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

Question

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 ...

Answer 1

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.

Answer 2

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.

Answer 3

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 $v=M^{-1}w$.

Answer 4

$\begin{bmatrix}Mv_1&Mv_2&...&Mv_n\end{bmatrix}=\begin{bmatrix}v_1&v_2&...&v_n\end{bmatrix}\begin{bmatrix}a_{11}&a_{12}&...&a_{1n}\\a_{21}&a_{22}&...&a_{2n}\\ \vdots&\vdots&\vdots&\vdots\\a_{n1}&a_{n2}&...&a_{nn}\end{bmatrix}$

The matrix on the right is just $M^T$.

Suppose you have a linear combination of the original basis u = [$v_1,v_2...v_n$]$[c_1, c_2 ... c_n]^T$

And suppose you're trying to find a linear combination of the new basis

u = [$Mv_1,Mv_2...Mv_n$]$[c'_1, c'_2 ... c'_n]^T$

We can rewrite this as

u = [$v_1,v_2...v_n$]$M^T[c'_1, c'_2 ... c'_n]^T$ =

[$v_1,v_2...v_n$]$([c'_1, c'_2 ... c'_n]M)^T$

So we can simply set $[c'_1, c'_2 ... c'_n] = [c_1, c_2 ... c_n]M^{-1}$