# General polynomial transformation (transformation matrices).

• Mutaja
In summary, the given polynomial in standard basis can be written as a column vector. By replacing x with (x-1) in the polynomial, the graph of the polynomial will shift/offset 1 to the right. To find the transformation matrix T, we can set up the equation T(p(x)) = p(x-1) and solve for T. This results in the transformation matrix T being equal to the matrix [-1, 1, 1; 1, -2, 0; 0, 1, 0]. Similarly, to find the transformation matrix Ta, we can set up the equation Ta(p(x)) = p(x-a) and solve for Ta. This results in the transformation matrix Ta being
Mutaja

## Homework Statement

A polynomial of degree two or less can be written on the form p(x) = a0 + a1x + a2x2.

In standard basis {1, x, x2} the coordinates becomes p(x) = a0 + a1x + a2x2 equivalent to ##[p(x)]_s=\begin{pmatrix}a0\\ a1\\ a2 \end{pmatrix}##.

Part a)

If we replace x with (x-1) in the polynomial, the graph to the polynomial will shift/offset 1 to the right.

Find the transformationmatrix, M, to the transformation

T(p(x)) = p(x-1)

so that

##T=\begin{pmatrix}a0\\ a1\\ a2 \end{pmatrix}## = ##M * \begin{pmatrix}a0\\ a1\\ a2 \end{pmatrix}##

part b)

Find the transformation matrix, Ma, to the transformation

Ta (p(x)) = p(x-a).

## Homework Equations

Transformation matrices, basic rules for dealing with matrices and polynomials. I'm assuming.

## The Attempt at a Solution

I understand polynomials, and I can solve problems where I'm given polynomials and then find transformation matrices or linear dependence/independence. But here I'm lost as to what I'm supposed to do. The a0, a1, a2 and (x-1) approach is confusing me.

Any guiding as to what I'm looking for as a starter would be gold here, although I fully understand and respect the forum rules that require me to have at least some work to show. I will continue to try out things and post any progress I make.

I know basic rules of polynomials fairly well, although it's a while since we went through that subject. I also know basic rules for matrices, transformation etc. that I went through with a few people on these forums as well. Also, as I stated above, if I'm given two polynomials and I'm asked to check if they can form a basis or something like that, I'm good to go.

I guess this is somewhat similar to what some people go through when they start basic algebra. Solving numbers - fine. When there's letters involved - that's when the problems start.

Any feedback is appreciated as always, thanks a lot in advance.

Last edited:
Mutaja said:

## Homework Statement

A polynomial of degree two or less can be written on the form p(x) = a0 + a1x + a2x2.

In standard basis {1, x, x2} the coordinates becomes p(x) = a0 + a1x + a2x2 equivalent to ##[p(x)]_s=\begin{pmatrix}a0\\ a1\\ a2 \end{pmatrix}##.

Part a)

If we replace x with (x-1) in the polynomial, the graph to the polynomial will shift/offset 1 to the right.

Find the transformationmatrix, M, to the transformation

T(p(x)) = p(x-1)

so that

##T=\begin{pmatrix}a0\\ a1\\ a2 \end{pmatrix}## = ##M * \begin{pmatrix}a0\\ a1\\ a2 \end{pmatrix}##

part b)

Find the transformation matrix, Ma, to the transformation

Ta (p(x)) = p(x-a).

## Homework Equations

Transformation matrices, basic rules for dealing with matrices and polynomials. I'm assuming.

## The Attempt at a Solution

I understand polynomials, and I can solve problems where I'm given polynomials and then find transformation matrices or linear dependence/independence. But here I'm lost as to what I'm supposed to do. The a0, a1, a2 and (x-1) approach is confusing me.

Any guiding as to what I'm looking for as a starter would be gold here, although I fully understand and respect the forum rules that require me to have at least some work to show. I will continue to try out things and post any progress I make.

I know basic rules of polynomials fairly well, although it's a while since we went through that subject. I also know basic rules for matrices, transformation etc. that I went through with a few people on these forums as well. Also, as I stated above, if I'm given two polynomials and I'm asked to check if they can form a basis or something like that, I'm good to go.

I guess this is somewhat similar to what some people go through when they start basic algebra. Solving numbers - fine. When there's letters involved - that's when the problems start.

Any feedback is appreciated as always, thanks a lot in advance.

Well, expand it. ##p(a_0+a_1 x + a_2 x^2)=a_0+a_1 (x-1)+ a_2 (x-1)^2=a_2-a_1+a_0+(a_1-2 a_2)x+a_2x^2##. In terms of column vector that's ##\begin{pmatrix}a_2-a_1+a_0\\ a_1-2 a_2\\ a2 \end{pmatrix}##. What matrix will turn your original column vector into that?

Dick said:
Well, expand it. ##p(a_0+a_1 x + a_2 x^2)=a_0+a_1 (x-1)+ a_2 (x-1)^2=a_2-a_1+a_0+(a_1-2 a_2)x+a_2x^2##. In terms of column vector that's ##\begin{pmatrix}a_2-a_1+a_0\\ a_1-2 a_2\\ a2 \end{pmatrix}##. What matrix will turn your original column vector into that?

Well, this seems to easy to be correct, but it's what I've done after looking at what makes sense and the relevant notes (supplied by my professor, so they should be accurate).

##T\begin{pmatrix}a_0\\ a_1\\ a_2 \end{pmatrix}## = ##\begin{pmatrix}a_2-a_1+a_0\\ a_1-2 a_2\\ a2 \end{pmatrix}##

##T\begin{pmatrix}a_0\\ a_1\\ a_2 \end{pmatrix}## = ##\begin{pmatrix}-1 & 1 & 1\\ 1 & -2 & 0\\ 0 & 1 & 0 \end{pmatrix}## * ##\begin{pmatrix}a_0\\ a_1\\ a_2 \end{pmatrix}##

So ##T\begin{pmatrix}a_0\\ a_1\\ a_2 \end{pmatrix}## will be the matrix that turns my original column vector into ##\begin{pmatrix}a_2-a_1+a_0\\ a_1-2 a_2\\ a2 \end{pmatrix}##.

Frankly, it does make some sense, because if I do the multiplication, I end up with the matrix you supplied - if I haven't done any silly mistakes just to get the answer I'm looking for (that I only see what I want to see if you understand).

Really appreciate that you take interest in this problem. Thanks.

Last edited:
Mutaja said:
Well, this seems to easy to be correct, but it's what I've done after looking at what makes sense and the relevant notes (supplied by my professor, so they should be accurate).

##T\begin{pmatrix}a_1\\ a_2\\ a_3 \end{pmatrix}## = ##\begin{pmatrix}a_2-a_1+a_0\\ a_1-2 a_2\\ a2 \end{pmatrix}##

##T\begin{pmatrix}a_1\\ a_2\\ a_3 \end{pmatrix}## = ##\begin{pmatrix}-1 & 1 & 1\\ 1 & -2 & 0\\ 0 & 1 & 0 \end{pmatrix}## * ##\begin{pmatrix}a_1\\ a_2\\ a_3 \end{pmatrix}##

So ##T\begin{pmatrix}a_1\\ a_2\\ a_3 \end{pmatrix}## will be the matrix that turns my original column vector into ##\begin{pmatrix}a_2-a_1+a_0\\ a_1-2 a_2\\ a2 \end{pmatrix}##.

Frankly, it does make some sense, because if I do the multiplication, I end up with the matrix you supplied - if I haven't done any silly mistakes just to get the answer I'm looking for (that I only see what I want to see if you understand).

Really appreciate that you take interest in this problem. Thanks.

Right approach, but what is ##a_3##?? Your column vector was ##\begin{pmatrix}a0\\ a1\\ a2 \end{pmatrix}##.

Dick said:
Right approach, but what is ##a_3##?? Your column vector was ##\begin{pmatrix}a0\\ a1\\ a2 \end{pmatrix}##.

Ah, yes, a careless mistake from me once again. Not quite used to writing matrices on here yet, and that combined with looking at notes with different examples makes it easy for me to do silly mistakes like that. I will try to keep them limited though, and thanks for pointing it out!

For my exercise b it's the same approach?

##p(a_0+a_1 x + a_2 x^2)=a_0+a_1 (x-a)+ a_2 (x-a)^2## = ... I didn't quite catch what you did here and with the column vector, unfortunately. It doesn't look too complicated though, so I'm sure I'll spot it after a couple of attempts.

Thanks a lot for your help, I really do appreciate it.

Mutaja said:
Ah, yes, a careless mistake from me once again. Not quite used to writing matrices on here yet, and that combined with looking at notes with different examples makes it easy for me to do silly mistakes like that. I will try to keep them limited though, and thanks for pointing it out!

For my exercise b it's the same approach?

##p(a_0+a_1 x + a_2 x^2)=a_0+a_1 (x-a)+ a_2 (x-a)^2## = ... I didn't quite catch what you did here and with the column vector, unfortunately. It doesn't look too complicated though, so I'm sure I'll spot it after a couple of attempts.

Thanks a lot for your help, I really do appreciate it.

What did you get for a matrix? And, yes, it's the same idea for exercise b. Just with 'a' instead of '1'. Try it out!

Dick said:
What did you get for a matrix? And, yes, it's the same idea for exercise b. Just with 'a' instead of '1'. Try it out!

I still ended up with the matrix ##\begin{pmatrix}-1 & 1 & 1\\ 1 & -2 & 0\\ 0 & 1 & 0 \end{pmatrix}## as seen in the now edited post #3.

I'm still not quite sure what you did ##p(a_0+a_1 x+a_2 x^2)=a_0+a_1 (x−a)+a_ 2(x−a)2 = ## [here]...

Mutaja said:
I still ended up with the matrix ##\begin{pmatrix}-1 & 1 & 1\\ 1 & -2 & 0\\ 0 & 1 & 0 \end{pmatrix}## as seen in the now edited post #3.

I'm still not quite sure what you did ##p(a_0+a_1 x+a_2 x^2)=a_0+a_1 (x−a)+a_ 2(x−a)2 = ## [here]...

I expanded the polynomial. E.g. ##(x-a)^2=x^2-2ax+a##. Then collect the terms of each power. The terms with no x in them will be the first entry of your new vector. The terms with one x will be the second, etc.

That matrix is wrong! If you multiply the vector by that matrix, the first entry is ##-a_0+a_1+a_2##, you want ##a_0-a_1+a_2##.

Dick said:
I expanded the polynomial. E.g. ##(x-a)^2=x^2-2ax+a##. Then collect the terms of each power. The terms with no x in them will be the first entry of your new vector. The terms with one x will be the second, etc.

That matrix is wrong! If you multiply the vector by that matrix, the first entry is ##-a_0+a_1+a_2##, you want ##a_0-a_1+a_2##.

I get it now. ##\begin{pmatrix}1 & -1 & 1\\ 1 & -2 & 0\\ 0 & 1 & 0 \end{pmatrix}## has to be the right one.

When I expand my polynomial for exercise b, I get:

##a_0-a_1 a-2a_2 a+a_2 a^2 + a_1 x + a_2 x^2## sorted by no x, x and ##x^2##.

When I want to put this into my matrix, how do I do that?

I want to do it like this - but then I completely ignore my ##a_2 a^2##:

##\begin{pmatrix}1 & -1a & 2a\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix}##

Mutaja said:
I get it now. ##\begin{pmatrix}1 & -1 & 1\\ 1 & -2 & 0\\ 0 & 1 & 0 \end{pmatrix}## has to be the right one.

It doesn't have to be. You successfully fixed the first row. Now fix the other two!

Mutaja said:
When I expand my polynomial for exercise b, I get:

##a_0-a_1 a-2a_2 a+a_2 a^2 + a_1 x + a_2 x^2## sorted by no x, x and ##x^2##.

When I want to put this into my matrix, how do I do that?

I want to do it like this - but then I completely ignore my ##a_2 a^2##:

##\begin{pmatrix}1 & -1a & 2a\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{pmatrix}##

##2 a_2 a## should be ##2 a_2 a x##. Check it. You should get three terms with no x, two with an x and one with an x^2. Those are the three entries in your result vector.

Dick said:
It doesn't have to be. You successfully fixed the first row. Now fix the other two!

##2 a_2 a## should be ##2 a_2 a x##. Check it. You should get three terms with no x, two with an x and one with an x^2. Those are the three entries in your result vector.

I'm so sorry for rushing through this. This is my final answer:

##\begin{pmatrix}1 & -1 & 1\\ 0 & 1 & -2\\ 0 & 0 & 1 \end{pmatrix}##

For my last matrix, is this correct?

##\begin{pmatrix}1 & -a & a^2\\ 0 & 1 & 2a\\ 0 & 0 & 1 \end{pmatrix}##

Mutaja said:
I'm so sorry for rushing through this. This is my final answer:

##\begin{pmatrix}1 & -1 & 1\\ 0 & 1 & -2\\ 0 & 0 & 1 \end{pmatrix}##

For my last matrix, is this correct?

##\begin{pmatrix}1 & -a & a^2\\ 0 & 1 & 2a\\ 0 & 0 & 1 \end{pmatrix}##

Almost, but you're still rushing. If you put a=1 into the second matrix you should get the first, right?

Dick said:
Almost, but you're still rushing. If you put a=1 into the second matrix you should get the first, right?

I'm rushing it as I write it from my paper onto this forum. Well, I guess, as it turns out, I've been rushing either way.

I see the lack of a minus sign in front of the 2a, and when I switched it from ##-2a_a a## to ##-2a_2 ax## I've somehow lost the minus sign.

Thanks a lot for all your help, Sir. I've learned so much the past days of being active on here. I should learn from what you're doing. Help with what I've already learned so I can fight back the constant threat of forgetting.

*inbefore new mistake*

## 1. What is a general polynomial transformation?

A general polynomial transformation is a mathematical process that involves transforming one coordinate system or function into another, using polynomial equations. It is a useful tool in various fields such as computer graphics, physics, and engineering.

## 2. How does a transformation matrix work?

A transformation matrix is a square matrix that contains the coefficients of a polynomial transformation. It is used to transform points from one coordinate system to another by multiplying the matrix with the coordinates of the point. The resulting coordinates are the transformed coordinates.

## 3. What are the advantages of using polynomial transformations?

Polynomial transformations offer a flexible and efficient way to transform coordinates or functions, as they can handle complex transformations that cannot be achieved with basic translation, rotation, and scaling operations. They also allow for non-linear transformations, which are often needed in scientific applications.

## 4. How do I determine the appropriate transformation matrix for a specific transformation?

The transformation matrix can be determined by solving a system of equations based on the desired transformation. The coefficients of the polynomial equations will form the elements of the matrix. Alternatively, there are online tools and software that can generate transformation matrices for specific transformations.

## 5. Are there any limitations to using polynomial transformations?

While polynomial transformations offer many advantages, they also have some limitations. They can be computationally intensive for higher degree polynomials, and they may introduce errors and distortion in the transformation process. Additionally, they may not be suitable for all types of transformations, such as those that involve non-polynomial functions.

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