if $f,g$ are linear transformations, find basis of $ \operatorname{im}(f) \cap \ker(f \circ g) $ and $ \operatorname{im}(g \circ f) + \ker(f) $

Question

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!

Answer 1

$$f(x_1,x_2,x_3)=(x_1+x_3)t^2+x_2=k_1t^2+k_2$$ 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.

Answer 2

One thing I learned early: the usual terminology here is an 'arbitary' polynomial rather than a 'random' one. 'Random' implies you want to bring in probability/measure theory, which you don't.

Now onto more serious matters. It is important that when talking about $ker(f \circ g)$ and $im(f)$ you recognise the space they actually live in, which is $\mathbb{R}[t]_2$. Your answers aren't in $\mathbb{R}[t]_2$ but $\mathbb{R}^3$, so they fail to answer the question and also cause confusion. It is vital that we correct your results to be subsets of the space they should be. For instance...

$im(f)= \{ax^2+b|a,b \in \mathbb{R}\}$, for instance. Can you do the same for $ker (f \circ g)$?

For the second question, can you identify $im(g \circ f)$ and $ker(f)$ as subspaces of $\mathbb{R}^3$?