Image of a basis forms a basis, if and only if matrix is invertible

Note that $\{ v_1, \ldots, v_n \}$ is a basis of $\mathbb{R}^n$ if and only if $\begin{bmatrix}v_1 \ldots v_n \end{bmatrix}$ is a nonsingular matrix.

We have

$$\begin{bmatrix} Mv_1, \ldots , Mv_n \end{bmatrix} = M\begin{bmatrix} v_1, \ldots , v_n \end{bmatrix} $$

$$\det\left(\begin{bmatrix} Mv_1, \ldots , Mv_n \end{bmatrix}\right) = \det(M)\det\left(\begin{bmatrix} v_1, \ldots , v_n \end{bmatrix} \right)$$

Hence $\begin{bmatrix} Mv_1, \ldots , Mv_n \end{bmatrix}$ is nonsingular if and only if $\det(M)$ is non-zero, that is if and only if $M$ is nonsingular.

Using the determinant is a very elegant solution. But if you can't use the determinant, you still can prove it.

Let us denote $\tilde{B_1}=[v_1~v_2~\ldots~v_n]$ and $\tilde{B}_2=[Mv_1~Mv_2~\ldots~Mv_n]$. Then your last equation says $\tilde B_2=\tilde B_1M$.

Hints:

To finish $B_2$ basis implies $M$ invertible:

Since $B_2$ is a basis, the matrix $\tilde B_2$ is invertible. On the RHS the matrix $\tilde B_1$ is invertible, because $B_1$ is a basis.

Therefore, you can manipulate the equation to get $M=\ldots$, where $\ldots$ is invertible, hence $M$ is invertible.

For $M$ invertible implies $B_2$ basis:

Suppose $B_2$ is not a basis, then you can find a nontrivial linear combintion of $B_2$ which is $0$. You can rewrite the equation and get a nontrivial solution of $Mx=0$. Hence $M$ is not invertible and we are done.

For M invertible

$$a_1Mv_1+a_2Mv_2+...+a_nMv_n=M(a_1v_1+a_2v_2+...+a_nv_n)=0 \iff a_1v_1+a_2v_2+...+a_nv_n=0 \iff a_i=0$$

therefore $Mv_i$ are linearly independent and form a basis.

For $B_2$ basis

• $\forall w\, \exists! v$ such that $w=Mv$ indeed $$w=a_1Mv_1+a_2Mv_2+...+a_nMv_n=M(a_1v_1+a_2v_2+...+a_nv_n)=Mv$$

then $M$ is invertible since exists $M^{-1}$ such that $v=M^{-1}w$.