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Suppose one is given a set of n basis for a vector space. Then one is given another set of n linearly independent vectors each of which is a linear combination of the vectors in the original basis. Does that immediately imply that this new set of vectors is a basis for the vector space? Thanks.

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Yes. In a vector space with a finite Basis, each linear independent set with the same number of elements is also a basis. See for example en.wikipedia.org/wiki/Steinitz_exchange_lemma –  martini Sep 10 '11 at 9:48
    
@martini, you could post your comment as an answer so it can be accepted. –  joriki Sep 10 '11 at 9:54
    
I will do so ... –  martini Sep 10 '11 at 9:55
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up vote 5 down vote accepted

Yes. In a vector space with a finite basis, each linear independent set with the same number of elements is also a basis. See for example http://en.wikipedia.org/wiki/Steinitz_exchange_lemma.

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The following exercises will greatly improve your understanding of the concepts related to your question above:

$\textbf{Question 1: }$ This question does not assume that dimension is well defined yet. Now suppose you have a basis $B$ of vectors $\{w_1 \ldots w_n\}$ of a vector space $V$ that has $n$ elements. Now suppose you have $n$ linearly independent vectors $\{v_1, v_2 \ldots v_n\}$ in $V$. Suppose you write down the matrix of the linear transformation $T$ that takes $w_1$ to $v_1$, $w_2$ to $w_n$, and $w_n$ to $v_n$. Your matrix will look like this:

$\left[\begin{array}{ccc} | & |& | \\ T(w_1) & \ldots & T(w_n) \\ | & | & | \end{array} \right].$

Recall that the $i-th$ column of this matrix is the coordinate vector of $w_i$ in the basis $B$. From just these facts, why does it follow that the vectors $v_1 \ldots v_n$ are also a basis for $V$?

The formula # of pivot variables + # of free variables = # of columns may be helpful.

$\textbf{Question 2: }$ Prove that if $U$ is a subspace of $V$ that is finite dimensional and that dim $U$ = dim $V$, then in fact $U$ = $V$.

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Thanks, D B. Is there another name for pivot and free variables? I have not encountered these terms before... –  hil Sep 10 '11 at 12:02
    
ah, I have managed to find a definition online. –  hil Sep 10 '11 at 12:07
    
Q1: Is it because if one writes $T$ in echelon form there are no rows of pure 0's because $\{ v_i'\}$ are linearly independent. Therefore, $T$is non-singular, hence invertible. So we could get any of the $w_i$'s as linear combinations of $v_i$'s. Hence the set of $v_i$'s must span $V$ and therefore is a basis for V? Q2: Is it because we can then choose the bases of $U$ and $V$ to coincide? –  hil Sep 10 '11 at 12:32
    
@hil linear independence just means injectivity. However what are the dimensions of the matrix? We get surjectivity once we get injectivity because your matrix is square. Now try question 2 –  user38268 Sep 10 '11 at 22:52
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