linear algebra question, help!!! stuck?

Question

i am not asking u to solve this problem for me,
i want to learn how to solve it. just tell me what i am supposed to do with this question.

Use a matrix representation to find bases for the kernel and and the image of the linear transformation T:P2-->P3
T(a+bx+cx^2)
=(a+2b-2c)+(2a+2b)x
+(-a+b-4c)x^2
+(3a+2b+2c)x^3

By the way what it means by matrix representation? is it just a matrix?

Answer 1

If you choose bases for the domain and codomain of the function, you can represent the vectors in both spaces as linear combinations of the basis vectors. The coefficients of these linear combinations can be written as column vectors, and linear operations on these column vectors are equivalent to linear operations on the underlying vector space. In particular, any linear transformation between the two spaces may be written as a matrix M, such that if v is the column vector representation of the vector f (chosen since in this case the underlying space is a space of functions), Mv = T(f). The practical upshot of all this is that for finite-dimensional spaces, you can actually forget about the underlying structure and simply apply the same theorems you would to R^n, thus making matrix algebra techniques applicable to finite-dimensional vector spaces in general.

So first we choose a basis for P₃ -- we choose {1, x, x², x³} (any four linearly independent vectors would suffice, but this is the most obvious choice). This also nice because the first three vectors generate the subspace P₂. Now, we find the transformations of the three basis vectors of P₂:

T(1) = 1 + 2x - x² + 3x³
T(x) = 2 + 2x + x² + 2x³
T(x²) = -2 - 4x² + 2x³

Using this, we now find the matrix M:

[1, 2, -2]
[2, 2, 0]
[-1, 1, -4]
[3, 2, 2]

Now that we have the transformation matrix, we can find the kernel of the matrix M, which will be precisely the vectors corresponding to the kernel of T. First we reduce M to rref:

[1, 2, -2]
[2, 2, 0] subtract 2*row I
[-1, 1, -4] add row I
[3, 2, 2] subtract 3*row I

[1, 2, -2]
[0, -2, 4] divide by -2
[0, 3, -6]
[0, -4, 8]

[1, 2, -2] subtract 2*row II
[0, 1, -2]
[0, 3, -6] subtract 3*row II
[0, -4, 8] add 4*row II

[1, 0, 2]
[0, 1, -2]
[0, 0, 0]
[0, 0, 0]

So clearly the kernel of M is span ([-2, 2, 1]). And what of the image? Clearly, it is spanned by {[1, 2, -1, 3], [2, 2, 1, 2], [-2, 0, -4, 2]}, but these vectors are not all linearly independent. However, the vectors [1, 2, -1, 3] and [2, 2, 1, 2] are linearly independent, and since the third vector can be written as a linear combination of them, they too span Im(M). So Im(M) = span ({[1, 2, -1, 3], [2, 2, 1, 2]}). Now, let us translate these vectors back to the original spaced P₂ and P₃:

Ker(T) = span ({-2+2x+x²})
Im (T) = span ({1+2x-x²+3x³, 2+2x+x²+2x³})

And we are done.

Answer 2

To find a matrix representation, choose a basis for P2, say {1,x,x^2} Find T(1) = 1 +2x-x^2+3x^3 (letting a =1,b=0,c=0 in your given formula. Now choose a basis for P3, say {1,x,x^2,x^3}. Express T(1) in terms of this basis. The coordinate vector would be (1,2,-1,3) from the above answer. Make this the first column of your matrix. Repeat for T(x) and T(x^2) to get the second and third columns. You will now have a matrix A that is a representation of T. The kernel of T is the nullspace of the matrix (the solution to Ax = 0) and the image of T is the column space of A. (the basis for the image of T is the linearly independent columns of A. These are the columns of A that have leading one's in the rref form of A.

Answer 3

in case you favor to get A*(x1,x2,x3)^T = (x2,x3,x1)^T you'll get A as follows: Row a million of A is the 2d row of the identity matrix, row 2 is the third row and row 3 is the first row. So A is the matrix 0 a million 0 0 0 a million a million 0 0 examine through multiplying A through (x1, x2, x3)^T. word: The T the following skill transpose, i.e., write this as a column vector.