Showing that a set of linear algebra statements are true

trap101
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V is the direct sum of W1 and W2 in symbols:

V = W1 (+) W2 if:

V = W1 + W2 and
W1 \cap W2 = {0}

Show that the following statements are equivalent:

1. V = W1 (+) W2

2. Every vector v\inV can be written uniquely as w1 + w2 where w1\inW1 and w2\inW2

3. V = W1 + W2 and for vectors w1\inW1 and w2\inW2 if w1 +w2 = 0 then w1 = w2

4. If \alpha1 is a basis for W1 and \alpha2 is a basis for W2 then
\alpha = \alpha1 \cup \alpha2

Attempt:

I'm thrown off here because I'm given the definition of a direct sum. So how can I even show that all these statements are equivalent? Would I start by assuming one of the statements is true? For example: Let's say statement 1 is true. that means...well it means exactly what statement 2 is. Maybe take a vector in W1 and one in W2 sum them together and create a vector in V? But then to do that I would have to define an exact vector space...
 
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trap101 said:
V is the direct sum of W1 and W2 in symbols:

V = W1 (+) W2 if:

V = W1 + W2 and
W1 \cap W2 = {0}

Show that the following statements are equivalent:

1. V = W1 (+) W2

2. Every vector v\inV can be written uniquely as w1 + w2 where w1\inW1 and w2\inW2

3. V = W1 + W2 and for vectors w1\inW1 and w2\inW2 if w1 +w2 = 0 then w1 = w2

4. If \alpha1 is a basis for W1 and \alpha2 is a basis for W2 then
\alpha = \alpha1 \cup \alpha2

Attempt:

I'm thrown off here because I'm given the definition of a direct sum. So how can I even show that all these statements are equivalent? Would I start by assuming one of the statements is true? For example: Let's say statement 1 is true. that means...well it means exactly what statement 2 is. Maybe take a vector in W1 and one in W2 sum them together and create a vector in V? But then to do that I would have to define an exact vector space...

Start by assuming that statement 1 is true. Then show that this implies that statement 2 must be true, which you can do by invoking the definition of a direct sum.

Next show that statement 2 being true implies that statement 3 is true. Then show that statement 3 being true implies that statement 4 is true. Finally, show that statement 4 implies statement 1.

You don't have to define a vector space. You don't mention it, but it seems to me that V is some arbitrary vector space and that W1 and W2 are subspaces of V.
 
Mark44 said:
Start by assuming that statement 1 is true. Then show that this implies that statement 2 must be true, which you can do by invoking the definition of a direct sum.

Next show that statement 2 being true implies that statement 3 is true. Then show that statement 3 being true implies that statement 4 is true. Finally, show that statement 4 implies statement 1.

You don't have to define a vector space. You don't mention it, but it seems to me that V is some arbitrary vector space and that W1 and W2 are subspaces of V.


But then what is it that I'm really doing beyond repeating the definition of a direct sum? Because it seems that those each just logically follow each other. Yes it is suppose to be an arbitrary vector space.
 
trap101 said:
But then what is it that I'm really doing beyond repeating the definition of a direct sum? Because it seems that those each just logically follow each other. Yes it is suppose to be an arbitrary vector space.

Yes, they do follow from each other. That's exactly what you are supposed to show. Each statement is different from the definition of direct sum. You have to show why they are equivalent.
 
Dick said:
Yes, they do follow from each other. That's exactly what you are supposed to show. Each statement is different from the definition of direct sum. You have to show why they are equivalent.

Ok, I'm still puzzled at how I can show it. Because if I didn't have the 4 statements for example. Starting with the definition of a direct sum, I would've just said that the definintion of a direct sum implies that every v\in V can be written uniquely as a combination of w1 and w2. I don't see how I'm "showing" this though.
 
trap101 said:
Ok, I'm still puzzled at how I can show it. Because if I didn't have the 4 statements for example. Starting with the definition of a direct sum, I would've just said that the definintion of a direct sum implies that every v\in V can be written uniquely as a combination of w1 and w2. I don't see how I'm "showing" this though.

Show HOW the implication works. You start with v=w1+w2=w1'+w2'. What property of the direct sum let's you say w1=w1' and w2=w2'? This isn't a hard proof by any means, but there is something to say.
 
Dick said:
Show HOW the implication works. You start with v=w1+w2=w1'+w2'. What property of the direct sum let's you say w1=w1' and w2=w2'? This isn't a hard proof by any means, but there is something to say.



Well the only properties I have are that the intersection of the two sets is {0}. Also each vector in the subspace is unique. I suppose by uniqueness:

there exists a unique vector in w1 and w2 s.t v is the combination of them?...Still not seeing it.
 
trap101 said:
Well the only properties I have are that the intersection of the two sets is {0}. Also each vector in the subspace is unique. I suppose by uniqueness:

there exists a unique vector in w1 and w2 s.t v is the combination of them?...Still not seeing it.

You don't HAVE uniqueness, you want to SHOW uniqueness. If v=w1+w2=w1'+w2' (w1, w1' in W1 and w2, w2' in W2) then w1-w1'=w2'-w2. So?
 
Dick said:
You don't HAVE uniqueness, you want to SHOW uniqueness. If v=w1+w2=w1'+w2' (w1, w1' in W1 and w2, w2' in W2) then w1-w1'=w2'-w2. So?


I'm going to go out on a limb and say 0=0, but I don't think that would make sense considering I'm trying to show uniqueness so I can't assume w1-w1' = 0 and the same for w2. And so my struggle continues with this sort of thing being my impediment from obtaining A's. sigh.
 
  • #10
trap101 said:
I'm going to go out on a limb and say 0=0, but I don't think that would make sense considering I'm trying to show uniqueness so I can't assume w1-w1' = 0 and the same for w2. And so my struggle continues with this sort of thing being my impediment from obtaining A's. sigh.

Which subspace is w1-w1' in? Same question for w2-w2'.
 
  • #11
Dick said:
Which subspace is w1-w1' in? Same question for w2-w2'.

w1-w1' is in W1 and w2-w2' is in W2. ==> W1 = W2 but that doesn't do anything.

unless:

v = (w1-w1') + (w2 - w2')
 
  • #12
trap101 said:
w1-w1' is in W1 and w2-w2' is in W2. ==> W1 = W2 but that doesn't do anything.

unless:

v = (w1-w1') + (w2 - w2')

So call u=w1-w1'=w2'-w2. From what you have said that must mean u is in W1 and W2. What's the only vector that's both in W1 and W2?
 
  • #13
Dick said:
So call u=w1-w1'=w2'-w2. From what you have said that must mean u is in W1 and W2. What's the only vector that's both in W1 and W2?

the only vector in W1 and W2 is the 0 vector. I'm still puzzled as to how this shows statement 2.
 
  • #14
trap101 said:
the only vector in W1 and W2 is the 0 vector. I'm still puzzled as to how this shows statement 2.

So u=0. Then w1-w1'=0 and w2'-w2=0. Remember what you want to prove?
 
  • #15
Dick said:
So u=0. Then w1-w1'=0 and w2'-w2=0. Remember what you want to prove?

As far as I remember I'm trying to prove statement 2. Then with that I'll prove statement 3, etc. But the fact that u = 0, how is this showing statement 2? I'm trying to go over the steps but I'm failing to see the connection
 
  • #16
trap101 said:
As far as I remember I'm trying to prove statement 2. Then with that I'll prove statement 3, etc. But the fact that u = 0, how is this showing statement 2? I'm trying to go over the steps but I'm failing to see the connection

Don't worry about the parts yet. You are trying to show 1)->2). You want to show that if v=w1+w2=w1'+w2', then w1=w1' and w2=w2'. I think you are practically there is if you string all the stuff we've been doing together.
 
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