prove: For any finite-dimension vector space $V$ with ordered basis $\beta$, $\phi_\beta$ is an isomorphism.

Question

For any finite-dimension vector space $V$ with ordered basis $\beta$, $\phi_\beta$ is an isomorphism.

My work: Let $\beta=(v_1,v_2, \dots ,v_n)$ be a n-dimensional ordered basis for $V$. To check $\phi_\beta$ is an isomorphism, we need to prove it is injective and surjective.

Let $x \in V$, then $x=a_1v_1+\dots+a_nv_n$ for $a_i \in F$.

Injective: we need to prove for all set of vectors $x$ in $V$, $\phi_\beta(x)$={$o_n$} Here, $\phi_\beta(x)=[x]_\beta$=$(a_1,\dots,a_n)^t=(0,\dots,0)^t$, thus $N(T)={0}$.

To prove surjective, since the linear mapping is defined $\phi_\beta:V \rightarrow F^n$, and that $dim(v)=dim(F^n)$, by the theorem ( Let $V$ and $W$ be vector spaces of equal dimensions, then $T$ is onto), we see it is surjective.

Hence, it is an isomorphism.

Can I leave like that?

Answer 1

Let $V$ be a $n$-dimensional vector space over the field $F$ and $\beta=(v_1,\dots,v_n)$ a basis.

Define the function $$ \phi_\beta : V \to F^n, \quad x=\sum_{i=1}^n x_i v_i \mapsto (x_1, \dots, x_n) \quad \text{for} \quad x_i \in F \,. $$

The function is an isomorphism iff $V, F^n$ are vector spaces, and $\phi_\beta$ is injective, and surjective.

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The vector space argument is given by definition.

Injective: check if $\phi_\beta$ only maps $0_V$ onto $0_{F^n}$. \begin{align} 0_{F^n} &\overset{!}{=} \phi_\beta(x) = (x_1, \dots, x_n) \\ \implies x_i &= 0_F \quad \forall i \\ \implies x &= 0_V \end{align} Thus, $\phi_\beta$ is injective.

Surjective: check that every vector in $F^n$ is mapped onto.

Let $x_{F^n}=(x_1, \dots, x_n) \in F^n$ be given. Define $x_V=\sum_{i=1}^n x_i v_i \in V$ and we get $\phi_\beta(x_V) = x_{F^n}$. Thus, $\phi_\beta$ is surjective and finally we showed it is an isomorphism.