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Find a linear transformation $T:P_{2}({R})\longrightarrow P_{4}(R)$ so that

$T(1) = x^4$

$T(x+x^2) = 1$

$T(x-x^2) = x+x^3$

I have only solved problems using standard basis, and now I have no idea on how to deal with this.

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  • $\begingroup$ Remember - $T$ is linear, so $T(u + v) = T(u) + T(v)$ $\endgroup$ Aug 20, 2015 at 22:43
  • $\begingroup$ Yeah, I could have set a linear system of equations to find $T(x)$ and $T(x^2)$, that is pretty clear now, thanks!! $\endgroup$ Aug 20, 2015 at 23:25

1 Answer 1

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You have that

$$x = \frac{(x+x^2) + (x-x^2)}{2}$$

So

$$T(x) = T\left( \frac{(x+x^2) + (x-x^2)}{2} \right) = \frac{T(x+x^2) + T(x-x^2)}{2}$$

$$ = \frac{1+x+x^3}{2}$$

And

$$T(x^2) = T(x+x^2-x) = T(x+x^2)-T(x)$$

$$=1 - \frac{1+x+x^3}{2} $$

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