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
ax3+bx2+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.
