Isomorphism Between External and Internal Direct Sums - Knapp Proposition 2.30

[Math Amateur]

I am reading Chapter 2: Vector Spaces over \(\displaystyle \mathbb{Q}, \mathbb{R} \text{ and } \mathbb{C}\) of Anthony W. Knapp's book, Basic Algebra.

I need some help with some issues regarding Theorem 2.30 (regarding an isomorphism between external and internal direct sums) on pages 59-60.

Theorem 2.27 and its proof read as follows:
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View attachment 2925

First Issue/QuestionIn the first paragraph of the proof given above, Knapp writes:

" ... ... If \(\displaystyle v\) is in \(\displaystyle V_1 \cap V_2\), then \(\displaystyle 0 = v + (-v)\) is a decomposition of the kind in (a), and the uniqueness forces \(\displaystyle v = 0 \). ... ... "

I am unsure of Knapp's reasoning but believe he is arguing as follows:

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If v' = -v we can write (by the definition of inverse and the commutativity of +) that

0 = v + v' = v + v'

Then we can regard

\(\displaystyle 0 = v + v'\) as one decomposition of the vector \(\displaystyle 0\) with \(\displaystyle v \in V_1\) and \(\displaystyle v' \in V_2\).

\(\displaystyle 0 = v' + v\) as one decomposition of the vector \(\displaystyle 0\) with \(\displaystyle v' \in V_1\) and \(\displaystyle v \in V_2\).

BUT ... there must be a unique decomposition of every vector in V, so we must have v = v' = 0 ... and therefore \(\displaystyle V_1 \cap V_2 = 0 \)

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Can someone please confirm that my reasoning is correct? (or otherwise point out the errors/shortcomings ...)

(Mind you I am concerned that at least something is wrong since every vector in V is supposed to have a unique decomposition, but I seem to be arguing the the vector 0 has no decomposition at all, unless 0 = 0 + 0 constitutes a decomposition ... ? )
Second Issue/QuestionIn the second paragraph of the proof above, Knapp writes:

" ... ... To see that it (the linear map, say \(\displaystyle L\)) is one-one, suppose that \(\displaystyle v_1 + v_2 = 0\). Then \(\displaystyle v_1 = - v_2\) shows that \(\displaystyle v_1\) is in \(\displaystyle V_1 \cap V_2\). By (b), this intersection is \(\displaystyle 0\). Therefore \(\displaystyle v_1 = v_2 = 0\), and the linear map in (c) is one-one. ... ... "

I cannot follow Knapp's reasoning in the above ...

My thoughts on how to show L is one-one are as follows:

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We have a linear map \(\displaystyle L\) such that

\(\displaystyle L( (v_1, v_2)) = v_1 + v_2\)

We need to show \(\displaystyle L\) is one-one i.e. we need to show that if \(\displaystyle L( (u_1, u_2)) = L( (v_1, v_2))\) then \(\displaystyle (u_1, u_2) = (v_1, v_2)\)

in the above, we have:

\(\displaystyle (u_1, u_2) \in V\) where \(\displaystyle u_1 \in V_1\) and \(\displaystyle u_2 \in V_2\)

and

\(\displaystyle (v_1, v_2) \in V\) where \(\displaystyle v_1 \in V_1\) and \(\displaystyle v_2 \in V_2\)

Now \(\displaystyle L( (u_1, u_2)) = u_1 + u_2
\)
and \(\displaystyle L( (v_1, v_2)) = v_1 + v_2\)

Now if \(\displaystyle L( (u_1, u_2)) = L( (v_1, v_2))
\)
then \(\displaystyle u_1 + u_2 = v_1 + v_2\) where \(\displaystyle u_1, v_1 \in V_1\) and \(\displaystyle u_2, v_2 \in V_2\)

So, then, \(\displaystyle u_1\) must equal \(\displaystyle v_1\) and \(\displaystyle u_2\) must equal \(\displaystyle v_2\), from which it follows that \(\displaystyle (u_1, u_2) = (v_1, v_2)\) and thus \(\displaystyle L\) is one-one.

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Can someone please confirm that my reasoning is correct? (or otherwise point out the errors/shortcomings ...)

Would appreciate some help

Peter

 

[Euge]

Your proofs are almost solid! For your first question, the only thing I criticize is what you write after "BUT..."; it would be more precise to say that $0 = 0 + 0$, so uniqueness forces $v = v' = 0$, which is equivalent to $v = 0$. In fact, in problems that require to show that a sum of vector spaces is direct, you usually show that if a sum of vectors is zero, then each vector is zero.

For your second question, there are two remarks I'd like to make. First, remember that if $L$ is a linear map of vector spaces, then $L$ is 1-1 if and only if ker($L$) = 0. Since your $L$ in this case is linear, it would be enough to show that $L(v_1, v_2) = 0$ implies $(v_1, v_2) = 0$. Second, just like in your first question, the only issue I have is at the end of your proof. In this one, you make the claim that $u_1 + u_2 = v_1 + v_2$ implies $u_1 = v_1$ and $u_2 = v_2$. The question is, how did you deduce that? Although the conclusion is correct, a reader going through your proof will most likely ask the same question. You didn't mention where you used $V_1\cap V_2 = 0$. Here's a way fix this. Note that $u_1 + u_2 = v_1 + v_2$ implies $u_1 - v_1 = v_2 - u_2$. Since $u_1, v_1\in V_1$, the difference $u_1 - v_1 \in V_1$; similarly, $v_2 - u_2\in V_2$. Since $u_1 - v_1$ and $v_2 - u_2$ are equal, this implies that both of them belong to $ V_1 \cap V_2$. But $V_1 \cap V_2 = 0$, which implies $u_1 - v_1 = 0$ and $v_2 - u_1 = 0$. Therefore $u_1 = v_1$ and $u_2 = v_2$.

 

[Math Amateur]

Your proofs are almost solid! For your first question, the only thing I criticize is what you write after "BUT..."; it would be more precise to say that $0 = 0 + 0$, so uniqueness forces $v = v' = 0$, which is equivalent to $v = 0$. In fact, in problems that require to show that a sum of vector spaces is direct, you usually show that if a sum of vectors is zero, then each vector is zero.

For your second question, there are two remarks I'd like to make. First, remember that if $L$ is a linear map of vector spaces, then $L$ is 1-1 if and only if ker($L$) = 0. Since your $L$ in this case is linear, it would be enough to show that $L(v_1, v_2) = 0$ implies $(v_1, v_2) = 0$. Second, just like in your first question, the only issue I have is at the end of your proof. In this one, you make the claim that $u_1 + u_2 = v_1 + v_2$ implies $u_1 = v_1$ and $u_2 = v_2$. The question is, how did you deduce that? Although the conclusion is correct, a reader going through your proof will most likely ask the same question. You didn't mention where you used $V_1\cap V_2 = 0$. Here's a way fix this. Note that $u_1 + u_2 = v_1 + v_2$ implies $u_1 - v_1 = v_2 - u_2$. Since $u_1, v_1\in V_1$, the difference $u_1 - v_1 \in V_1$; similarly, $v_2 - u_2\in V_2$. Since $u_1 - v_1$ and $v_2 - u_2$ are equal, this implies that both of them belong to $ V_1 \cap V_2$. But $V_1 \cap V_2 = 0$, which implies $u_1 - v_1 = 0$ and $v_2 - u_1 = 0$. Therefore $u_1 = v_1$ and $u_2 = v_2$.

Thanks Euge ... thanks in particular for the critique/commentary on the proofs ... it is really helpful to me as I try to get a full understanding of the concepts I am dealing with and a full understanding of a rigorous proof for each Proposition/Theorem ...

Just reflecting on what you have said ...

Thanks again ...

Peter

 

[Math Amateur]

Your proofs are almost solid! For your first question, the only thing I criticize is what you write after "BUT..."; it would be more precise to say that $0 = 0 + 0$, so uniqueness forces $v = v' = 0$, which is equivalent to $v = 0$. In fact, in problems that require to show that a sum of vector spaces is direct, you usually show that if a sum of vectors is zero, then each vector is zero.

For your second question, there are two remarks I'd like to make. First, remember that if $L$ is a linear map of vector spaces, then $L$ is 1-1 if and only if ker($L$) = 0. Since your $L$ in this case is linear, it would be enough to show that $L(v_1, v_2) = 0$ implies $(v_1, v_2) = 0$. Second, just like in your first question, the only issue I have is at the end of your proof. In this one, you make the claim that $u_1 + u_2 = v_1 + v_2$ implies $u_1 = v_1$ and $u_2 = v_2$. The question is, how did you deduce that? Although the conclusion is correct, a reader going through your proof will most likely ask the same question. You didn't mention where you used $V_1\cap V_2 = 0$. Here's a way fix this. Note that $u_1 + u_2 = v_1 + v_2$ implies $u_1 - v_1 = v_2 - u_2$. Since $u_1, v_1\in V_1$, the difference $u_1 - v_1 \in V_1$; similarly, $v_2 - u_2\in V_2$. Since $u_1 - v_1$ and $v_2 - u_2$ are equal, this implies that both of them belong to $ V_1 \cap V_2$. But $V_1 \cap V_2 = 0$, which implies $u_1 - v_1 = 0$ and $v_2 - u_1 = 0$. Therefore $u_1 = v_1$ and $u_2 = v_2$.

Thanks again Euge ... just a clarification ... ...

You write:

" ... ... the only thing I criticize is what you write after "BUT..."; it would be more precise to say that $0 = 0 + 0$, so uniqueness forces $v = v' = 0$, which is equivalent to $v = 0$. ... ... "

Can you be explicit about why exactly $0 = 0 + 0$ and uniqueness forces $v = v' = 0$ and hence $v = 0$?

What is the exact formal argument?

Peter

 

[Euge]

Thanks again Euge ... just a clarification ... ...

You write:

" ... ... the only thing I criticize is what you write after "BUT..."; it would be more precise to say that $0 = 0 + 0$, so uniqueness forces $v = v' = 0$, which is equivalent to $v = 0$. ... ... "

Can you be explicit about why exactly $0 = 0 + 0$ and uniqueness forces $v = v' = 0$ and hence $v = 0$?

What is the exact formal argument?

Peter

What I had was a formal argument, but I'll put in more detail. In $V_1 \oplus V_2$, if $u_1 + u_2 = v_1 + v_2$ with $u_1,\, v_1 \in V_1$ and $u_2,\, v_2\in V_2$, then $u_1 = v_1$ and $u_2 = v_2$. This is what I mean by uniqueness. Since $0\in V_1,\, 0\in V_2$, $0 = 0 + 0$ and $0 = v + v'$, we must have $v = 0$.