2
$\begingroup$

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?

$\endgroup$
4
  • 1
    $\begingroup$ The proof for surjectivity is wrong. Just because the dimensions of domain and codomain equal, it need not be surjective. You need to show that the dimensions of domain, codomain and image equal each other. $\endgroup$
    – jack
    Feb 8 '20 at 19:14
  • 1
    $\begingroup$ You should define $\phi_\beta$ before you start with your proof. $\endgroup$
    – jack
    Feb 8 '20 at 19:18
  • $\begingroup$ is injective part ok? Also I forgot to add that there is a definition beforehand that defines $\phi_{\beta}$ $\endgroup$
    – Beacon
    Feb 8 '20 at 19:20
  • 1
    $\begingroup$ the injective part is not as clear and precise as it could be. i am writing one right now $\endgroup$
    – jack
    Feb 8 '20 at 19:25
2
$\begingroup$

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.


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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.