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: $$\textsf{V}$$ 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 $$\textsf{V}=\left\oplus\cdots\oplus\left$$.

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 $$\textsf{V}=\left\oplus\cdots\oplus\left$$ 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 $$\textsf{V}$$.

• Consider these: definition of "Span"; definition of "direct sums"; the concrete form of each $\langle v_j \rangle$. For the converse, translate everything to linear combinations. – xbh Oct 31 at 10:55
• do you want to do it the pedestrian way? because you could jump a few fences by considering the map induced by a set of $m$ vectors $k^{m} \to V$. Was just considering to write down the pedestrian way, but it got too fidelly for me. – Enkidu Oct 31 at 13:23

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!

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. $$\blacktriangleright$$

• Sorry that I have to be a little nitpicky, but sadly the direct sum of subspaces is not defined, it is a sum of subspaces with a certain property. So sadly the thing with forcing it into uniqueness does not work out. You actually need to proof that their intersections vanish, which is precisely what males proving the n-fold direct some so tideous, as one needs to check all possible intersections of the sums of the linear hulls – Enkidu Nov 3 at 22:15
• @Enkidu I don't quite follow… Nevertheless I updated and added the proof of equivalence. – xbh Nov 4 at 0:37
• you are using the internal direct sum as if it comes with the uniqueness, but it doesn't, it is an ordinary sum with an additional property! which you never really proved! – Enkidu Nov 4 at 11:20
• @Enkidu Maybe you could take a look at the textbook Linear Algebra Done Right [3rd edition] Definition 1.40. The OP never mentioned what s/he learned about $\oplus$ in the context of Linear Algebra, so I just based my answer on what I have learned. – xbh Nov 4 at 12:02
• @Enkidu Fine. Garrett Birkhoff, Saunders Mac Lane, A Survey of Modern Algebra, Section 7.8, p.195, paragraph 4, definition of 2 subspaces; Section 7.8, p. 198, Exercises #11~#13, definition of several subspaces. Also, in the proof of $\implies$ part I don't need to "check" the uniqueness anymore, since Section 7.8, p. 193, Thm. 14 did that for me. – xbh Nov 4 at 13:53