# Prove that every vector $v \in V = V_1 \oplus V_2$ can be written uniquely as $v = v_1 + v_2$, where $v_1 \in V_1$ and $v_2 \in V_2$.

My example: Prove that every vector $v \in V = V_1 \oplus V_2$ can be written uniquely as $v = v_1 + v_2$, where $v_1 \in V_1$ and $v_2 \in V_2$.

My solution: For the direct sum $V_1 \oplus V_2$ we have that $V_1 \cap V_2 = \{0\}$. Let assume $V$ is a finite vector space with the basis $P = \{u_1,...,u_n\}$. Let $N = \{u_1,...,u_k\}$ be the basis of $V_1$ and let $M=\{u_{k+1},...,u_n\}$ be the basis of $V_2$.

Because $v_1 \in V_1$ we can write it as the linear combination of the vectors from $N$ and $v_2 \in V_2$ is possible to write as the linear combination of the vectors from $M$.

We have that $v =\sum_{i=1}^n l_iu_i = \sum_{i=1}^k r_iu_i +\sum_{i=k+1}^n s_iu_i = v_1 + v_2$. For the linear independence we require that the scalars $l,r,s$ equal to zero if $v=0$.

I don't know if I proved anything, any suggestions please?

• are you sure the problem statement is correct? Shouldn't it say 'for unique $v_1, v_2$'? – Student Mar 6 '17 at 15:59
• Do you mean "uniquely written as.."? Otherwise there is nothing to prove because the statement is trivial – tofurind Mar 6 '17 at 16:00
• Yes, it is for unique $v_1$ and $v_2$, I will edit it. – Leif Mar 6 '17 at 16:00
• You don't need to work with bases. If $v_1 + v_2 = w_1 + w_2$ then $$v_1 - w_1 = v_2 - w_2 \in V_1 \cap V_2$$ so that both differences are zero. – Umberto P. Mar 6 '17 at 16:01
• Thank you. Sometimes I am not very efficient. – Leif Mar 6 '17 at 16:03

## 1 Answer

Because of the definition of $\oplus$, we know that $V = V_1 + V_2$, hence each $v \in V$ can be written as $v = v_1 + v_2$ where $v_1 \in V_1$ and $v_2 \in V_2$. Let us prove that this can be done in a unique way.

Let $v = v_1 + v_2 = w_1 + w_2$ for $v_i, w_i \in V_i$. Then we have that $$v_1 - w_1 = w_2 - v_2.$$ Note that $v_1 - w_1 \in V_1$, hence $w_2 - v_2 \in V_1$. Analogously, we have that $v_1 - w_1 = w_2 - v_2 \in V_2$. Therefore we have that $$v_1 - w_1 \in V_1 \cap V_2 = \{0\},$$ so $v_1 = w_1$. Analogously $v_2 = w_2$.