Show that $v_1,…,v_n$ is a basis of $\textsf{V}$ if and only if $\textsf{V}=\left<v_1\right>\oplus\cdots\oplus\left<v_n\right>$

Question

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 $\textsf{V}$.

Answer 1

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!

Answer 2

Proof. $\blacktriangleleft$ $\boxed\implies$ If $(v_j)_1^n$ is a basis, then for each $v \in V$, there exists a unique list of scalars $(c_j)_1^n$ such that $v = \sum_1^n c_j v_j \in \bigoplus _1^n \langle v_j\rangle$, hence $V = \sum_1^n \langle v_j \rangle$. Since the expression is unique for every $v \in V$, by the defnition of direct sums, $\sum_1^n \langle v_j \rangle$ is a direct sum.

$\boxed\impliedby$ Suppose $0 = \sum_1^n c_j v_j$ for a list of scalars, then according to $c_j v_j \in \langle v_j \rangle$ for every $j$ and the definition of direct sums, $c_j = 0$ for all $j$. Hence $(v_j)_1^n$ is linearly independent. The dimension formula implies $\dim V = \sum_1^n \dim ( \langle v_j \rangle) = n$, thus $(v_j)_1^n$ shall be a basis. $\blacktriangleright$

Appendix

Given vector space $V$, and subspaces $V_j [j =1, \dots, m]$,

$V = V_1 \oplus V_2 \oplus \cdots \oplus V_m$ is defined to be $(1)V = \sum_1^m V_j$ and $(2)$ for each $v\in V$, there are unique vectors $v_j \in V_j$ such that $v = \sum_1^m v_j$.

This is equivalent to

$V = V_1 \oplus V_2 \oplus \cdots \oplus V_m$ is defined to be $(1)V = \sum_1^m V_j$ and $(2)$ for each $k\in\{1,2,\dots,m\}$, $V_k \cap \sum_{j \neq k} V_j = \{0\}$.

Proof. $\blacktriangleleft$ $\boxed\implies$ If the expression is unique for every $v \in V$, then for $v_k \in V_k \cap \sum_{j \neq k} V_j$, there is some $u_j \in V_j [j \neq k]$ such that $v_k = \sum_{j \neq k} u_j$, hence $0 =-v_k + \sum_{j \neq k} u_j$. The uniqueness forces all $u_j$ and $v_k$ be $0$. Therefore $V_k \cap \sum_{j \neq k} = 0$ as we want.

$\boxed\impliedby$ Suppose for each $k$, $V_k \cap \sum_{j \neq k } V_j = 0$. If $0 = \sum_1^m v_j$, then $v_k = - \sum_{j \neq k} \in V_k \cap \sum_{ j \neq k } V_j = \{0\}$, so $v_k =0$ for all $k \in \{1,2,\dots,m\}$. Therefore $0$ has a unique expression $0=0+\dots + 0$. For any other $v \in V$, if $v = \sum_1^m u_j = \sum_1^m w_j$ where $u_j, w_j \in V_j$ for all $j$, then $0 = \sum_1^m (u_j - w_j)$. Since each $u_j - w_j \in V_j$ as well, the uniqueness forces $u_j = w_j$ for all $j$. Equivalently, each decomposition of $v$ is unique. $\blacktriangleright$