finite dimensional vector space, n linearly independent vectors will span the space, and therefore form a basis.
– Dave
May 28, 2016 at 15:39
• Are $y_2, \ldots, y_n$ vectors in the first space $V$? And yes to the second question. May 28, 2016 at 15:58
• @Dave Thanks. If there is a theorem that a vector space Q that the dimension is n can have any set of n linearly independent vectors k1,k2...kn as its basis, how can we ensure that k1,k2...kn doesn't form a vector space other than Q but not Q? I have this question because two different set of n linearly independent vectors can form two different vector space with dimension n. It seems hard to ensure k1,k2...kn must form Q. May 29, 2016 at 3:13
• @KelvinS If you take any $m$ linearly independent vectors of a vector space $Q$ of dimension $n$, then (1) $m \le n$ (2) the space spanned by these linearly independent vectors is always an $m$-dimensional subspace of $Q$. If $m = n$, then you get an $n$-dimensional subspace of $Q$, and that must be $Q$ itself! May 29, 2016 at 5:02

To the first question, the answer is true, but true if we see the vector spaces $V$ and $S$ as isomorphic structures. See that we can have $V = span((1 \ 0 \ 0)^t,(0 \ 1 \ 0)^t)$ and $S = span((1 \ 0 \ 0)^t,(0 \ 0 \ 1)^t)$ and aren't the same space.