# Is $Tv=\mathcal{M}(v)$ an isomorphism from $V$ onto $\mathbf{F}^{n,1}$

Is the Following Proof Correct ?

Theorem. Given that $v_1,v_2,...,v_m$ is a basis for $V$ then the linear transformation $T:V\to\mathbf{F}^{n,1}$ defined by $$T(v)=\mathcal{M}(v)$$ is an isomorphism of $V$ onto $\mathbf{F}^{n,1}$ where $\mathcal{M}$ is defined such that $$\mathcal{M}(v)=\begin{pmatrix}c_1\\c_2\\c_3\\.\\.\\.\\c_n\end{pmatrix}$$ when $v=c_1v_1+c_2v_2+...+c_nv_n$ for the above mentioned basis.

Proof. We prove that $T$ is invertible by proving that $T$ is injective and surjective.

It is evident that $T(0)=\mathcal{M}(0)$ is an $n-$by$-1$ matrix with all entries equal to $0$ now assume that $Tv=\mathcal{M}(v)=0$ where $v$ is an arbitrary $v\in V$ and $0$ is equivalent to $\mathcal{M}(0)$, using the given bases we may express $v$ as $\beta_1v_1+\beta_2v_2+...+\beta_nv_n$ for some $\beta_1\beta_2,...,\beta_n\in\mathbf{F}$ consequently $$\mathcal{M}(v)=\begin{pmatrix}\beta_1\\\beta_2\\.\\.\\.\\\beta_n\end{pmatrix}=\begin{pmatrix}0\\0\\.\\.\\.\\0\end{pmatrix}=\mathcal{M}(0)$$ therefore $\beta_1=0,\beta_2=0,...\beta_n=0$ implying that $v=0$, moreover since $v$ was arbitrary it follows that $\operatorname{null}T=\{0\}$ indicating that $T$ is injective.

Now consider an arbitrary $A=\begin{pmatrix}\alpha_1\\\alpha_2\\.\\.\\.\\\alpha_n\end{pmatrix}\in\mathbf{F}^{n,1}$ evidently $w=(\alpha_1v_1+\alpha_2v_2+...+\alpha_nv_n)\in V$ and $\mathcal{M}(w)=A$ therefore $T$ is surjective

Taking the above two conclusions together we may conclude that $T$ is invertible and therefore is an isomorphism from $V$ onto $\mathbf{F^{n,1}}$

$\blacksquare$

Yes, your proof is correct. I would say, however, that the statement "it is evident that $T(0)=\mathcal{M}(0)$ is an $n-$by$-1$ matrix with all entries equal to $0$" is unnecessary.