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

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?

#### Attachments

• 8.3 KB Views: 360

Related Calculus and Beyond Homework Help News on Phys.org
vela
Staff Emeritus
Homework Helper

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?
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.

$$\vec{u} + \vec{v} = \begin{bmatrix} 2x - y \\ x \\ y \\ -x \end{bmatrix} + \begin{bmatrix} 2a - b \\ a \\ b \\ -a \end{bmatrix}.$$ 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.

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
Homework Helper

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.

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
Staff Emeritus
Homework Helper

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.

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

Is it asking if S is in R^4?

vela
Staff Emeritus
Homework Helper

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?

Hmmm. I think the answer is no.

c1[2 1 0 -1]^t + c2[-1 0 1 0]=[a b c d]^t

Writing 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 d

Row 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
Staff Emeritus
Homework Helper

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.

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

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
Staff Emeritus
Homework Helper