[Linear Algebra] Sum & Direct Sum of Subspaces

[iJake]

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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?

 

[andrewkirk]

The answer to (a) is correct.

The answer to (b) is correct, but the statement that S+T=T should read S+T=S. Note that S contains T.

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

It means that the two subspaces only intersect at 0, so the sum is direct.

You haven't said what the sum is in (c). What is it?

By the way, in 2D and 3D cases it's often easier to work visually than algebraically - although where a formal proof is required, you'd better check whether your lecturer will accept a visual proof. To work visually, form a visual image of each subspace. In the above cases, the subspaces are
- two planes for (a)
- a line and a plane for (b)
- a line and a plane for (c)
Regardless of whether you submit a visual or algebraic proof, the visual acts as a check on your answer and can show you what answer you need to find an algebraic proof for.

The space generated by the sum of a 1D space (a line) and a 2D space (a plane) will be a 3D space unless the plane contains the line, in which case it will just be that plane. The space generated by two 2D spaces (planes) will be at least 3D unless the spaces are identical.
Since the only 3D space contained in $\mathbb R^3$ iis $\mathbb R^3$ itself, that must be the answer in such cases.

The sum will be direct unless the line contains the plane, because a line that is not contained in a plane can intersect that plane at at most one point, and that point must be zero, since all subspaces intersect at zero.

 

[iJake]

Thank you for your excellent answer! For (c), would it be true that the sum is $\mathbb R^3$?

Also, would it be true that I can also determine whether or not the sum is direct using the cardinality of the Basis I calculated for each subspace? That is, if $\# B > dim \mathbb R^n$, I can immediately know that the sum is not direct.

Thank you again.

 

[MidgetDwarf]

I think Axler: Linear Algebra Done Right, explains this topic very well.

 

[iJake]

I'll check it out, thanks!

 

[andrewkirk]

Thank you for your excellent answer! For (c), would it be true that the sum is $\mathbb R^3$?

Also, would it be true that I can also determine whether or not the sum is direct using the cardinality of the Basis I calculated for each subspace? That is, if $\# B > dim \mathbb R^n$, I can immediately know that the sum is not direct.

Thank you again.

There are two bases: one each for S and T. If the cardinalities add to more than 3 the sum cannot be direct. That enables us to conclude in (a) that the sum is not direct. But the converse does not apply. It is possible for the sum to not be direct even though the sum of basis cardinalities is 3 or less. Case (b) is an example of that.

 

[fresh_42]

For (c), would it be true that the sum is $\mathbb R^3$?

Have you tried to visualize this case? One possibility is to us a coordinate system like this one:

E.g. if you have a plane and a straight, then the question is, whether we can find three linear independent vectors, two in the plane, one for the straight, which are not necessarily orthogonal anymore. As all are expected to be subspaces, they all have the origin in common, which rules out that the straight and the plane can be in parallel. So the only question is, whether the plane contains the straight or not. Similar considerations can be made for two planes or two straights. I recommend to draw your examples, to see how the algebraic and the geometric descriptions relate to each other.