Determining a subspace of polynomials with degree 3

  • Thread starter cuttlefish
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  • #1
cuttlefish
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Homework Statement


Determine which of the following are subspaces of P3:
a) all polynomials a0+a1x+a2x^2+a3x^3 where a0=0
b) all polynomials a0+a1x+a2x^2+a3x^3 where a0+a1+a2+a3=0
c) all polynomials a0+a1x+a2x^2+a3x^3 for which a0, a1, a2, a3 are integers
d) all polynomials of the form a0+a1x where a0 and a1 are real numbers

Homework Equations


in order for U={v in Z: v=(a0+a1x+a2x^2+a3x^3)} to be a subspace, these polynomials should fulfil:
1) 0 vector included in U
2) x in U, y in U, x+y also in U
3) c (some constant) in U, x in U, c*x also in U

sorry for writing "in", I'm not sure how to get the sign and I'm not even sure what it's called in English. In Swedish you say "belongs to".

The Attempt at a Solution



a) I'm pretty sure that this is fulfils all of the conditions. Because a0=0, the zero vector is included when x=0 then a0+a1*0+a2*0+a3*0=0 is true.

b) Here is where I begin to get a little confused. My problem is mainly with the concept of the null vector. My question is whether "0 vector included in U) means that the set of polynomials U somewhere contains a polynomial that, for some value of x, gives the value 0, or whether it means that U contains a polynomial equal to zero specifically for x=0. The way I've solved this problem so far is to say:

a0+a1x+a2x^2+a3x^3=0 occurs when x=1 since we then get a0+a1+a2+a3=0 in accordance with the restriction imposed in the problem.

But is this the correct way to think of this problem?

c) This is where my above reasoning begins to confuse me. If I have a0+a1x+a2x^2+a3x^3, all of whose coefficients are integers, one could say that a0 was the integer 0 and then for x=0 the null vector would be realized. But is it important that the coefficients don't receive a specific value like that, that they be just general constants?

d) I'm not even sure how to go about this one.

What's really need is for someone to point out what I should be checking. I know it should be obvious from the conditions of a subspace but I guess I'm just not really sure what they mean.
 

Answers and Replies

  • #2
HallsofIvy
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Homework Statement


Determine which of the following are subspaces of P3:
a) all polynomials a0+a1x+a2x^2+a3x^3 where a0=0
b) all polynomials a0+a1x+a2x^2+a3x^3 where a0+a1+a2+a3=0
c) all polynomials a0+a1x+a2x^2+a3x^3 for which a0, a1, a2, a3 are integers
d) all polynomials of the form a0+a1x where a0 and a1 are real numbers

Homework Equations


in order for U={v in Z: v=(a0+a1x+a2x^2+a3x^3)} to be a subspace, these polynomials should fulfil:
1) 0 vector included in U
2) x in U, y in U, x+y also in U
3) c (some constant) in U, x in U, c*x also in U

sorry for writing "in", I'm not sure how to get the sign and I'm not even sure what it's called in English. In Swedish you say "belongs to".
What you wrote is perfectly correct. Either "(is) in" or "belongs to" are good. For the symbols, use "[ itex ]x\in U[ /itex ]" without the spaces: [itex]x\in U[/itex]. (Double click on that formula, or any formula, to see the code used.)

The Attempt at a Solution



a) I'm pretty sure that this is fulfils all of the conditions. Because a0=0, the zero vector is included when x=0 then a0+a1*0+a2*0+a3*0=0 is true.
The "zero vector", when your vectors are functions, is the function the is identically 0, 0 for all x. The "zero vector" is in this set because you can take a1= a2= a3= 0 which still satisfies "a0= 0".

b) Here is where I begin to get a little confused. My problem is mainly with the concept of the null vector. My question is whether "0 vector included in U) means that the set of polynomials U somewhere contains a polynomial that, for some value of x, gives the value 0, or whether it means that U contains a polynomial equal to zero specifically for x=0. The way I've solved this problem so far is to say:
Neither! As I said above, it means the polynomial that is 0 for all x. That requires that all coefficients be 0. Does such a polynomial satisfy "a0+ a1+ a2+ a3= 0"?

a0+a1x+a2x^2+a3x^3=0 occurs when x=1 since we then get a0+a1+a2+a3=0 in accordance with the restriction imposed in the problem.

But is this the correct way to think of this problem?
No, it is not. The "0 vector" is the vector \(\displaystyle 0+ 0x+ 0x^2+ 0x^3\) and that satisfies the condition.

c) This is where my above reasoning begins to confuse me. If I have a0+a1x+a2x^2+a3x^3, all of whose coefficients are integers, one could say that a0 was the integer 0 and then for x=0 the null vector would be realized. But is it important that the coefficients don't receive a specific value like that, that they be just general constants?
Again, the question is whether the specific 0 polynomial, [itex]0+ 0x+ 0x^2+ 0x^3[/itex] satisfies the conditions- which it does: 0 is an integer.

You should also find that [itex](a0+ a1x+ a2x^2+ a3x^3)+ (b0+ b1x+ b2x^2+ b3x^3)[/itex], the sum of sum of two such vectors, is [itex](a0+ b0)+ (a1+ b1)x+ (a2+ b2)x^2+ (a3+ b3)x^3[/itex] and the sum of two integers is an integer.

However, what about the scalar product: [itex](1/2)(1+ x+ x^2+ x^3)[/itex]?

d) I'm not even sure how to go about this one.
Exactly the same way. Now the 0 "vectors" is 0+ 0x. Is 0 a real number?
What about (a+ bx)+ (c+ dx)= (a+ c)+ (b+d)x where a, b, c, d are real numbers? Are a+ c and b+ d real numbers? What about a(b+ cx)= ab+ acx where a, b, and c are real numbers. Are ab and ac real numbers?

What's really need is for someone to point out what I should be checking. I know it should be obvious from the conditions of a subspace but I guess I'm just not really sure what they mean.
You only difficulty seems to be understanding exactly what the "zero vector" is. In general, it is the "additive identity"- that is, 0 is defined by the property that 0+ v= v for any vector v. For function spacex, just being equal to 0 at some values of x is not enough. In order that (O+ f)(x)= f(x) for all x (I am using "O" for the 0 function), we must have O(x)+ f(x)= f(x) so that O(x)= 0 for all x.
 
  • #3
cuttlefish
13
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Thanks! That was exactly what I was hoping to get.
 

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