Basis for $\mathbb R^n$ and invertible $n\times n$ matrix Let $s = \{v_1, v_2, v_3, ... , v_n\}$ be a basis for $\mathbb R^n$. Suppose $B$ is an invertible $n\times n$ matrix, show that $w = \{ Bv_1, Bv_2, Bv_3, ...,Bv_n\}$ is also a basis. I know that you just need to show that $BS$ is also invertible because it then implies that $w$ is linearly independant. Any ideas how to do this?
 A: Given that $\mathcal S$ is a basis for $\mathbb R^n$ and $B$ is an invertible $n \times n$ matrix, we want to show that $\mathcal W$ is also a basis for $\mathbb R^n$. Since $\mathcal W$ has the correct dimension, it suffices to show that $\mathcal W$ is a linearly independent set. To this end, suppose that there exist $c_1, \ldots, c_n \in \mathbb R$ such that:
$$
c_1 (B\vec v_1) + \cdots + c_n (B\vec v_n) = \vec 0
$$
We want to show that $c_1 = \cdots = c_n = 0$. To this end, notice by the linearity of matrix multiplication that:
$$
B(c_1 \vec v_1 + \cdots + c_n \vec v_n) = \vec 0
$$
But since $B$ is invertible, we may left-multiply both sides by $B^{-1}$ to obtain:
$$
c_1 \vec v_1 + \cdots + c_n \vec v_n = \vec 0
$$
But since $\mathcal S$ is a basis for $\mathbb R^n$, recall that $\mathcal S$ must be linearly independent. Hence, it follows that $c_1 = \cdots = c_n = 0$, as desired. $~~\blacksquare$
A: Suppose that $w_i = B v_i, i\in \{1,2, ...,n\}$. Now suppose that one of the vectors of new basis is linearly dependant so it doesn't form basis. $w_j = \sum_k\alpha_k w_k$. So then:
$$
v_i = B^{-1}w_i \implies v_j = B^{-1}(\sum_k \alpha_k w_k ) \implies v_j = \sum_k \alpha_k v_k
$$
And this is with contradiction with statement that vectors $v_i$ forms basis.
