# Homework Help: Linear transformation from given matrices - bases unknown

1. Jul 21, 2009

### phyzmatix

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

Find a linear transformation

$$T: P_2 \rightarrow M_{22}$$

such that

$$T(1+x)=\left[\begin{array}{cc}1&0\\0&0\end{array}\right]$$

$$T(x+x^2)=\left[\begin{array}{cc}0&1\\1&0\end{array}\right]$$

$$T(1+x^2)=\left[\begin{array}{cc}0&0\\0&1\end{array}\right]$$

3. The attempt at a solution

In all the examples I have access to the linear transformation is already defined, the bases given and the question is to find the matrix (easy enough).

For this question, I could've reversed this approach had I known what the bases were. However, the bases aren't given and I'm stumped.

Could someone offer me a hint here please?

Thanks!
phyz

2. Jul 21, 2009

### Dick

T(ax^2+bx+c)=aT(x^2)+bT(x)+cT(1). If you knew what T(x^2), T(x) and T(1) were it would be easy to write down a formula for T acting on a general element of P2. Can you find them from the given information?

3. Jul 22, 2009

### phyzmatix

Hi Dick!

I think I know where you're going, but not sure how to get there. Let's see if I understand what you're saying though and give it a shot:

From what you've given me
$$T(ax^2+bx+c)=aT(x^2)+bT(x)+cT(1)$$

I thought that we could then say

$$T(1)+T(x)=\left[\begin{array}{cc}1&0\\0&0\end{array}\right]$$

$$T(x)+T(x^2)=\left[\begin{array}{cc}0&1\\1&0\end{array}\right]$$

$$T(1)+T(x^2)=\left[\begin{array}{cc}0&0\\0&1\end{array}\right]$$

However, I think we'll still be acting legit when swapping rows 1 and 2 of the second equation so that

$$T(1)+T(x)=\left[\begin{array}{cc}1&0\\0&0\end{array}\right]$$

$$T(x)+T(x^2)=\left[\begin{array}{cc}1&0\\0&1\end{array}\right]$$

$$T(1)+T(x^2)=\left[\begin{array}{cc}0&0\\0&1\end{array}\right]$$

If we then subtract equation 3 from 1, we get

$$T(1)+T(x)-Y(1)-T(x^2)=\left[\begin{array}{cc}1&0\\0&0\end{array}\right]-\left[\begin{array}{cc}0&0\\0&1\end{array}\right]$$

$$T(x)-T(x^2)=\left[\begin{array}{cc}1&0\\0&-1\end{array}\right]$$

which the same as equation 2 except for the minus sign. If all of what I've done so far is still ok and if my reasoning still holds, this means that

$$T(1)=\left[\begin{array}{cc}0&0\\0&0\end{array}\right]$$

$$T(x)=\left[\begin{array}{cc}1&0\\0&0\end{array}\right]$$

$$T(x^2)=\left[\begin{array}{cc}0&0\\0&1\end{array}\right]$$

Having done all of this I have to admit that
1. I don't know if what I've done is mathematically acceptable and
2. if everything up until this point is ok, what do I do next?

Thanks for the help!

4. Jul 22, 2009

### Dick

It's perfectly mathematically acceptable. Now just add things up T(ax^2+bx+c)=aT(x^2)+bT(x)+c=?? What's the resulting matrix?

5. Jul 29, 2009

### phyzmatix

Hi Dick!

Sorry about the slow response time, but I've been rather busy this past week.

I believe our final answer is:

$$T(ax^2+bx+c)=aT(x^2)+bT(x)+cT(1)$$

$$=a\left[\begin{array}{cc}0&0\\0&1\end{array}\right]+b\left[\begin{array}{cc}1&0\\0&0\end{array}\right]+c\left[\begin{array}{cc}0&0\\0&0\end{array}\right]$$

$$=\left[\begin{array}{cc}b&0\\0&a\end{array}\right]$$

But I'm not so sure about the notation, do we write it as

$$\left[T(p(x))\right]=\left[\begin{array}{cc}b&0\\0&a\end{array}\right]$$

?

6. Jul 29, 2009

### Dick

How about just writing T(ax^2+bx+c)? I think that's a little clearer.

7. Jul 29, 2009

### phyzmatix

Ok cool. Once again thank you for your help!