linear algebra:linearly independent sets?

Question

prove that the set {1,x,x^2} is linearly independent?
my confusion is
as we know X^2 -2x +1=0 or x^2+2x+1=0 then how i can show that
a(1)+b(x)+c(x^2)=0 implies a=b=c=0 for linearly independent. a,b,c are elements of field.
similarly how i show {1+x, 1+x+x^2} is linearly independent.

Answer 1

a(1) + b(x) + c (x²) = 0 for all x (and not for one particular x)

if x = 0 then a = 0

bx + cx² = 0

if x = 1, then b + c = 0
if x = - 1, then -b + c = 0
(we can choose others values for x than 0, 1 and -1)

so
c = b and c = -b
b = -b
b = 0
c = 0

a = b = c = 0

Answer 2

Suppose a(1) + b(x) + c(x^2) = 0.

Note carefully that this is an equality of *polynomials*, since the elements in your vector space are polynomials. This means that a(1) + b(x) + c(x^2) = 0 for *every* value of x.

In particular, if x = 0, x = 1, or x = -1, then we find

a = 0
a + b + c = 0
a - b + c = 0

If you solve this system of equations, you will discover that the only solution is a = b = c = 0.

---

Suppose a(1 + x) + b(1 + x + x^2) = 0. Then a + ax + b + bx + bx^2 = 0, or regrouping terms, (a + b) + (a + b)x + bx^2 = 0.

This is an equality of polynomials, so it holds for every x. In particular, plugging in x = 0, x = 1, and x = -1, we get

(a + b) = 0
(a + b) + (a + b) + b = 0
(a + b) - (a + b) + b = 0

If you solve this system of equations, then you will discover that the only solution is a = b = 0.

Answer 3

It follows directly from the definitions. Let F is a field, F[x] - the ring of polynomials of 1 variable over F, which can be considered as a vector space, its zero (neutral) element (null-vector) is the so-called zero polynomial /often noted 0 - the same way as 0 - zero element of F/:
0 = 0 + 0*x + 0*x^2 + 0*x^3 + ..., which is the ONLY polynomial of indefinite degree /You can consider it either of arbitrary degree with all-zero coefficients, or, as some authors write in the books, deg(0) = -infinity, there is a reason in it/. Also remind when 2 polynomials over F are identical: by definition they should be of equal degree and all corresponding coefficients to be equal. The problem here is purely psychological due to ambiguous notation we usually use - the right side in
a(1)+b(x)+c(x^2) = 0, You've written above, is looking like the number 0, or the zero element of the field, but should be considered as the zero polynomial, rewrite it as
a(1)+b(x)+c(x^2) = 0 + 0x + 0x^2 and apply the definition, You obtain a = b = c = 0 as required.

The same approach can be applied to the second example, I leave it to Yourself to convince.

Answer 4

as we know X^2 -2x +1=0 or x^2+2x+1=0

No, we don't know that, that is false

x is a transcendental element. It DOESN'T satisfy any equation of the form:
a_n x^n+...+a_1 x+ a_0=0, when at least one of coefficients is non-zero.

Answer 5

Your answer to T would not make experience - you reported the vectors are no longer scalar multiples of one yet another so as that they are linearly based. i desire you meant they're scalar multiples (with factor 2.5) so as that they are linearly based. a series with extra beneficial than 3 vectors of three components is often linearly based. With scalar multiples of three linearly self reliant vectors you additionally could make the different vector, so which you won't be able to have 4 self reliant vectors.