- #1

iJake

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⇒

[/B]

Calculate ##S + T## and determine if the sum is direct for the following subspaces of ##\mathbf R^3##

a) ## S = \{(x,y,z) \in \mathbf R^3 : x =z\}##

## T = \{(x,y,z) \in R^3 : z = 0\}##

b) ## S = \{(x,y,z) \in \mathbf R^3 : x = y\}##

## T = \{(x,y,z) \in \mathbf R^3 : x = y = z\}##

c) ## S = \{(x,y,z) \in \mathbf R^3 : x = y = 0\}##

## T = \{(x,y,z) \in \mathbf R^3 : x = z\}##

Not much to say here.

OK, so in this first case, I will first include my own approach to the problem (using a method I reached with someone who was helping me) and then I will include a classmate's approach which seems simpler. My own approach:

Note that I use B to refer to the Basis of a subspace

##v \in S+T \iff \exists s \in S, t \in T | v = s + t##

##S : (x,y,z) = (x,y,x) = x \cdot (1,0,1) + y \cdot (0,1,0)##

##B_S = \{(1,0,1) , (0,1,0)\}##

##T : (x,y,z) = (x,y,0) = x \cdot (1,0,0) + y \cdot (0,1,0)##

##B_T = \{(1,0,0) , (0,1,0)\}##

To find their intersection, ##v \in S∩T ⇒ ##

##v \in S ⇒ v = a \cdot (1,0,1) + b \cdot (0,1,0)##

##v \in T ⇒ v = α \cdot (1,0,0) + β \cdot (0,1,0)##

##a \cdot (1,0,1) + b \cdot (0,1,0) = α \cdot (1,0,0) + β \cdot (0,1,0)##

## b = \alpha = 0, a = \beta##

As ##b=0 ⇒ v = a \cdot (0,1,0)##, the span of ##v## is ##(0,1,0) \in S∩T ⇒## the sum is not direct

My classmate's approach (I assume starting here without repeating the conditions of the problem will be fine):

##(a,b,c) = (x,y,x) + (x', y', 0)##

##x + x' = a ⇒ x = a - c##

##y + y' = b ⇒ y = b - y'##

##x = c ⇒ x = c##

##\forall (a,b,c) \in \mathbf R^3##

##(a,b,c) = (c, b-y', c) + (a-c, y', 0)##

##S+T = \mathbf R^3##

He then writes that the term "b-y'" indicates the sum is not direct. I do not have a strong grasp on the differences between our methods, besides the fact that I took the sets down to their basis and worked from there.

Now, I will try to be a bit more agile with my argument and avoid repeating trivial steps,

b) ##B_S = \{(1,1,0) , (0,0,1)\}##

##B_T = \{(1,1,1)\}##

##v \in S∩T ⇒ ##

##v \in S ⇒ v = a \cdot (1,1,0) + b \cdot (0,0,1)##

##v \in T ⇒ v = \alpha \cdot (1,1,1)##

##a \cdot (1,1,0) + b \cdot (0,0,1) = \alpha \cdot (1,1,1)##

##(a,a,b) = (α,α,α) ⇒ a = b = α##

##S + T = T##

Not quite sure about my conclusion here, but I think it also indicates the sum is not direct.

c) ##B_S = \{(0,0,1)\}##

##B_T = \{(1,0,1) , (0,1,0)\}##

##v \in S∩T ⇒##

##v \in S ⇒ v = a \cdot (0,0,1)##

##v \in T ⇒ v = \alpha \cdot (1,0,1) + \beta (0,1,0)##

But I'm a little confused here. What does this last result mean, if it implies that all my variables are 0?

## Homework Statement

[/B]

Calculate ##S + T## and determine if the sum is direct for the following subspaces of ##\mathbf R^3##

a) ## S = \{(x,y,z) \in \mathbf R^3 : x =z\}##

## T = \{(x,y,z) \in R^3 : z = 0\}##

b) ## S = \{(x,y,z) \in \mathbf R^3 : x = y\}##

## T = \{(x,y,z) \in \mathbf R^3 : x = y = z\}##

c) ## S = \{(x,y,z) \in \mathbf R^3 : x = y = 0\}##

## T = \{(x,y,z) \in \mathbf R^3 : x = z\}##

## Homework Equations

Not much to say here.

## The Attempt at a Solution

OK, so in this first case, I will first include my own approach to the problem (using a method I reached with someone who was helping me) and then I will include a classmate's approach which seems simpler. My own approach:

Note that I use B to refer to the Basis of a subspace

##v \in S+T \iff \exists s \in S, t \in T | v = s + t##

##S : (x,y,z) = (x,y,x) = x \cdot (1,0,1) + y \cdot (0,1,0)##

##B_S = \{(1,0,1) , (0,1,0)\}##

##T : (x,y,z) = (x,y,0) = x \cdot (1,0,0) + y \cdot (0,1,0)##

##B_T = \{(1,0,0) , (0,1,0)\}##

To find their intersection, ##v \in S∩T ⇒ ##

##v \in S ⇒ v = a \cdot (1,0,1) + b \cdot (0,1,0)##

##v \in T ⇒ v = α \cdot (1,0,0) + β \cdot (0,1,0)##

##a \cdot (1,0,1) + b \cdot (0,1,0) = α \cdot (1,0,0) + β \cdot (0,1,0)##

## b = \alpha = 0, a = \beta##

As ##b=0 ⇒ v = a \cdot (0,1,0)##, the span of ##v## is ##(0,1,0) \in S∩T ⇒## the sum is not direct

My classmate's approach (I assume starting here without repeating the conditions of the problem will be fine):

##(a,b,c) = (x,y,x) + (x', y', 0)##

##x + x' = a ⇒ x = a - c##

##y + y' = b ⇒ y = b - y'##

##x = c ⇒ x = c##

##\forall (a,b,c) \in \mathbf R^3##

##(a,b,c) = (c, b-y', c) + (a-c, y', 0)##

##S+T = \mathbf R^3##

He then writes that the term "b-y'" indicates the sum is not direct. I do not have a strong grasp on the differences between our methods, besides the fact that I took the sets down to their basis and worked from there.

Now, I will try to be a bit more agile with my argument and avoid repeating trivial steps,

b) ##B_S = \{(1,1,0) , (0,0,1)\}##

##B_T = \{(1,1,1)\}##

##v \in S∩T ⇒ ##

##v \in S ⇒ v = a \cdot (1,1,0) + b \cdot (0,0,1)##

##v \in T ⇒ v = \alpha \cdot (1,1,1)##

##a \cdot (1,1,0) + b \cdot (0,0,1) = \alpha \cdot (1,1,1)##

##(a,a,b) = (α,α,α) ⇒ a = b = α##

##S + T = T##

Not quite sure about my conclusion here, but I think it also indicates the sum is not direct.

c) ##B_S = \{(0,0,1)\}##

##B_T = \{(1,0,1) , (0,1,0)\}##

##v \in S∩T ⇒##

##v \in S ⇒ v = a \cdot (0,0,1)##

##v \in T ⇒ v = \alpha \cdot (1,0,1) + \beta (0,1,0)##

But I'm a little confused here. What does this last result mean, if it implies that all my variables are 0?

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