# General polynomial transformation (transformation matrices).

1. Feb 19, 2014

### Mutaja

1. The problem statement, all variables and given/known data

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).

2. Relevant equations

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

3. 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: Feb 19, 2014
2. Feb 19, 2014

### Dick

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?

3. Feb 20, 2014

### Mutaja

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: Feb 20, 2014
4. Feb 20, 2014

### Dick

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

5. Feb 20, 2014

### Mutaja

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.

6. Feb 20, 2014

### Dick

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!

7. Feb 20, 2014

### Mutaja

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]...

8. Feb 20, 2014

### Dick

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 $a_0-a_1+a_2$.

9. Feb 20, 2014

### Mutaja

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 $a_2 a^2$:

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

10. Feb 20, 2014

### Dick

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.

11. Feb 20, 2014

### Mutaja

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?

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

12. Feb 20, 2014

### Dick

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

13. Feb 20, 2014

### Mutaja

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*