I have attached the problem.part a) show that S is a subspace of

[charlies1902]

I have attached the problem.

part a) show that S is a subspace of R4

I have to show the following 3 conditions
0 vector is in S
if U and V are in S, then U+V is in S
If V is in S, then cV is in S where c is a scalar


if s and t=0 which are real #s, then the 0 vector is in R4, thus S is in R4.
If U=[2x-y, x, y, -x]^t and V=[2a-b, a, b, -a]^t then U+V still has four rows, thus S is in R4
If V=[2a-b, a, b, -a]^t, then cV still has four rows, thus S is in R4.

Is that the right procedure for part a?

 

[vela]

I have attached the problem.

part a) show that S is a subspace of R4

I have to show the following 3 conditions
0 vector is in S
if U and V are in S, then U+V is in S
If V is in S, then cV is in S where c is a scalarif s and t=0 which are real #s, then the 0 vector is in R4, thus S is in R4.
If U=[2x-y, x, y, -x]^t and V=[2a-b, a, b, -a]^t then U+V still has four rows, thus S is in R4
If V=[2a-b, a, b, -a]^t, then cV still has four rows, thus S is in R4.

Is that the right procedure for part a?

No, it isn't.

If a vector $\vec{x}$ is in $S$, you know that it can be written as
$$\vec{x} = \begin{bmatrix} 2s - t \\ s \\ t \\ -s \end{bmatrix}$$ for some value of $s$ and $t$. Conversely, if a vector can be written in that form, it is in S.

As you noted, if s=t=0, then $\vec{x} = 0$. Therefore, you can conclude that 0 is in $S$. That's what you wanted to show, not that 0 is in ℝ⁴ nor that $S$ is in ℝ⁴, which doesn't make sense.

Now if you add your $\vec{u}$ and $\vec{v}$, you get
$$\vec{u} + \vec{v} = \begin{bmatrix} 2x - y \\ x \\ y \\ -x \end{bmatrix} + \begin{bmatrix} 2a - b \\ a \\ b \\ -a \end{bmatrix}.$$ Can you show that that sum can be written in the form
$$\begin{bmatrix} 2c - d \\ c \\ d \\ -c \end{bmatrix}$$ for some value of $c$ and $d$? If you can, then you can conclude that $\vec{u}+\vec{v} \in S$.

I'll leave the last part for you to figure out on your own.

 

[charlies1902]

Okay I defined U=(2x-y, x, y, -x)^t, Where x&y are real
V=(2a-b, a, b, -a)^t Where a&b are real
U and V are vectors in the subspace S.

I added U+V=(2x-y+2a-b, x+a, y+b, -x-a)^t = [2(x+a)-(y+b), x+a, y+b, -(x+a)] Which is in the form of S, thus U+V is in S.

Is the above correct so far?Last condition, if V is in S, then cV must be in S.
cV=V=(c2a-cb, ca, cb, -ac)^t
Now, this is where I get stuck again. How does this show that cV is in V? Is it because multiply each component by a scalar c still keeps it in the same form of S?

 

[HallsofIvy]

Okay I defined U=(2x-y, x, y, -x)^t, Where x&y are real
V=(2a-b, a, b, -a)^t Where a&b are real
U and V are vectors in the subspace S.

I added U+V=(2x-y+2a-b, x+a, y+b, -x-a)^t = [2(x+a)-(y+b), x+a, y+b, -(x+a)] Which is in the form of S, thus U+V is in S.

Is the above correct so far?

Yes, that is good.

Last condition, if V is in S, then cV must be in S.
cV=V=(c2a-cb, ca, cb, -ac)^t
Now, this is where I get stuck again. How does this show that cV is in V? Is it because multiply each component by a scalar c still keeps it in the same form of S?

Basic algebra: c2a- cb= 2(ca)- (cb) so (c2a- cb, ca, cb, -ac)= (2x- y, x, y, -x) with x= ca and y= cb.

 

[charlies1902]

I just looked in the answer in the back of the book, they seemed to have done it in a much easier way.

They said S=span{[2 1 0 -1]^t, [-1 0 1 0]}. therefore S is a subspace.

 

[vela]

If you've already proved in your class that the span of a set of vectors is a subspace, then, sure, you can go ahead and use that result.

 

[charlies1902]

Okay thanks. Also what is part d) is S=R^4 asking?

Is it asking if S is in R^4?

 

[vela]

No, it's asking if S is exactly the same as ℝ⁴. Is every vector in S in ℝ⁴ and is every vector in ℝ⁴ in S?

 

[charlies1902]

Hmmm. I think the answer is no.

c1[2 1 0 -1]^t + c2[-1 0 1 0]=[a b c d]^tWriting this in "augmented" matrix form gives (each space denotes a new element in the matrix:
2 -1 a
1 0 b
0 1 c
-1 0 dRow reducing this into RREF gives
1 0 b/2
0 1 2b-a
0 0 c-2b+a
0 0 d+b

Thus S doesn't equal all of R^4 because the system is consistent only if c-2b+a=d+b=0

Is this the right approach?

 

[vela]

That's a valid argument. An aside: note if you solve for everything in terms of b and c, you get
\begin{align*}
a &= 2b-c \\
b &= b \\
c &= c \\
d &= -b
\end{align*} which should look familiar.

If you know about dimensions, you could also make an argument on that ground as well. Or you could find a vector in ℝ⁴ that's not in S. There are a variety of ways to show the two spaces aren't the same.

 

[charlies1902]

Is every vector in S in ℝ⁴ and is every vector in ℝ⁴ in S?

I'm thinking about this statement, although it wasn't asked. In my RREF form of the matrix c-2b+a=0 and b+d=0.
Is it true if I say this:
All of the vectors in S are in R^4, but not all of the vectors in R^4 are in S because c-2b+a=0 and b+d=0 has to be satisfied.

 

[HallsofIvy]

Or you could pick a vector, say (1, 1, 0, 0), that doesn't satisfy those equations and show that it is NOT in the subspace.

 

[vela]

I'm thinking about this statement, although it wasn't asked.

Actually, the statement was what was asked. When you say two sets are equal, it means that everything in one set is in the other. If you can show that one set contains elements not in the other, the sets aren't equal. That's what you're doing here. You're showing that ℝ⁴ contains vectors that aren't in S.

In my RREF form of the matrix c-2b+a=0 and b+d=0.
Is it true if I say this:
All of the vectors in S are in R^4, but not all of the vectors in R^4 are in S because c-2b+a=0 and b+d=0 has to be satisfied.

Yes.