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Find a basis of the image and kernel of the following linear transformation: $T:\mathbb{R}^3$ $\rightarrow$ $ P_2 (\mathbb{R})$ defined by $T(a,b,c) = (a-b) + (b-c)x + (a-c)x^2$


I'm pretty sure I've found a basis of the kernel which would be $(1,1,1)$, but I'm not too sure how to get the basis of the image. This also needs to be done without the use of the dimension theorem.

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    $\begingroup$ Look at what kinds of polynomials appear in the image, in particular at their coefficients. Can you see how, for a polynomial in the image, the coefficient of $x^2$ is determined by the other two coefficients? $\endgroup$ – mdp Nov 10 '14 at 16:02
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Notice that

$$T(a,b,c)=a(1+x^2)+b(-1+x)+c(-x-x^2)\\=(a-c)(1+x^2)+(b-c)(-1+x)$$ so

$$T(a,b,c)\in \operatorname{span}(1+x^2,-1+x)$$ and by the rank-nullity theorem we see that the rank of $T$ is $2$ so $(1+x^2,-1+x)$ is a basis for the image of $T$.

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  • $\begingroup$ I am guessing (and it really is just a guess) that the "dimension theorem" that isn't supposed to be used is another name for the rank-nullity theorem. But on the other hand, it isn't so hard to show that this set is independent directly. $\endgroup$ – mdp Nov 10 '14 at 16:09
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Once you've found a basis for the kernel, you are almost done for a basis of the image. In general the image is generated by images of the standard basis $e_1,e_2,e_3$(or indeed by the images of any basis of the domain vector space). The only difficulty is that those images might be linearly dependent. But a linear dependence between $f(e_1),f(e_2),f(e_3)$ is given by coefficients $c_1,c_2,c_3$ (not all zero) such that $c_1f(e_1)+c_2f(e_2)+c_3f(e_3)=0$, and since $f$ is linear this is the same as $f(c_1e_1+c_2e_2+c_3e_3)=0$, and since $c_1e_1+c_2e_2+c_3e_3$ is just the vector with coordinates $(c_1,c_2,c_3)$ this is the same as $(c_1,c_2,c_3)\in\ker(f)$. But you have computed that this happens precisely when $$(c_1,c_2,c_3)=c(1,1,1)=(c,c,c)$$ for some $c\in\Bbb R$. That means you have the linear dependency obtained for $(c_1,c_2,c_3)=(1,1,1)$, which is $f(e_1)+f(e_2)+f(e_3)=0$, and no other linear dependencies (except multiples of this same relation).

Since all of $f(e_1),f(e_2),f(e_3)$ occur in the relation, you can use the relation to express one of them as linear combination of the others, which shows you can drop that generator. So as a basis of the image you could take $\{f(e_1),f(e_2)\}$ or $\{f(e_1),f(e_3)\}$ or $\{f(e_2),f(e_3)\}$, whichever you like best. (There are many other bases not using these polynomials, but the is no reason to choose them.)

In general, for each vector $v$ in the basis of the kernel, you can express the image of one basis vector (for which the coefficient in $v$ is nonzero) in terms of the other images, allowing to drop that basis vector. Altogether from your generating set for the image you drop as many (carefully chosen) vectors as the dimension of the kernel, to get a basis of the image.

Note that in the example you can immediately see the linear independence between the vectors left over, but in more complicated situations you might not. The argument given however shows that since you used all elements of the kernel (as relations allowing to drop a image generator) it is guaranteed that those generators that are left over will be linearly independent (without the vectors dropped, there are no dependencies left)

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