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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

" ... ... 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:

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

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 ... ? )

" ... ... 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

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

__In the first paragraph of the proof given above, Knapp writes:__**First Issue/Question**" ... ... 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:

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

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 ... ? )

__In the second paragraph of the proof above, Knapp writes:__**Second Issue/Question**" ... ... 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

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