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Let $U = P_2(\mathbb{R})$, $W = \left\{A ∈ M_2(\mathbb{R}) ~|~ A~ \text{is symmetric}\right\}$. These are all vector spaces over $\mathbb{R}$. Their standard bases are:

$A = {1, x, x^2} ⊂ P_2(\mathbb{R})$

$C = \left\{ I=\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, J=\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, K=\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} \right \} ⊂ M_2(\mathbb{R})$

Let $S: U \longrightarrow W$ be a linear transformation given by the following rules:

$1+x \mapsto \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} \tag{1}$

$1-x \mapsto \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}\tag{2}$

$x-x^2 \mapsto \begin{bmatrix} 0 & -1 \\ -1 & 0 \end{bmatrix}\tag{3}$

Find $\left[S\right]_{C \leftarrow A}$

Workings:

I'm not too sure now how to go about this now.

I believe that I need to find $S(1), S(x)$ and S(x^2)$ And then find the basis of each.

But I don't know what $S(1)$ or anything would be.

I guess $S(1) = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$.

But I'm not sure.

Any help will be appreciated.

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I made some edits in your original posts in order to simplify the writing of my answer.

You have $S(2)=S[(1-x)+(1+x)]=S(1-x)+S(1+x)=4I+4K$. Hence $S(1)=2I+2K$.

By a similar way $S(2x)=S[(1+x)-(1-x)]=2J$. Hence $S(x)=J$.

Finally $S(x-x^2)=S(x)-S(x^2)=-J$. Hence $S(x^2)=2J$.

Conclusion, the matrix you're looking for is

$$\begin{pmatrix} 2 & 0 & 0\\ 0 & 1 & 0\\ 2 & 0 & 2 \end{pmatrix}$$ written in the basis you defined in your question.

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  • $\begingroup$ Okay I see what you are saying. Thanks. $\endgroup$
    – HypestHype
    Nov 1 '15 at 18:38
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Hint: Express $1$ from the data: $$1=\frac12(1+x+1-x).$$ Similarly for $x$ and $x^2$.

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hint

You need to express each of $1,x,x^2$ in terms of $1-x, 1+x, x-x^2$.

For example, $$1= \frac{1-x}{2}+ \frac{1+x}{2}+0(x-x^2)$$ Now you get $S(1)$ by using the linearity. $$S(1)= \frac{1}{2}S(1-x)+ \frac{1}{2}S(1+x)+0S(x-x^2).$$

So $$S(1)=\begin{bmatrix}2&0\\0&2\end{bmatrix}=2I+2K$$

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  • $\begingroup$ So $S(1) = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix}$ which the basis would be $\begin{bmatrix} 2 \\ 0 \\ 0 \\ 2 \end{bmatrix}$ $\endgroup$
    – HypestHype
    Nov 1 '15 at 18:25
  • $\begingroup$ @HypestHype no you have to express $S(1)$ in terms of the basis $\{I,J,K\}$. I have added that to my answer. $\endgroup$
    – Anurag A
    Nov 1 '15 at 18:31
  • $\begingroup$ Oh. Someone else added the $\left\{I,J,K\right\}$ so I never even noticed. $\endgroup$
    – HypestHype
    Nov 1 '15 at 18:34
  • $\begingroup$ I added $I,J,K$ in order to simplify the notations. ;-). $\endgroup$ Nov 1 '15 at 18:38
  • $\begingroup$ @mathcounterexamples.net that was indeed very helpful. Saved some unnecessary typing :-) $\endgroup$
    – Anurag A
    Nov 1 '15 at 18:39

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