if $f$ is linear transformation, is proof that $f(U)$ is linear subspace and find its dim Given is Linear transformation $f \in L(\mathbb R^3, \mathbb R[t]_3)$ such that: 
$\ker f = span([1,1,1]^T)  $ 
Given is also $U = span([3,1,-1]^T,[-1,1,3]^T)$ 
Proof that $f(U)$ is linear subspace in $\mathbb R[t]_3$ and find its dim
My try
$$ f(U) = \left\{ f(u) | u \in U \right\} $$
Take a random $u \in U$. Then
$$ u = \alpha[3,1,-1]^T + \beta[-1,1,3]^T$$
We know that $f$ is linear transformation so:
$$f(u) = \alpha f \left([3,1,-1]^T \right) + \beta f\left([-1,1,3]^T\right)$$
I check if the scalar multiplies or adds vectors from the subspace derives from $f(U)$ 
$$ \gamma f(u) = (\alpha\gamma) f \left([3,1,-1]^T \right) + (\beta\gamma) f\left([-1,1,3]^T\right)$$ 
and
$$f(u)+f(u') = (\alpha + \alpha') f \left([3,1,-1]^T \right) + (\beta + \beta') f\left([-1,1,3]^T\right) $$
ok. So this is linear subspace.
Now I am going to find dim.
But $[3,1,-1]^T,[-1,1,3]^T $ are linearly independent. So 
$$ \dim f(U) = 2 $$
What is the problem?
The problem is that the correct answer should be
$$ \dim f(U) = 1 $$
I suspect that it comes to $\ker f = span([1,1,1]^T)  $ because $\ker f \in U $. But why does it matter? I do not think that this is any theorem on that situation
 A: While $(3, 1, -1)$ and $(-1, 1, 3)$ are linearly independent, there is no guarantee that $f(3, 1, -1)$ and $f(-1, 1, 3)$ (which span $f(U)$) are linearly independent. What if $f$ maps one (or both!) to $0$?
Think about it this way: consider the restriction $f|_U$ of the linear map $f$ to the set $U$. This is now a linear map from $U$ to $\mathbb{R}^3$. What would the kernel be for this linear transformation? It should be the kernel of $f$ intersected with $U$. What is that intersection? Is it trivial? Is it not trivial?
Note that the range of $f|_U$ is simply $f(U)$. Using the rank-nullity theorem on $f|_U$ (remembering it's now defined on a $2$-dimensional space $U$), you can compute the dimension of this space using the dimension of the kernel.
A: $f(\begin{bmatrix}3&1&-1\end{bmatrix}^T),f(\begin{bmatrix}-1&1&3\end{bmatrix}^T)$ might not be linearly independent.
Notice that $\begin{bmatrix}3\\1\\-1\end{bmatrix}=2\begin{bmatrix}1\\1\\1\end{bmatrix}-\begin{bmatrix}-1\\1\\3\end{bmatrix}$
$f(u) =\alpha f(\begin{bmatrix}3&1&-1\end{bmatrix}^T)+\beta f(\begin{bmatrix}-1&1&3\end{bmatrix}^T)\\=\alpha f(2\begin{bmatrix}1&1&1\end{bmatrix}^T-\begin{bmatrix}-1&1&3\end{bmatrix}^T)+\beta f(\begin{bmatrix}-1&1&3\end{bmatrix}^T)\\=(\beta-\alpha)f(\begin{bmatrix}-1&1&3\end{bmatrix}^T)$
Since $\begin{bmatrix}-1&1&3\end{bmatrix}^T\notin\ker(f),f(\begin{bmatrix}-1&1&3\end{bmatrix}^T)\ne0$. Thus, $\dim f(U)=1$. 
