Prove that the vectors $v_1,v_2,\ldots,v_k \operatorname{span}R^n$ if and only if $[v_1]_B,[v_2]_B,\ldots,[v_k]_B \operatorname{span}R^n$. From section on Change of Basis

$\longrightarrow$ Assume the vectors $v_1,v_2,\ldots,v_k\operatorname{span}R^n$, we must show that $[v_1]_B,[v_2]_B,\ldots,[v_k]_B\operatorname{span}R^n$.  We can write the vectors in $R^n$ as linear combinations of the coordinate vectors $[v_1]_B,[v_2]_B,\ldots,[v_k]_B$.
This is as far as I can get.  I have trouble proving things, I don't know how to.  I can't seem to prove any of the exercises I am faced with.  Are there any strategies to learn proofs?
 A: First you need to make sure you are clear on definitions.

"The vectors ${\bf v}_1,{\bf v}_2,\ldots,{\bf v}_k$ span ${\Bbb R}^n$" means: for every vector ${\bf w}$ in ${\Bbb R}^n$ there exist scalars $\lambda_1,\lambda_2,\ldots,\lambda_k$ such that
  $$\lambda_1{\bf v}_1+\lambda_2{\bf v}_2+\cdots+\lambda_k{\bf v}_k={\bf w}\ .$$



The coordinate vector of ${\bf v}$ in ${\Bbb R}^n$ with respect to the ordered basis $B=\{{\bf b}_1,{\bf b}_2,\ldots,{\bf b}_n\}$ for ${\Bbb R}^n$ is the vector
  $[{\bf v}]_B=(x_1,x_2,\ldots,x_n)$ such that
  $${\bf v}=x_1{\bf b}_1+x_2{\bf b}_2+\cdots+x_n{\bf b}_n\ .$$

Next you will need a result about the connections between coordinate vectors.
Lemma.  If $B$ is an ordered basis for ${\Bbb R}^n$ and ${\bf v}_1,{\bf v}_2$ are vectors in ${\Bbb R}^n$ and $\lambda_1,\lambda_2$ are scalars, then
$$[\lambda_1{\bf v}_1+\lambda_2{\bf v}_2]_B
  =\lambda_1[{\bf v}_1]_B+\lambda_2[{\bf v}_2]_B\ .$$
Proof.  Write $[{\bf v}_1]_B=(x_1,\ldots,x_n)$ and $[{\bf v}_2]_B=(y_1,\ldots,y_n)$.  By definition this means that
$${\bf v}_1=x_1{\bf b}_1+\cdots+x_n{\bf b}_n\quad\hbox{and}\quad
  {\bf v}_2=y_1{\bf b}_1+\cdots+y_n{\bf b}_n\ .$$
Therefore
$$\lambda_1{\bf v}_1+\lambda_2{\bf v}_2
  =(\lambda_1x_1+\lambda_2y_1){\bf b}_1+\cdots
    +(\lambda_1x_n+\lambda_2y_n){\bf b}_n$$
and so
$$\eqalign{[\lambda_1{\bf v}_1+\lambda_2{\bf v}_2]_B
  &=(\lambda_1x_1+\lambda_2y_1,\ldots,\lambda_1x_n+\lambda_2y_n)\cr
  &=\lambda_1(x_1,\ldots,x_n)+\lambda_2(y_1,\ldots,y_n)\cr
  &=\lambda_1[{\bf v}_1]_B+\lambda_2[{\bf v}_2]_B\cr}$$
which proves the result claimed.  An almost identical proof shows that the same relationship holds for any linear combination of $k$ vectors.
Now we can begin the proof that you need.  Note that it is an "if and only if" statement, so there are two things to prove.
First suppose that
$$[{\bf v}_1]_B,\ldots,[{\bf v}_k]_B\quad\hbox{span}\quad {\Bbb R}^n\ .
  \tag{$*$}$$
We have to prove that ${\bf v}_1,\ldots,{\bf v}_k$ span ${\Bbb R}^n$.  So, let ${\bf w}$ be in ${\Bbb R}^n$; then $[{\bf w}]_B$ is also in ${\Bbb R}^n$.  From $(*)$, there exist scalars $\lambda_1,\ldots,\lambda_k$ such that
$$[{\bf w}]_B=\lambda_1[{\bf v}_1]_B+\cdots+\lambda_k[{\bf v}_k]_B\ .$$
By the lemma we have
$$[{\bf w}]_B=[\lambda_1{\bf v}_1+\cdots+\lambda_k{\bf v}_k]_B$$
and so
$${\bf w}=\lambda_1{\bf v}_1+\cdots+\lambda_k{\bf v}_k\ .$$
We have shown that any ${\bf w}$ in ${\Bbb R}^n$ can be written as a linear combination of ${\bf v}_1,\ldots,{\bf v}_k$, and so by definition, these vectors span ${\Bbb R}^n$.
Secondly, suppose that
$${\bf v}_1,\ldots,{\bf v}_k\quad\hbox{span}\quad {\Bbb R}^n\ ;\tag{$*{*}$}$$
we have to prove that $[{\bf v}_1]_B,\ldots,[{\bf v}_k]_B$ span ${\Bbb R}^n$.  So, let ${\bf x}=(x_1,\ldots,x_n)$ be in ${\Bbb R}^n$ and write
$${\bf w}=x_1{\bf b}_1+\cdots+x_n{\bf b}_n\ .$$
By assumption $(*{*})$, there exist scalars $\lambda_1,\ldots,\lambda_k$ such that
$${\bf w}=\lambda_1{\bf v}_1+\cdots+\lambda_k{\bf v}_k\ .$$
Putting all these facts together, using the lemma and the definitions, we have
$$\eqalign{{\bf x}
  &=[{\bf w}]_B\cr
  &=[\lambda_1{\bf v}_1+\cdots+\lambda_k{\bf v}_k]_B\cr
  &=\lambda_1[{\bf v}_1]_B+\cdots+\lambda_k[{\bf v}_k]_B\ .\cr}$$
We have shown that any ${\bf x}$ in ${\Bbb R}^n$ can be written as a linear combination of $[{\bf v}_1]_B,\ldots,[{\bf v}_k]_B$, and so by definition, these vectors span ${\Bbb R}^n$.  This completes the proof.
A: Assuming that $\left[ \vec{x} \right]_B$ is the coordinate vector with respect to $\vec{x}$, here is a small hint. The map $\left[ \cdot \right]_B$ is linear, i.e., for $\alpha \in \mathbb{R}$ and $\vec{v} , \vec{w} \in \mathbb{R}^n$
$$ \left[ \alpha \vec{v} + \vec{w} \right]_B = \alpha \left[ \vec{v} \right]_B + \left[ \vec{w} \right]_B $$
This should get you somewhere, if not, just comment and I can keep leading you in the right direction. The above equation is sometimes called the Linearity of Coordinates.
I might also help to know that $\left[ \cdot \right]_B$ is a one-to-one map.
Here is a edit to finish why the above works.
Let us assume the set of vectors $ \{ \vec{v}_i \} $ span $\mathbb{R}^n$. Then for a vector $\vec{t} \in \mathbb{R}^n$, we have
$$ t = c_1 \vec{v}_1 + c_2 \vec{v}_2 + \cdots + c_n \vec{v}_n \; \; \; \; \; c_i \in \mathbb{R} $$
Then,
$$ \left[ \vec{t} \right]_B = \left[ c_1 \vec{v}_1 + c_2 \vec{v}_2 + \cdots + c_n \vec{v}_n \right]_B $$
$$ = c_1 \left[ \vec{v}_1 \right]_B + c_2 \left[ \vec{v}_2 \right]_B + \cdots + c_n \left[ \vec{v}_n \right]_B $$
where the equality comes from the linearity of coordinates. We see that for any arbitrary vector $\vec{t} \in \mathbb{R}^n$, $\left[ \vec{t} \right]_B$ can be written as a linear combination of the vectors $ \{ \left[ \vec{v}_i  \right]_B \}$ Thus, these span $\mathbb{R}^n$.
One can go further to proof that if $\{ \vec{v}_i \}$ are a basis, then so are $\{ \left[ \vec{v}_i \right]_B \}$.
A: More generally, let $E$ be a vector space of dimension $n$ over the scalar field $F$, and let $(e_1, ..., e_n)$ be the canonic basis of $E$.
By definition, the vectors $(a_1, .., a_k)$ are called linearly independent if
$$
\forall (\lambda_1, .., \lambda_k)\in F^k,\quad \sum_i \lambda_i a_i = 0 \implies \lambda_i = 0\ \forall i \tag{1}
$$
Now let's assume that $(v_1,...,v_n)$ and $(b_1, ..., b_n)$ are both basis of $E$ (ie the subspace they generate is $E$, ie they "span" $E$). Each $v_k$ can be decomposed on the canonic basis $(e_1,..,e_n)$, meaning that it can be expressed as a linear combination of these vectors.
Now, if $(b_1,..., b_n)$ is a basis of $E$, then each $v_k$ can also be expressed as a linear combination of these vectors. This is what you wrote "$[v_k]_B$":
$$
\forall k,\quad \exists (\mu_{k,1}, ..., \mu_{k,n})\neq (0,...,0)\in F^n\ \big/\ v_k = [v_k]_B = \sum_i \mu_{k,i} b_i \tag{2}
$$

The proof then proceeds by showing that the family $([v_1]_B,...,[v_n]_B)$ is linearly independent, and since it is maximal (ie contains $n$ vectors in a vector space of dimension $n$) it forms a basis. To do that, let's assume that they are not linearly independent. Then by $(1)$, we could find $(\lambda_1, ..., \lambda_n)\neq(0,...,0)$ such that
$$
\sum_k\lambda_k [v_k]_B = 0 = \sum_k\lambda_k \sum_i \mu_{k,i} b_i = \sum_i \left( \sum_k \lambda_k\mu_{k,i} \right) b_i
$$
which would mean that we found a null linear combination of $b_i$s, proving that they are not linearly independent, which contradicts our assumptions. 
The reverse is exactly the same since we can swap $(b_1,...,b_n)$ with $(e_1,...,e_n)$.
A: So this might be a more appropriate solution given what you've learned, if you're doing a usual intro linear algebra course:
We know that a collection of vectors $\mathbf{v}_1,\mathbf{v}_2,...,\mathbf{v}_k$ span $\mathbb{R}^n$ if and only if the matrix $V = [\mathbf{v}_1 \; \mathbf{v}_2 \; ... \;\mathbf{v}_k]$ is onto, which means every equation of the form $V\mathbf{x} = \mathbf{b}$, where $\mathbf{b} \in \mathbb{R}^n$, has some solution $\mathbf{x}$ in $\mathbb{R}^m$. 
Now, what about the vectors $[\mathbf{v}_1]_{B},[\mathbf{v}_2]_{B},...,[\mathbf{v}_k]_{B}$? We know that we can obtain a vector in $B$ coordinates by multiplying by a certain matrix, namely $[\mathbf{v}_i]_{B} = P_{B}^{-1}\mathbf{v}_i$, where $P_{B}$ denotes the matrix whose columns are the basis vectors in $B$. ($P_{B}$ is the 'change-of-coordinates' matrix).
It follows that the matrix $\tilde{V} = [[\mathbf{v}_1]_{B} \; [\mathbf{v}_2]_{B} \; ... \;[\mathbf{v}_k]_{B} ]$ is given by the equation $\tilde{V} = P_{B}^{-1}V$. You now need to show that $\tilde{V}$ is onto, so pick $\mathbf{b} \in \mathbb{R}^n$. We want to solve the equation:
\begin{align}& \tilde{V}\mathbf{x} = \mathbf{b} \\
\Leftrightarrow & P_{B}^{-1}V\mathbf{x} = \mathbf{b} \\
\Leftrightarrow & V\mathbf{x} = P_{B}\mathbf{b} \end{align}
We know that the last equation has a solution since $V$ is onto, which completes the proof.
