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

• MHB
• Math Amateur
In summary, the conversation discusses issues and questions regarding Theorem 2.30 in Anthony W. Knapp's book Basic Algebra, specifically regarding an isomorphism between external and internal direct sums. The first question asks for confirmation of a reasoning involving the uniqueness of decompositions, while the second question asks for clarification on how to show that a linear map is one-one. The expert suggests some corrections and improvements to the reasoning presented in the conversation, including clarifying the use of $V_1 \cap V_2 = 0$ and showing that $u_1 + u_2 = v_1 + v_2$ implies $u_1 = v_1$ and $u_2 = v_2$
Math Amateur
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MHB
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:
View attachment 2924
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 $$\displaystyle V_1 \cap V_2 = 0$$

-----------------------------------------------------------------

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:

----------------------------------------------------------------

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.

-----------------------------------------------------------------

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

Would appreciate some help

Peter

Last edited:
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 $u_2 = v_2$.

Euge said:
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

Euge said:
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

Peter said:
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 $v = 0$.

## 1. What is the meaning of isomorphism in the context of external and internal direct sums?

Isomorphism refers to a one-to-one correspondence or mapping between two mathematical structures that preserves their structural properties. In the context of external and internal direct sums, isomorphism means that there is a bijective linear map between the two, which preserves their vector space structure.

## 2. What is an external direct sum?

An external direct sum is a way of constructing a new vector space from two or more given vector spaces. It is denoted by ⊕ and combines the elements of the individual vector spaces into a larger, direct sum space.

## 3. What is an internal direct sum?

An internal direct sum is a decomposition of a given vector space into a direct sum of two or more subspaces. It is denoted by ⊕ and represents the direct sum of the subspaces, with no overlap between them.

## 4. How does Knapp Proposition 2.30 relate to isomorphism between external and internal direct sums?

Knapp Proposition 2.30 states that for a given vector space V, if there exists an internal direct sum decomposition of V into subspaces U1, U2,...,Un, then there exists an isomorphism between V and the external direct sum of these subspaces (U1 ⊕ U2 ⊕ ... ⊕ Un).

## 5. What is the significance of isomorphism between external and internal direct sums in linear algebra?

The isomorphism between external and internal direct sums is a fundamental concept in linear algebra as it allows for the study and manipulation of vector spaces through their subspaces. It also provides a useful tool for understanding the structure of vector spaces and their various decompositions, which has applications in many areas of mathematics and physics.

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