MHB Isomorphism Between External and Internal Direct Sums - Knapp Proposition 2.30

[Math Amateur]

I am reading Chapter 2: Vector Spaces over $$\mathbb{Q}, \mathbb{R} \text{ and } \mathbb{C}$$ 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:
View attachment 2924
View attachment 2925

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

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

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

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

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

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

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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 $$L$$) is one-one, suppose that $$v_1 + v_2 = 0$$. Then $$v_1 = - v_2$$ shows that $$v_1$$ is in $$V_1 \cap V_2$$. By (b), this intersection is $$0$$. Therefore $$v_1 = v_2 = 0$$, 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 $$L$$ such that

$$L( (v_1, v_2)) = v_1 + v_2$$

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

in the above, we have:

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

and

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

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

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

So, then, $$u_1$$ must equal $$v_1$$ and $$u_2$$ must equal $$v_2$$, from which it follows that $$(u_1, u_2) = (v_1, v_2)$$ and thus $$L$$ 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 $v = 0$.