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Let $T(f(t))$= $\begin{bmatrix} f(0) & f(1)\\ f(2) & f(3) \end{bmatrix}$ from $P_2$ to $\mathbb{R}^{2\times 2}$. To show that it is not an isomorphism, I need to show that either kernel of the transformation is not equal to the zero element only, or that the image is not the whole target space. I am struggling in showing that either of these is false, dealing with polynomials in transformations is very counter-intuitive. Thanks for help!

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Do you know the dimension of $P_2$? What about $\mathbb{R}^{2\times 2}$? –  AppliedSide Feb 13 '11 at 3:10
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Is it well-known that $P_2$ stands for the vector space of polynomials of degree at most two? -I would have written $\mathbb{R}_2[t]$ instead. Assuming so, Nick's comment solves the question in one minute. –  a.r. Feb 13 '11 at 7:43
    
Having the $2\times2$ matrices as target space is somewhat misleading since $T$ doesn't behave well for the multiplication (as a matter of fact $P_2$ is not even closed under polynomial multiplication). Thus $T$ is more naturally a map $P_2\rightarrow{\Bbb R}^4$ and its not being an isomorphism should be even more obvious now. –  Andrea Mori Feb 13 '11 at 10:01

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A polynomial of degree at most 2 is determined by its values at three points. So if $f(0)=f(1)=f(2)=0$, then $f(3)=0$ also. Thus $T$ is not surjective.

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I assumed someone asking this question wasn't comfortable with dimensional arguments, so gave an elementary solution. I've shown $T$ is not surjective because its image doesn't contain (for example) $\begin{bmatrix} 0 & 0\\ 0 & 1 \end{bmatrix}$. –  Chris Eagle Feb 13 '11 at 13:21

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