Basis of a subspace Given a basis $\left[\begin{array}{cccc}
v_{1} & v_{2} & \ldots & v_{m}\end{array}\right]$ of a subspace of dim m where $v_{i}\in R^{n}$ how can we find different basis of the same subspace?
 A: Take any invertible $m\times m$ matrix with entries in $\mathbb{R}$, and interpret its columns  as linear combinations of your basis elements. For example, if we take the $2$-dimensional subspace of $\mathbb{R}^3$ which has as a basis $\{v_1=(2,1,0),v_2=(-1,1,0)\}$, then taking the invertible $2\times 2$ matrix $\bigl(\begin{smallmatrix} 2 & 3 \\ 4  & 7 \end{smallmatrix}\bigr)$, the new basis of the subspace which this matrix gives is
$$\Bigl\{2v_1+4v_2=(-2,6,0),\;\;3v_1+7v_2=(-1,10,0)\Bigr\}$$
Every basis of the subspace can be produced in this way.
A: Let $w_{1}=a_{1}v_{1}+a_{2}v_{2}+\dots a_{m}v_{m}$. Then $v_{1}$ is dependent on $\left[\begin{array}{cccc}
w_{1} & v_{2} & \ldots & v_{m}\end{array}\right]$. $w_{1},v_{2}\dots v_{m}$ are all linearly independent. Hence, $\left[\begin{array}{cccc}
w_{1} & v_{2} & \ldots & v_{m}\end{array}\right]$ forms a basis.
Let $w_{2}=b_{1}w_{1}+b_{2}v_{2}+\dots b_{m}v_{m}$. Then $v_{2}$ is dependent on $\left[\begin{array}{cccc}
w_{1} & w_{2} & \ldots & v_{m}\end{array}\right]$.
In this way, by replacing all $v_{i}$ by $w_{i}$, we can form a new basis. 
