# Subspace and basis

1. Oct 27, 2012

### pyroknife

The problem is attached.
I'm having problems with parts a and c, well maybe not part a (probably just need to check if I did this part right. I'm just not sure if I'm wording part a right.

Anyways for part a I must prove it's a subspace so I must satisfy 3 conditions:
1) 0 is in S
2) if U and V are in S, then U+V must be in S
3) if V is in S, then fV is in S for some scalar f.

for 1)
By inspection if a=b=c=0 then 0 is in S

for 2)
if U is of the form:
a1-b1 a1
b1+c1 a1-c1
and V is of the form:
a2-b2 a2
b2+c2 a2-c2
then U+V=
a1+a2-b1-b2 a1+a2
b1+b2+c1+c2 a1+a2-c1-c2

Thus U+V is in S. <<< Can I say this?

for 3)
fV=
f(a2-b2) f(a2)
f(b2+c2) f(a2-c2)

Thus fV is in S. <<< Can I say this?

For part c, I don't even know where to begin. Can someone give me a hint?

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Last edited: Oct 27, 2012
2. Oct 27, 2012

### LCKurtz

For both of those I would suggest a little more. For example, for 3 I would multiply the f into the expressions and then add: Let a = fa2, b =fb2, c =fc2 and show you get the form given in the problem.
Set the expressions in the given matrix with a,b,c 's equal to the given matrix and see if the equations are consistent. You have four equations in 3 unknowns so it may or may not work.

3. Oct 27, 2012

### pyroknife

I skipped a step in there (factored) because I was being lazy about typing it out on here.
for 3)
fV=
fa2-fb2 fa2
fb2+fc2 fa2-fc2
=
f(a2-b2) f(a2)
f(b2+c2) f(a2-c2)
which is of the same form as S.

On one of my previous assignments, this is what I said (applied to the context of this problem):

For 2) Since a1 and b1 are both in R, then a1, b1, c1, a2, b2, c2 are in R, then a1+a2-b1-b2, a1+a2, b1+b2+c1+c2, a1+a2-c1-c2 is still in R.

For 3)

since a2, b2, c2 are in R, then fa2-fb2, fa2, fb2+fc2, fa2-fc2 is still in R.

Sorry, I meant I didn't know how to do part "c" not part "b."

4. Oct 27, 2012

### LCKurtz

Technically, no, it isn't. The first one is if you let a = fa2, b= fb2 and c =fc2

Write out your matrix$$\begin{bmatrix} a-b & a\\b+c & a-c \end{bmatrix}$$as the sum of matrices, where the first one contains only the a terms, the second only b terms and the third only the c terms. Then factor the letters out of each and see whether the matrices you have are linearly independent.

5. Oct 27, 2012

### Clandry

oops

6. Oct 27, 2012

### pyroknife

Sorry, what do you mean by that?

Thanks.

7. Oct 27, 2012

### LCKurtz

I mean if you let a = fa2, b= fb2 and c =fc2 in your matrix above in blue, it will be in the form
$$\begin{bmatrix} a-b & a\\b+c & a-c \end{bmatrix}$$
but the one in red wouldn't be. Additionally, when writing up your argument you should have it written that fV = the red one = the blue one, in that order and then note the blue one is in the right form.

8. Oct 28, 2012

### pyroknife

Yeah I i screwed that order up, but I am still not sure why the blue, rather than the red, resembles the original form more. Like if I were to look at the given S it seems that red would be more intuitive.

On second thought, I can see the blue resembling it pretty closely as well.

But how do you decide which one is better? It seems other one would work. Would it have been wrong if i had it in the red form where the f is factored from the terms?

9. Oct 28, 2012

### LCKurtz

I would certainly take points off for that. Look at the upper left element. In the blue example it is a difference of two things. In the red one it is a product of two things. While they represent the same number, they are not the same form of expression. In fact, it is the crux of the argument that when multiplying the matrix by f, which multiplies into each element and gives the red matrix, it can be rearranged by the distributive law to get the blue matrix, which gives the correct form to be an element of the subspace. That step should not be glossed over or misunderstood.

You should also note that although I have used (3) in our discussion, many of the same criticisms apply to your proof for (2) and also need attention.

Last edited: Oct 28, 2012