Polynomial vector space

[lackyluk]

Good morning.
First of all, sorry for my English.

I have this exercise which is giving me a lot of confusion

a) Determine if [MATH]W={(r+s)x3+(r+t)x2+(s−t)x+(r+t)∣r,s,t∈R}⊆R3[x][/MATH] is a linear subspace. If yes, determine a Basis for it.
b) Complete the basis found in a) to a basis of W.

My attempt:

a) Yes is a subspace (let alone this, its not the problem). A basis B is [MATH](1,1,0,1),(1,0,1,0) [/MATH]b) I complete it to W with [MATH](0,0,1,0)[/MATH].

But more than something is wrong evidently.
Can someone give me the solution, maybe even with a little explanation, so I can examine it understanding where I have confusion.

Thanks a lot.

 

[HallsofIvy]

Your English is excellent! Far better than my (put any language here- possibly even English!). I will say that your mathematical notation, which is independent of language, is abominable, for reasons I point out!

"Determine if
W=(r+s)x^3+(r+t)x^2+(s−t)x+(r+t)
is a linear subspace"

A "linear subspace" of what? In order to talk about a "subspace" you first have to have a "space"! I am going to assume that the problem Is simply to determine if this is a vector space.
(Strictly speaking, it isn't! What you have written is a single vector. You mean "the set of all vectors of the form W=...")

I don't know what you mean when you say "A basis B is (1,1,0,1),(1,0,1,0)(1,1,0,1),(1,0,1,0)". All vectors in the set are polynomials of degree 3 or less, not ordered quadruples of numbers! I could guess that you are representing the cubic \(\displaystyle ax^3+ bx^2+ cx+ d\) as (a, b, c, d). That's perfectly valid but you need to say so!

To find a basis let one of r, s, t be 1 and the others 0:
r= 1, s= t= 0: x^3+ x^2+ 1
s= 1, r= t= 0: x^3+ x^2+ x
t= 1, r= s= 0: x^2- x+ 1
Now, you have to show that this is a basis: that it spans the space and these are independent. That is spans the space follows from the fact that any vector in the space can be written in the form
(r+s)x^3+(r+t)x^2+(s−t)x+(r+t)= r(x^3+ x^+ 1)+ s(x^3+ x^2+ x)+ t(x^2- x+ 1).

The fact that they are independent follows from the fact that no linear combination of x^3+ x^2+ x and x^2- x+ 1 only will give x^3+ x^2+ 1.

The last part is "Complete the basis found in a) to a basis of W."
That makes no sense at all! The "basis found in a" is "a basis of W"!
In (a) W was referred to as a "subspace" of some larger space so I suspect should be "complete the basis" to a basis of this larger space. Of course, it is never said what this larger space is! I suspect it is the space of all cubic polynomials. If that is the case any such polynomial can be written ax^3+ bx^2+ cx+ d for any numbers, a, b, and c. The polynomials in W are (r+s)x^3+(r+t)x^2+(s−t)x+(r+t). That would already be a basis for the entire space if, for any a, b, c, and d, we can find r, s. and t such that a= r+ s, b= r+ t, c= s- t, and d= r+ t. (The fact that this is four equations to solve for three unknowns suggests that this is not true.)
a= r+ s so s= a- r. b= r+ t so t= b- r. Then c= s- t= a-r- (b-r)= a- b. Of course, a, b, c, and d can be any numbers. It might not happen that "c= a- b". Further, d= r+ t= r+ (b- r)= b. d might not be equal to b. Such a case would be a= b= 1, c= 1 (so not a- b= 0) and d= 2 (so not b): x^3+ x^2+ x+ 2. Adding that to the previous polynomials gives a basis.

 

[lackyluk]

First of all thanks for your reply and time and sorry for the abomination.
This is the actual text of the exercise in my own language (italian btw).



And that was my best attempt to translate it in English.

So yes I would have add ''is a linear subspace of R3[x]'', so you are right, sorry again.

Now, for the point a), an explanation of what I tried and wrote on my first post is:

[MATH]W = (r+s)x^3+(r+t)x^2+(s−t)x+(r+t) = r(1,1,0,1) + s(1,0,1,0) + t(0,1,-1,1) = W [/MATH]
So a span (I think ''span'' is the term for what I want to say) of [MATH]W[/MATH] is [MATH]<(1,1,0,1), (1,0,1,0), (0,1,-1,1)>[/MATH].
But the third vector is a linear combination of the other two, so a basis [MATH]B[/MATH] for [MATH]W[/MATH] are the first two vector alone [MATH]B = (1,1,0,1), (1,0,1,0) = (x^3 + x^2 + 1, x^3 + x) [/MATH]
On the point b),
Im sorry if it make no sense to you but is literally what the exercise ask me to solve.

'b) Complete the basis B found on the previous point to a basis of W'

And I have to admit, this is it the point that made me so confuse...

So My attempt was:
[MATH]W[/MATH] have dimension 3 because of the three free variable r, s, t.
Its Basis [MATH]B[/MATH] have dimension 2.

So I added a third not dependant vector to complete it to three...
There we are with [MATH](1,1,0,1), (1,0,1,0), (0,0,1,0) = (x^3 + x^2 + 1, x^3 + x, x) [/MATH]