MHB Direct Sum Property: Proving Uniqueness

[Sudharaka]

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.

 

[caffeinemachine]

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]

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. :)