# Linear Transformation

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• February 15th 2011, 08:50 AM
Darkprince
Linear Transformation
Hello, we have recently started studying linear transformation in university and I couldn't understand them well (we have done three lectures about them)
And I have to do the following question:

Let f:R[X] -> R[X] be the linear transformation sending a polynomial P(X) to f(P(X))= P(X+1) - P(X).

a) Let f4: R4[X] -> R[X] be the linear transformation induced by restriction of f to the R-vector space of polynomial of degree at most 4. Determine the kerne and the image of f4. (4 is a subscript)

b)Answer part (a) again with f4 repaced by fn (n any non-negative integer).
Deduce that for any P in R[X], there exists Q in R[X] such that f(Q)=P

c) Let Q in R[X] and let S in R[X] be a polynomial such that f(S)=Q. Show that any other solutions of the equation f(P)=Q can be written P=S+S' with S' in ker(f).

I don't know even how to start this problem and how to continue it.
I know the definitions of image, kernel etc but I find it difficult to put the theory into a practical problem. Any help of how to face such types of problems and maybe any general guidance for linear transformations would be appreciated!!!

• February 15th 2011, 09:14 AM
FernandoRevilla
Quote:

Originally Posted by Darkprince
Let f:R[X] -> R[X] be the linear transformation sending a polynomial P(X) to f(P(X))= P(X+1) - P(X).

a) Let f4: R4[X] -> R[X] be the linear transformation induced by restriction of f to the R-vector space of polynomial of degree at most 4. Determine the kerne and the image of f4. (4 is a subscript)

Start finding $f_4(x^n)\;\;(n=0,1,2,3,4)$ then, you'll have the matrix of $f_4$ with respect to the canonical basis of $\mathbb{R}_4[x]$ . Now, to find the kernel and the image is just routine.

Fernando Revilla
• February 15th 2011, 09:23 AM
TheEmptySet
Quote:

Originally Posted by Darkprince
Hello, we have recently started studying linear transformation in university and I couldn't understand them well (we have done three lectures about them)
And I have to do the following question:

Let f:R[X] -> R[X] be the linear transformation sending a polynomial P(X) to f(P(X))= P(X+1) - P(X).

a and b) Let f4: R4[X] -> R[X] be the linear transformation induced by restriction of f to the R-vector space of polynomial of degree at most 4. Determine the kerne and the image of f4. (4 is a subscript)

b)Answer part (a) again with f4 repaced by fn (n any non-negative integer).
Deduce that for any P in R[X], there exists Q in R[X] such that f(Q)=P

c) Let Q in R[X] and let S in R[X] be a polynomial such that f(S)=Q. Show that any other solutions of the equation f(P)=Q can be written P=S+S' with S' in ker(f).

I don't know even how to start this problem and how to continue it.
I know the definitions of image, kernel etc but I find it difficult to put the theory into a practical problem. Any help of how to face such types of problems and maybe any general guidance for linear transformations would be appreciated!!!

So for part a let
$\{1,x,x^2,x^3,x^4 \}$ be your basis of $\mathbb{R}_{4}[x]$

Now lets see what happens to the basis elements
$f(1)=1-1=0$
$f(x)=(x+1)-x=1$
$\displaystyle f(x^2)=(x+1)^2-x^2=(x+1-x)(x+1+x)=2x+1=\binom{2}{1}x+\binom{2}{2}1$
$\displaystyle f(x^3)=(x+1)^3-x^3=\binom{3}{1}x^2+\binom{3}{2}x+\binom{3}{3}$
$\displaystyle f(x^4)=(x+1)^4-x^4=\binom{4}{1}x^3+\binom{4}{2}x^2+\binom{4}{3}x^ 1+\binom{4}{4}$
$\displaystyle f(x^n)=\sum_{k=1}^{n}\binom{n}{k}x^{n-k}$

Now with this, what basis elements are in the kernel? what about the image?

Hint: this can be represented as a matrix

$A=\begin{bmatrix}
0 & 1 & 1 & 1 & 1 \\
0 & 0 & 2 & 3 & 4 \\
0 & 0 & 0 & 3 & 6 \\
0 & 0 & 0 & 0 & 4
\end{bmatrix}$

Can you finish from here?

Edit: too slow
• February 15th 2011, 09:41 AM
Darkprince
Thanks for the responses! After the obtained matrix, to find the kernel should I set the matrix multiplied by [x y z e] = 0?
I am not pretty sure about how to continue. I think I understood the first part, but again I am a bit of confused. Thanks for the answers again!
• February 15th 2011, 10:28 AM
FernandoRevilla
$\ker f_4\equiv \begin{bmatrix}
0 & 1 & 1 & 1 & 1 \\
0 & 0 & 2 & 3 & 4 \\
0 & 0 & 0 & 3 & 6 \\
0 & 0 & 0 & 0 & 4\\
0 & 0 & 0 & 0 & 0
\end{bmatrix}\begin{bmatrix}{x_0}\\{x_1}\\{x_2}\\{ x_3}\\{x_4}\end{bmatrix}=\begin{bmatrix}{0}\\{0}\\ {0}\\{0}\\{0}\end{bmatrix}$

where $(x_0,x_1,x_2,x_3,x_4)^t$ are the coordinates of a generic polynomial $p(x)\in \mathbb{R}[x]$ with respect to $B=\{1,x,x^2,x^3,x^4\}$ . Solving, you'll obtain $\ker f_4=\mathbb{R}$ .

Fernando Revilla
• February 15th 2011, 10:38 AM
Darkprince
kerf4 = R, since our system of equations has an infinite number of solutions?
And after that how do we proceed to find the image? I know that the image of a linear transformation, i.e im(T)={w in W such that w= T(v) for some v in V}
Sorry for asking so many questions, I am struggling out to understand about the linear transformations generally, since I am not yet familiar with them!
• February 15th 2011, 11:56 AM
FernandoRevilla
Quote:

Originally Posted by Darkprince
kerf4 = R, since our system of equations has an infinite number of solutions?

The solutions are $(x_0,x_1,x_2,x_3,x_4)=(\lambda,0,0,0,0)\;\;(\lambd a\in\mathbb{R})$ so, the polynomials in the kernela are $p(x)=\lambda$ .

Quote:

And after that how do we proceed to find the image? I know that the image of a linear transformation, i.e im(T)={w in W such that w= T(v) for some v in V}

Any vector $y=(y_0,y_1,y_2,y_3,y_4)^t$ of the image has the form:

$\begin{bmatrix}{y_0}\\{y_1}\\{y_2}\\{y_3}\\{y_4}\e nd{bmatrix}=\begin{bmatrix}
0 & 1 & 1 & 1 & 1 \\
0 & 0 & 2 & 3 & 4 \\
0 & 0 & 0 & 3 & 6 \\
0 & 0 & 0 & 0 & 4\\
0 & 0 & 0 & 0 & 0
\end{bmatrix}\begin{bmatrix}{x_0}\\{x_1}\\{x_2}\\{ x_3}\\{x_4}\end{bmatrix}$

so, you can express $y=x_0C_1+x_1C_2+x_2C_3+x_3C_4+x_4C_5$ and the columns of the matrix span $\textrm{Im}f$ .

Fernando Revilla
• February 15th 2011, 03:05 PM
Darkprince
Thanks again!
So is there a general rule for finding the kernel and image of a linear transformation? How did we choose our base in the first time? We are said about a polynomial of degree at most 4 that's why we chose x, x^2, x^3, x^4 for our base. Why did we included 1 in our base?
Also why did we take y=(y0,y1,y2,y3,y4)^t and not just (y0,y1,y2,y3,y4)?
Finally any guidance about parts b,c (which I did not know how to handle) would be helpful and appreciated and any recommendation about any reference that will help me put in practise linear transformations!
Thanks again and sorry for bothering!
• February 15th 2011, 11:48 PM
FernandoRevilla
Quote:

Originally Posted by Darkprince
So is there a general rule for finding the kernel and image of a linear transformation?

If the corresponding vector spaces have finite dimension, one method is to use the matrix equation, but not the only one.

Quote:

How did we choose our base in the first time? We are said about a polynomial of degree at most 4 that's why we chose x, x^2, x^3, x^4 for our base. Why did we included 1 in our base?

More important than the problem you posted:

Exercise Prove that $B_n=\{1,x,x^2,\ldots,x^n\}$ are linearly independent and span $\mathbb{R}_n[x]$ . As a consequence, $B_n$ is a basis of $\mathbb{R}_n[x]$ .

Quote:

Also why did we take y=(y0,y1,y2,y3,y4)^t and not just (y0,y1,y2,y3,y4)?

Usually, we write the coordinates of vectors as columns but it is not obligatory.

Quote:

Finally any guidance about parts b,c (which I did not know how to handle) would be helpful

Follow TheEmptySet outline using $B_n$ .

Quote:

and appreciated and any recommendation about any reference that will help me put in practise linear transformations!

Have you a text book or class notes? . At any rate, an important advise:

It seems you are tryng to solve a problem without knowing the corresponding theory (bad, very bad method). Study previously it and ask your particular doubts.

Fernando Revilla
• February 16th 2011, 04:40 AM
Darkprince
Thanks again for all the information. I will again revise the theory for this exercise.
• February 21st 2011, 08:13 AM
Darkprince
Hello all again! Sorry for bothering!
After reviewing my theory and trying again to do the exercise I want to ask the following:
for part a kernel is R, image is {1,x,x^2,x^3} and in part b kernel is R and image is {1,x,x^2,...,x^(n-1)} since any vector in the image can be written as a combination of the image of the basis vectors??
Are my answers and my reasoning correct?

Also how do I deduce that for any P in R[X], there exists Q in R[X] such that f(Q)=P?

Thanks all again!
• February 21st 2011, 08:55 AM
FernandoRevilla
Quote:

Originally Posted by Darkprince
for part a kernel is R, image is {1,x,x^2,x^3} and in part b kernel is R and image is {1,x,x^2,...,x^(n-1)} since any vector in the image can be written as a combination of the image of the basis vectors??
Are my answers and my reasoning correct?

Right.

Quote:

Also how do I deduce that for any P in R[X], there exists Q in R[X] such that f(Q)=P?

If $P\in \mathbb{R}_{n-1}[x]$ consider $Q\in\mathbb{R}_n[x]$ such that $f_n(Q)=P$ .

Fernando Revilla
• February 21st 2011, 09:14 AM
Darkprince
Part b says about P and Q in R[x], so we can choose different degrees for P and Q and so the following reasoning holds:
If Q in Rn[x] then there is a P in Rn-1[x] such that f(Q)=P, since our linear transformation reduces the degree of our polynomial by one?
Is it just that the reasoning?

For part c since f is linear, then f(S+S')=f(S)+f(S')=Q+f(S') and then how do I proceed? How do I deduce that S' is in ker(f)?

Thanks again for all, appreciated all the given help!
• February 21st 2011, 10:56 AM
Darkprince
Sorry again, I deduced part c by the following way:

Let P=S+S', then f(S+S')=Q, then f(S)+f(S')=Q. Since f(S)=Q then f(S') must be zero so S' must be a constant polynomial, so S' is in Ker f

I think I am correct, a check would be appreciated and a confirmation for my reasoning for part b in my above post.

Thanks again for all, sorry for bothering! Cheers!
• February 21st 2011, 11:07 AM
FernandoRevilla
Quote:

Originally Posted by Darkprince
Part b says about P and Q in R[x], so we can choose different degrees for P and Q and so the following reasoning holds:
If Q in Rn[x] then there is a P in Rn-1[x] such that f(Q)=P, since our linear transformation reduces the degree of our polynomial by one?
Is it just that the reasoning?

Right if you change $P$ by $Q$ .

Quote:

For part c since f is linear, then f(S+S')=f(S)+f(S')=Q+f(S') and then how do I proceed? How do I deduce that S' is in ker(f)?

If $f(S)=Q$ and $f(P)=Q$ then, $0=f(P)-f(S)=f(P-S)=0$ . This means that $P-S\in \ker f$ so, $P-S=S'\in \ker f$ . That is, $P=S+S'$ with $S'\in\ker f$

Fernando Revilla
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