Suppose $B_1=\{v_1,v_2,…,v_n\}$ is a basis of $\mathbb{R}^n$, and $M$ is an $n*n$ matrix. Prove that $B_2=\{Mv_1,Mv_2,…,Mv_n\}$ is also a basis of $\mathbb{R}^n$ if and only if $M$ is invertible.
Following is what I have so far:
Assume $B_2$ is basis of $\mathbb{R}^n$.
Then, $B_2$ is a set of linearly independent vectors, and $B_2$ spans $\mathbb{R}^n$.
Since $B_1$ is also a basis of $\mathbb{R}^n$, then any element(vector) of $B_2$ is a linear combination of elements(vectors) of $B_1$ and vice-versa.
$Mv_1= a_{11}v_1+a_{21}v_2+…+a_{n1}v_n$ , where $a_{11},a_{21},…,a_{n1}\in \mathbb{R}$
Likewise, $Mv_2= a_{12}v_1+a_{22}v_2+…+a_{n2}v_n$ , where $a_{12},a_{22},…,a_{n2}\in \mathbb{R}$
$\begin{bmatrix}Mv_1&Mv_2&…&Mv_n\end{bmatrix}=\begin{bmatrix}v_1&v_2&…&v_n\end{bmatrix}\begin{bmatrix}a_{11}&a_{12}&…&a_{1n}\\a_{21}&a_{22}&…&a_{2n}\\ \vdots&\vdots&\vdots&\vdots\\a_{n1}&a_{n2}&…&a_{nn}\end{bmatrix}$
Not sure what to do next …
Best Answer
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.