Basis for Range of Linear transformation

[charlies1902]

The problem is attached. The problem is "find a basis for the range of the linear transformation T."



p(x) are polynomials of at most degree 3. R(T)={p''+p'+p(0) of atmost degree 2}

This is pretty much as far as I got. I'm not sure how to do the rest.


I'm thinking of picking a random function, let's say:
ax^3+bx^2+cx+d

p'(x)=3ax^2+2bx+c
p''(x)=6ax+2b

p"+p'+p(0)=(6ax+2b)+(3ax^2+2bx+c)+(d)

The book says the answer is {1, x, x^2}. Is it because in the line above, you can see that x^2, x are in there and 1 is there in the form of d?

 

[Bacle2]

Have you tried seeing the effect of R(T) on a basis for P3?

 

[Clandry]

What does the transformation due to each elmeent in the basis?

 

[charlies1902]

Yes, I did, but I am not getting there answer.

Here is my modified procedure (ignore original post, i didn' tknow what i was doing).A basis for P3 is {1, x, x^2, x^3}

T(1)=0+0+1=1
T(x)=0+1+0
T(x^2)=2+2x+0
T(x^3)=6x+3x^2+0
This is basically
T(1)=1+0x+0x^2+0x^3
T(x)=1+0x+0x^2+0x^3
T(x^2)=2+2x+0x^2+0x^3
T(x^3)=0+6x+3x^2+0x^3
Where these coefficients represent the column for the transformation matrix:
1 1 2 0
0 0 2 6
0 0 0 3
0 0 0 0

rref of this matrix is
1 1 0 0
0 0 1 0
0 0 0 1
0 0 0 0

There's a pivot in the 1st, 3rd, and 4th column which would correspond to: 1, x^2, x^3 as the basis.How did they get 1, x, x^2?

 

[Mark44]

Yes, I did, but I am not getting there answer.

Here is my modified procedure (ignore original post, i didn' tknow what i was doing).


A basis for P3 is {1, x, x^2, x^3}

T(1)=0+0+1=1
T(x)=0+1+0
T(x^2)=2+2x+0
T(x^3)=6x+3x^2+0




This is basically
T(1)=1+0x+0x^2+0x^3
T(x)=1+0x+0x^2+0x^3
T(x^2)=2+2x+0x^2+0x^3
T(x^3)=0+6x+3x^2+0x^3
Where these coefficients represent the column for the transformation matrix:
1 1 2 0
0 0 2 6
0 0 0 3
0 0 0 0

rref of this matrix is
1 1 0 0
0 0 1 0
0 0 0 1
0 0 0 0

There's a pivot in the 1st, 3rd, and 4th column which would correspond to: 1, x^2, x^3 as the basis.


How did they get 1, x, x^2?

You have:
T(x³) = 3x² + 6x
T(x²) = 2x + 2
T(x) = 1
T(1) = 1

Every output "vector" is a linear combination of 1, x, and x².

 

[charlies1902]

You have:
T(x³) = 3x² + 6x
T(x²) = 2x + 2
T(x) = 1
T(1) = 1

Every output "vector" is a linear combination of 1, x, and x².

Yes, but why does my rref form of the matrix say it's 1, x, x^3?

The pivot columns correspond to {1, x,x^3}.Wait...I think I got this problem confused with another problem. I put the coefficients in the columns instead of rows.

So the matrix should look like this
1 1 2 0
0 0 2 6
0 0 0 3
0 0 0 0 excepted it should be transposed. Is that right?

 

[Bacle2]

Notice there is no multiple of x^3 in the image of T, so your matrix should

not correspond to {1,x,x^3}..

 

[charlies1902]

Notice there is no multiple of x^3 in the image of T, so your matrix should

not correspond to {1,x,x^3}..

Yeah, I did my matrix wrong (did it in columns instead of rows.