# Basis for Range of Linear transformation

1. Oct 29, 2012

### 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?

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2. Oct 29, 2012

### Bacle2

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

3. Oct 29, 2012

### Clandry

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

4. Oct 29, 2012

### 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?

5. Oct 29, 2012

### Staff: Mentor

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.

6. Oct 29, 2012

### charlies1902

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?

Last edited: Oct 29, 2012
7. Oct 29, 2012

### 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}..

8. Oct 29, 2012

### charlies1902

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