I have done this task, however, my solutions are suspiciously strange. Could someone verify if my reasoning is okay? Possibly something to advise …
If all is ok (I hope), this post would be nice pattern for future visitors
Task
Given are linear transformations:
$f \in L(\mathbb R^3, \mathbb R[t]_2) $
$$ \vec{x} = [x_1,x_2,x_3]^T \rightarrow f(\vec{x})(t)=(x_1+x_3)t^2 + x_2$$
and
$g \in L(\mathbb R[t]_2,\mathbb R^3) $
$$p \in \mathbb R[t]_2 \rightarrow g(p) = [p(-1),p'(0),p(1)]^T $$
Find basis of subspace:
a) $ \operatorname{im}(f) \cap \ker(f \circ g) $
b) $ \operatorname{im}(g \circ f) + \ker(f) $
My solution
a)
Take a random polynomial:
$$p(t) = at^2 + bt + c $$
$$p(1) = a + b + c $$
$$p(-1) = a – b +c $$
$$p'(0) = b $$
so $ g(p) = [a-b+c,b,a+b+c]^T $
Ok now we are looking for $ f\circ g $
$$f(g(p)) = 2(a+c)t^2 + b $$
Ok, now I want its kernel:
$$f(g(p)) = 0 \leftrightarrow a = – c \wedge b = 0$$
so $\ker(f\circ g) = \operatorname{span}[1,0,-1]^T $
I need also $ \operatorname{im}(f)$ so $$ \operatorname{im}(f) = \operatorname{span}([1,0,0]^T,[0,1,0]^T)$$
but
$$ ([1,0,0]^T,[0,1,0]^T,[1,0,-1]^T $$
are linearly independent so $ \operatorname{im}(f) \cap \ker(f \circ g) = 0 $
b)
Now I am looking for $ \operatorname{im}(g \circ f)$
$$g(f(\vec{x})) = [x_1+x_2+x_3,0,x_1+x_2+x_3]^T = \operatorname{span}([1,0,1]^T) = \operatorname{im}(g \circ f)$$
Now $ker(f)$
$$ f(\vec{x})(t)=(x_1+x_3)t^2 + x_2 = 0 \leftrightarrow x_1 = -x_3 \wedge x_2 = 0$$
so
$$\ker(f) = \operatorname{span}([1,0,-1]^T)$$
Ok now we are looking for:
$$ \operatorname{im}(g \circ f) + \ker(f) = \operatorname{span}([1,0,-1]^T,[1,0,1]^T) $$
but $[1,0,-1]^T,[1,0,1]^T $ are linearly independent so
$ \operatorname{span}([1,0,-1]^T,[1,0,1]^T) $ is a basis of $ \operatorname{im}(g \circ f) + \ker(f) $
Thanks for your time!
Best Answer
$$f(x_1,x_2,x_3)=(x_1+x_3)t^2+x_2=k_1t^2+k_2$$ So $\text{Im}(f)$ is spanned by $\{1,x^2\}=\{(1,0,0),(0,0,1)\}$. Clearly, $\ker(f\circ g)=\text{span}\{(1,0,-1)\}\subset\text{span}\{(1,0,0),(0,0,1)\}=\text{Im}(f)$, so the answer to part $(a)$ is $\ker(f\circ g)$.
Answer to part $(b)$ is correct.