Linear algebra

  • Thread starter hartigan83
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  • #1
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I am in a problem seminar class and I have not taken Linear Algebra in over 4 years so I am having alot of problems with this. Please help....:eek:

Homework Statement


Let P be the set of all polynomials with real coefficients and of degree less than 3. Thus,
P = {f:f(x)= a(sub0) +a(sub1)x +a(sub2)x^2, a(sub i ) is in the reals}
P is a vector spaces over the field of Reals under the usual opperations of addition and scalar multiplication of polynomials.

Let V = {f element of P : f(-2) =f(1)}

Find a basis for V .

Homework Equations



Not sure

I know that to be a basis, that the set must be linearly independent and span V.:confused:

The Attempt at a Solution



I proved in the first part of this problem that V is a subspace of P :smile:

I also said that in order for f(-2) = f(1),
a(sub0) - 2a(sub1) +4a(sub2) = a(sub0) + a(sub1) + a(sub2)
-3a(sub1) + 3a(sub2) so
a(sub1) = a(sub2)
not sure if this is right or not....
so I randomly chose 2 different elements of this set. and showed that they were linearly independent.
the elements I chose were v(sub 1) (x)= 1 + 3x +3x^2 and v(sub 2 ) (x) = 0 +1x + 1x^2
Now I found a theorem in my linear algebra book that says;
"Let H be a subspace of a finite-dimentional vector space V. Any linearly independent set in H can be expanded, if necessary to a basis for H. Also H is finit dimentional and dim H < or = dim V "

Now I know that dim P = 3
Thus dim V < or = 3. and since the set of two numbers I used above were linearly independent, they should be able to be expanded to form the basis, but my text does not tell me how to do this....:frown:

Am I going about this wrong? I dont really understand how to find span V either.

Please point me in the right direction!!!
 
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Answers and Replies

  • #2
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hartigan83,

Your answer is excellent.
I think you should, as a further exercice, take an (any) element of V and see how you can express it as a linear combinaition of v(sub 1) and v(sub 2).

In other words: find out a and b such that

v(given) = a v(sub 1) + b v(sub 2) (for any x)

Michel
 
  • #3
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I would start representing the polynomials by the coefficients

So you are getting the space to be (represented by) the subspace of vectors

(x,y,z) where x=y. Find a basis of this space.
 
Last edited:
  • #4
HallsofIvy
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The set of all polynomials of degree less that or equal to 3 is
[tex]\{ax^3+ bx^2+ cx+ d\}[/tex]
which has dimension 4. As gammamc said, you can think of that as simply {(a,b,c,d)}. Your subspace is defined by the single equation f(-2)= f(1) which should reduce the dimension by 1: your subspace has dimension 3 so you are looking for 3 basis vectors.
f(-2)= -8a+ 4b-2c+ d and f(1)= a+ b+ c+ d. The requirement that f(-2)= f(1) is -8a+ 4b- 2c+ d= a+ b+ c+ d or -9a+ 3b- 3c= 0 which reduces to -3a+ b- c= 0 or c= -3a+ b. Here's a method I like to use: take each coefficient equal to 1 in turn, the other 0 if possible and use that to solve for the remaining coefficients. Here if we know a and b, we can solve for c:
If a= 1, b= 0, d= 0, what is c? If a= 0, b= 1, d= 0, what is c? If a= 0, b= 0, d= 1, what is c? Those give you the three basis vectors.
 
  • #5
AKG
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The set of all polynomials of degree less that or equal to 3 is
[tex]\{ax^3+ bx^2+ cx+ d\}[/tex]
which has dimension 4.
It's the polynomials of degree less than 3.
 
  • #6
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So you are getting the space to be (represented by) the subspace of vectors

(x,y,z) where x=y. Find a basis of this space.
why (x,y,z) where x=y, wouldny it be y=z?

and I am not really sure how to find a basis other than picking random elements like I did above...
 
  • #7
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hartigan83,

Your answer is excellent.
I think you should, as a further exercice, take an (any) element of V and see how you can express it as a linear combinaition of v(sub 1) and v(sub 2).

In other words: find out a and b such that

v(given) = a v(sub 1) + b v(sub 2) (for any x)

Michel
thanks for the encouragement, but I dont feel like I am getting anywhere closer to finding a basis...:frown:
so pick another random element such as (4,5,5) and find out how to write it in terms of v(sub1) and v(sub2)?

4 + 5x + 25x^2 =a(1 + 3x +3x^2) + b(0 +1x + 1x^2)
Now I see that a=4 since we need to have a constant of 4 so I get
4 + 5x + 25x^2 =4(1 + 3x +3x^2) + b(0 +1x + 1x^2)
4 + 5x + 25x^2 =(4 + 12x +12x^2) + b(0 +1x + 1x^2)
this leaves me with
-7x+13x^2 = b(0+x+x^2)
If b=7, this will get rid of the 7x but it will leave me with 6x^2
If b=13 this will get rid of the 13x^2 but will leave me with -20x...:grumpy:
What did I do?
Does this mean that this new number is also linearly independent? and can be added to the set?
 
  • #8
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It's the polynomials of degree less than 3.
Thanks I was confused for a minute thinking that I was supposed to include degree 3 too.....
 
  • #9
AKG
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P has dimension 3. V is a subspace of P, so it's dimension is less than or equal to 3. (By the way, do you see why V really is a subspace, and not just some subset?) V is clearly a proper subspace, i.e. it is not all of P, because not every element of P has the same value at -2 and at 1, e.g. the polynomial f(x) = x. Hopefully your book also tells you that (for finite dimensional vector spaces) the dimension of V equals 3 iff V = P, otherwise the dimension of V is strictly less than 3. So V has dimension 0, 1, or 2. You've found a linearly independent subset of V containing 2 vectors. Your book should also tell you that this means that the dimension of V must be greater than or equal to 2. By this point, you should be able to deduce that the dimension of V must be exactly 2, and hence your linearly independent pair of vectors must also be a basis for V, i.e. it does span V.

This is sort of an abstract way to show that you've found a basis. The more concrete way would be to show directly that the set you've found is a spanning set, i.e. show exactly how to express an arbitrary vector in V as a linear combination of v1 and v2. Now you've determined, I assume correctly, that V can be defined by the condition a1 = a2, so an arbitrary element of V looks like

b + ax + ax2

which we'll just write as (b,a,a). You have (1,3,3) and (0,1,1). Find c and d such that

(b,a,a) = c(1,3,3) + d(0,1,1)

You get equations:

b = c
a = 3c + d
a = 3c + d

right? Now it's really easy to solve for c and d in terms of a and b, so you're done.
thanks for the encouragement, but I dont feel like I am getting anywhere closer to finding a basis...
so pick another random element such as (4,5,5) and find out how to write it in terms of v(sub1) and v(sub2)?

4 + 5x + 25x^2 =a(1 + 3x +3x^2) + b(0 +1x + 1x^2)
Now I see that a=4 since we need to have a constant of 4 so I get
4 + 5x + 25x^2 =4(1 + 3x +3x^2) + b(0 +1x + 1x^2)
4 + 5x + 25x^2 =(4 + 12x +12x^2) + b(0 +1x + 1x^2)
this leaves me with
-7x+13x^2 = b(0+x+x^2)
If b=7, this will get rid of the 7x but it will leave me with 6x^2
If b=13 this will get rid of the 13x^2 but will leave me with -20x...
What did I do?
Does this mean that this new number is also linearly independent? and can be added to the set?
You made two mistakes here. One is that you started with (4,5,5) but then started working with 4 + 5x + 25x2. (4,5,5) is of course just 4 + 5x + 5x2. The main mistake you made though is that you don't want to find how to express one particular element of V as a linear combination of v1 and v2. You want to show how any arbitrary element can be expressed as a linear combination of your two basis vectors.
 
  • #10
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hartigan83,

I could not undersand clearly the meaning of your last post.
However, maybe this reminder could help:

two polynoms are equal for all values of x
if and only if their coefficients are equal

Your set V of polynoms is defines as {a0+a1x+x1x²}
So, if you take

P1 = 1
P2 = x + x²

you have a basis for V,
and any polynomial in V is a combination of P1 and P2.
 
  • #11
19
0
P has dimension 3. V is a subspace of P, so it's dimension is less than or equal to 3. (By the way, do you see why V really is a subspace, and not just some subset?) V is clearly a proper subspace, i.e. it is not all of P, because not every element of P has the same value at -2 and at 1, e.g. the polynomial f(x) = x. Hopefully your book also tells you that (for finite dimensional vector spaces) the dimension of V equals 3 iff V = P, otherwise the dimension of V is strictly less than 3. So V has dimension 0, 1, or 2. You've found a linearly independent subset of V containing 2 vectors. Your book should also tell you that this means that the dimension of V must be greater than or equal to 2. By this point, you should be able to deduce that the dimension of V must be exactly 2, and hence your linearly independent pair of vectors must also be a basis for V, i.e. it does span V.

This is sort of an abstract way to show that you've found a basis. The more concrete way would be to show directly that the set you've found is a spanning set, i.e. show exactly how to express an arbitrary vector in V as a linear combination of v1 and v2. Now you've determined, I assume correctly, that V can be defined by the condition a1 = a2, so an arbitrary element of V looks like

b + ax + ax2

which we'll just write as (b,a,a). You have (1,3,3) and (0,1,1). Find c and d such that

(b,a,a) = c(1,3,3) + d(0,1,1)

You get equations:

b = c
a = 3c + d
a = 3c + d

right? Now it's really easy to solve for c and d in terms of a and b, so you're done.
thanks soooooo much!
You are right the second way is sooooo much eisier!
 
  • #12
HallsofIvy
Science Advisor
Homework Helper
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It's the polynomials of degree less than 3.
:blushing: One of these days, I really have to learn to read!

Of course, that makes it easier: the set is {ax2+ bx+ c}.
That space has dimension 3 and one equation will reduce that by 1: we are seeking a basis for a subspace of dimension 2 and so 2 independent vectors in that subspace.

f(-1)= f(2) means a- b+ c= 4a+ 2b+ c or 3a+ 3b= 0: b= -a.
Taking a= 1, c= 0 gives, with b= -1, x2- x.
Taking a= 0, c= 1, gives, with b= 0, 1.

The two "vectors", x2- x and 1 form a basis for that subspace: any quadratic polynomial of the form f(x)= a(x2- x)+ c has the property that f(-1)= f(2) and, conversely, any quadratic polynomial having that property can be written in that form.
 

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