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\}$
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 $0$?
Reminder: a sum $S+T$ is direct if (by definition) $S\cap T=\{0\}$. Equivalently, $S+T$ is direct if and only if $\dim(S+T)=\dim(S)+\dim(T)$.
For the first case, $ S $ and $ T $ are of dimension $2 $, if they were in direct sum they would span a subspace of dimension $4 $, in $\Bbb R^3 $, that's impossible.
For the second case, $T$ is a line contained in the plane $S$. The sum is never direct in such a case, unless the subspace contained in the other is trivial.
In the last case, you were able to prove that the intersection is $\{0\}$, so $S$ and $T$ are in direct sum (note that since $S$ is a line, it is in direct sum with a subspace $W$ if and only if it is not contained in $W$).