Linear algebra coordinates, basis?
So i have a problem
there is a vector space K[x]3 (3 is subscript) of polynomials over K
B={1+2x , x+3x^2-x^3, -2x^2+4x^3, 1-x+2^3}
I had to first prove B is a basis...which i'm confident i did.
Second part says
determine [x^2]B (B is subscript). (hint, use coordinates relative to a more natural basis)
So the example problems don't really deal with the x^2 issue. I'll give this an attempt.
the more natural basis to deal with
[1,0,0,0] [0,1,0,0] ...... and so on
The 4 polynomials i have listed, lets call them, a,b,c, and d.
Woudl i just plug x^2 into those polynomials and place them into a vector......set the vector =
t* [1,0,0,0] + y[0,1,0,0] + z(0,0,1,0] + s*[0,0,0,1]
and then solve.
And then....isn't that just equal to the original equation's...sticking x^2 into them??
Comments
If I understand the second part correctly, we're looking for coordinate [x^2]B = <t, y, z, s> such that:
t (1 + 2x) + y (x+3x^2-x^3) + z (-2x^2+4x^3) + s (1-x+2^3) = x^2
I don't know how a more natural basis would help, but I would just expand the above and solve the resulting system of EQs:
t + s = 0
2 t + y - s = 0
3 y - 2 z = 1
-y + 4 z - 2 s = 0
And then say that [x^2]B = < -2/13, 6/13, 5/26, 2/13 >
to instruct that they variety a foundation, it suffices to instruct that they are linearly autonomous, on account that they seem to be a sequence of two components in a 2-dimensional area. assume cv_1 + dv_2 = b. Then, utilising our dot products: v_1.(cv_1 + dv_2) = v_1.b ===> c(v_1.v_1) + d(v_1.v_2) = v_1.b ===> c * a million + d * 0 = v_1.b ... (they are 2 of the given dot products) ===> c = v_1.b further, dot producting with v_2, we get: d = v_2.b If b = 0, then all dot products with it are equivalent to 0, and the equation above is cv_1 + dv_2 = 0. So, we've basically shown that: c = v_1.0 = 0 d = v_2.0 = 0 for this reason, v_1 and v_2 are linearly autonomous. we've additionally got here across a prevalent variety for the coefficients. If we desire the coefficients for (a, b), we basically compute: c = v_1.(a, b) = (x_1, x_2).(a, b) = ax_1 + bx_2 d = v_2.(a, b) = (y_1, y_2).(a, b) = ay_1 + by_2 wish that facilitates!