# [SOLVED]Direct Sum Property

#### Sudharaka

##### Well-known member
MHB Math Helper
Hi everyone,

I encountered this question and thought about it several hours. I am writing down my answer. I would greatly appreciate if somebody could find a fault in my answer or else confirm it is correct.

Problem:

Let $$V_1,\,\cdots,\,V_k$$ be subspaces in a vector space $$V$$, $$V=V_1+\cdots+V_k$$. Show that the sum is direct iff there is at least one $$v\in V$$ such that $$v=v_1+\cdots+v_k$$ where $$v_i\in V_i$$, in a unique way.

My Solution:

If $$V=V_1\oplus \cdots \oplus V_k$$ then by the definition of the direct product each $$v\in V$$ can be uniquely written as $$v=v_1+\cdots+v_k$$ where $$v_i\in V_i$$. So the forward implication is clearly true.

Now let us prove the reverse direction. There exist and element $$v\in V$$ such that $$v=v_1+\cdots+v_k$$ where $$v_i\in V_i$$, in a unique way. Suppose that the sum is not a direct sum. Then there exist at least one element $$x\in V$$ such that $$x$$ has two different representations,

$x=x_1+\cdots+x_k\mbox{ and }x=x'_1+\cdots+x'_k$

where $$x_i,\,x'_i\in V_i$$. Now there exist some $$v_0\in V$$ such that, $$v=x+v_0$$. Hence,

$v=x_1+\cdots+x_k+v_0\mbox{ and }v=x'_1+\cdots+x'_k+v_0$

Now since we can write $$v_0=v_1^0+\cdots+v_k^0$$, we get,

$v=(x_1+v^0_1)+\cdots+(x_k+v^0_k)\mbox{ and }v=(x'_1+v^0_1)+\cdots+(x'_k+v^0_k)$

Note that, $$x_i+v_i^0\in V_i$$ and $$x'_i+v_i^0\in V_i$$.

But since $$v$$ has a unique representation $$v=v_1+\cdots+v_k$$ we have,

$v_1=x_1+v^0_1=x'_1+v^0_1$

$v_2=x_2+v^0_2=x'_2+v^0_2$

and generally,

$v_i=x_i+v^0_i=x'_i+v^0_i$

Therefore,

$x_i=x'_i\mbox{ for all }i$

Therefore we arrive at a contradiction. The sum must be direct.

Last edited:

#### caffeinemachine

##### Well-known member
MHB Math Scholar
Hi everyone,

I encountered this question and thought about it several hours. I am writing down my answer. I would greatly appreciate if somebody could find a fault in my answer or else confirm it is correct.

Problem:

Let $$V_1,\,\cdots,\,V_k$$ be subspaces in a vector space $$V$$, $$V=V_1+\cdots+V_k$$. Show that the sum is direct iff there is at least one $$v\in V$$ such that $$v=v_1+\cdots+v_k$$ where $$v_i\in V_i$$, in a unique way.

My Solution:

If $$V=V_1\oplus \cdots \oplus V_k$$ then by the definition of the direct product each $$v\in V$$ can be uniquely written as $$v=v_1+\cdots+v_k$$ where $$v_i\in V_i$$. So the forward implication is clearly true.

Now let us prove the reverse direction. There exist and element $$v\in V$$ such that $$v=v_1+\cdots+v_k$$ where $$v_i\in V_i$$, in a unique way. Suppose that the sum is not a direct sum. Then there exist at least one element $$x\in V$$ such that $$x$$ has two different representations,

$x=x_1+\cdots+x_k\mbox{ and }x=x'_1+\cdots+x'_k$

where $$x_i,\,x'_i\in V_i$$. Now there exist some $$v_0\in V$$ such that, $$v=x+v_0$$. Hence,

$v=x_1+\cdots+x_k+v_0\mbox{ and }v=x'_1+\cdots+x'_k+v_0$

Now since we can write $$v_0=v_1^0+\cdots+v_k^0$$, we get,

$v=(x_1+v^0_1)+\cdots+(x_k+v^0_k)\mbox{ and }v=(x'_1+v^0_1)+\cdots+(x'_k+v^0_k)$

Note that, $$x_i+v_i^0\in V_i$$ and $$x'_i+v_i^0\in V_i$$.

But since $$v$$ has a unique representation $$v=v_1+\cdots+v_k$$ we have,

$v_1=x_1+v^0_1=x'_1+v^0_1$

$v_2=x_2+v^0_1=x'_2+v^0_1$

and generally,

$v_i=x_i+v^0_1=x'_i+v^0_1$
A typo here. It should be $v_i=x_i+v^0_i=x'_i+v^0_i$
The rest is correct.

#### Sudharaka

##### Well-known member
MHB Math Helper
A typo here. It should be $v_i=x_i+v^0_i=x'_i+v^0_i$
The rest is correct.
Yeah, I was typing it too fast, didn't check much for typos. I have edited it in the original post. Thank you very much for the confirmation and pointing out the typo. I really appreciate it. I am confident that this and the several other questions I posted recently are correct, but just want to get the opinion of everybody. I have a exam coming up and these are from a sample test.