MHB Direct Sum of n Vector Spaces Over F - Knapp Proposition 2.31 - Pages 61-62

[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 2.31 (regarding the direct sum of n vector spaces) on pages 61-62.

Theorem 2.31 and its accompanying text read as follows:
View attachment 2928
View attachment 2929
Question 1

In the text above (under the statement of the Proposition) we find the following statement:

"... ... Notice that the second condition in (b) is stronger than the condition that $$V_i \cap V_j = 0 $$ for all $$i \ne j$$. ... ... "

I am unable to demonstrate that this is the case ... ... Can someone please explain exactly why the second condition in (b) is stronger than the condition that $$V_i \cap V_j = 0 $$ for all $$i \ne j$$?Question 2

In the text above (under the statement of the Proposition) we find the following statement:

"... ... Figure 2.3 illustrates how the condition that $$V_i \cap V_j = 0 $$ for all $$i \ne j$$ can be satisfied even though (b) is not satisfied and even though the vector subspaces do not therefore form a direct sum ... ... "

Again, I do not follow this at all ... can someone please explain how Figure 2.3 illustrates the condition mentioned but does not meet (b) etc ...?

I would really appreciate some clarification ... ...

Peter***EDIT***

Just thinking some more about question 2 above, if the lines in Figure 2.3 are the 1-dimensional spaces $$V_1, V_2 \text{ and } V_3$$, then I guess it is actually clear (how did I miss it ... ... :-( ... ... ) from the diagram that $$V_i \cap V_j = 0 $$ for all $$i \ne j$$ since each pair only meet in one point ... but then is it really $$V_i \cap V_j = 0 $$ or is it $$V_i \cap V_j = {X} $$ where X is the point of intersection ... and why do the second condition of (b) fail? (unless of course that Knapp is saying that (b) fails because the first condition of (b) is not met since $$V_1 + V_2 + V_3 \ne V$$?

 

[Euge]

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 2.31 (regarding the direct sum of n vector spaces) on pages 61-62.

Theorem 2.31 and its accompanying text read as follows:
View attachment 2928
View attachment 2929
Question 1

In the text above (under the statement of the Proposition) we find the following statement:

"... ... Notice that the second condition in (b) is stronger than the condition that $$V_i \cap V_j = 0 $$ for all $$i \ne j$$. ... ... "

I am unable to demonstrate that this is the case ... ... Can someone please explain exactly why the second condition in (b) is stronger than the condition that $$V_i \cap V_j = 0 $$ for all $$i \ne j$$?Question 2

In the text above (under the statement of the Proposition) we find the following statement:

"... ... Figure 2.3 illustrates how the condition that $$V_i \cap V_j = 0 $$ for all $$i \ne j$$ can be satisfied even though (b) is not satisfied and even though the vector subspaces do not therefore form a direct sum ... ... "

Again, I do not follow this at all ... can someone please explain how Figure 2.3 illustrates the condition mentioned but does not meet (b) etc ...?

I would really appreciate some clarification ... ...

Peter***EDIT***

Just thinking some more about question 2 above, if the lines in Figure 2.3 are the 1-dimensional spaces $$V_1, V_2 \text{ and } V_3$$, then I guess it is actually clear (how did I miss it ... ... :-( ... ... ) from the diagram that $$V_i \cap V_j = 0 $$ for all $$i \ne j$$ since each pair only meet in one point ... but then is it really $$V_i \cap V_j = 0 $$ or is it $$V_i \cap V_j = {X} $$ where X is the point of intersection ... and why do the second condition of (b) fail? (unless of course that Knapp is saying that (b) fails because the first condition of (b) is not met since $$V_1 + V_2 + V_3 \ne V$$?

Hi Peter,

If the second condition in (b) holds, then consider the intersection $ V_i\cap V_j $ where $i \neq j $. Note that $V_j$ may be viewed as a subspace of

$ V_1 + \cdots + V_{i-1} + V_{i+1} + \cdots + V_n$.

So $V_i \cap V_j$ may be viewed as a subspace of

$V_i \cap (V_1 + \cdots + V_{i-1} + V_{i+1} + \cdots + V_n) $.

Since the above term is zero, so is $ V_i \cap V_j $. This answers your fIrst question.

In Figure 2.3, each line passes through the origin (in fact they must since vector spaces must contain $0$). Consider two of the three lines. Each line is spanned by a vector from the origin, and the two vectors are linearly independent (since the lines are not parallel). So the two vectors span the plane. This means that the sum of the two lines is R^2. So the third line has nontrivial intersection with the sum of the two lines. This answers your second question.

 

[Math Amateur]

Hi Peter,

If the second condition in (b) holds, then consider the intersection $ V_i\cap V_j $ where $i \neq j $. Note that $V_j$ may be viewed as a subspace of

$ V_1 + \cdots + V_{i-1} + V_{i+1} + \cdots + V_n$.

So $V_i \cap V_j$ may be viewed as a subspace of

$V_i \cap (V_1 + \cdots + V_{i-1} + V_{i+1} + \cdots + V_n) $.

Since the above term is zero, so is $ V_i \cap V_j $. This answers your fIrst question.

In Figure 2.3, each line passes through the origin (in fact they must since vector spaces must contain $0$). Consider two of the three lines. Each line is spanned by a vector from the origin, and the two vectors are linearly independent (since the lines are not parallel). So the two vectors span the plane. This means that the sum of the two lines is R^2. So the third line has nontrivial intersection with the sum of the two lines. This answers your second question.

Thanks Euge ... that is an extremely clear and helpful post ... much appreciated ...

Peter