Question: $\textsf{V}$ is a vector space over a field $\textsf{K}$. Let $v_1,v_2,\dots,v_n$ be vectors in $\textsf{V}$. Show that $v_1,…,v_n$ is a basis of $\textsf{V}$ if and only if $\textsf{V}=\left<v_1\right>\oplus\cdots\oplus\left<v_n\right>$.
Since its an if and only if statement I must prove both sides are true.
[From right to left]
Assume $v_1,…,v_n$ is a basis then I must show that $\textsf{V}=\left<v_1\right>\oplus\cdots\oplus\left<v_n\right>$ is true. I know that if $v_1,…,v_n$ is a basis then by definition of a basis we know that $v_1,…,v_n$ is linearly independent and that the span is $\textsf{V}$.
From here I am not really sure how to proceed. I was thinking if $v_1,…,v_n$ is linearly independent then you cant form a linear combination of each of them so the intersection between each one is $\{0\}$ and therefore the direct sum of all of the vectors in the basis gives us $\textsf{V}$.
Best Answer
As mentioned above, lets jump some fences:
consider the map $$\varphi:k^n \to V \\ e_i \mapsto v_i$$ Since $\{e_i\}_i$ is a basis this is a well defined map.
Now this map is surjective if and only if for all $v \in V$ there is an element $w \in k^n$ s.t. $$\varphi(w)=\varphi(\sum_{i=1}^n w_i e_i )= \sum_{i=1}^n w_i \varphi(e_i) = \sum_{i=1}^n w_i v_i=v$$
which is just the definition of a generating set.
This means $\varphi$ is surjective if and only if $v_1,...,v_n$ defines a generating set.
Furthermore $\varphi$ is injective if and only if $$\varphi(w)=\varphi(\sum_{i=1}^n w_i e_i )= \sum_{i=1}^n w_i \varphi(e_i)= \sum_{i=1}^n w_i v_i=0 \implies w=0\iff \forall i\; w_i=0$$ but this is just the definition of linear independence. So \varphi is injective if and only if $v_1,...,v_n$ is linearly independent.
so $v_1,....,v_n$ is a basis if and only if $\varphi$ is an isomorphism . but that means, since $k^n = ke_1\oplus ... \oplus ke_n$ that $V \cong \varphi(ke_1)\oplus ... \oplus \varphi(ke_n)$. And now $\varphi (ke_i)=\langle v_i \rangle$ and we are done!