Let $T:\textbf{R}^{4}\to\textbf{R}$ be a linear transformation. Determine $\dim\text{Im}(T)$ and $\dim\ker(T)$. I asked a question on linear maps a few days ago and the community was extremely helpful in allowing me to understand the topics of Linear Maps. I am coming back with a new question in hopes that I can figure out how to get this problem done, as I'm struggling with a lot of the definitions and just calculations that needed to be done.
For reference, this is the problem:
Here is what I understand about these topics in-so far:

*

*A basis for a vector space V is a set of vectors A in V such that A is linearly independent and spans V.


*The linear map is defined by having properties: t(u+v)=t(u)+t(v) (linearity) and it holds under scalar multiplication like at(u) = t(au).


*The image of, x, under a linear transformation is the effect of the transformation on vector x. Does this mean that image is: T([$e_1,e_2,e_3,e_4$]) = [$e_1+e_2+e_3+e_4$]?


*The kernel is the set of vectors that map to the 0 vector?
With this knowledge, how could I begin to approach this problem / solve the question at hand. Many of the answers can be true here.
 A: In order to solve the proposed problem, it is useful to remind the image of a linear transformation $T:\textbf{R}^{4}\to\textbf{R}$ is spanned by $T(\mathcal{B})$, where $\mathcal{B}$ is a basis for $\textbf{R}^{4}$.
In the present case, it has been given that $\mathcal{B} = \{e_{1},e_{2},e_{3},e_{4}\}$. Hence it can be concluded that:
\begin{align*}
\text{Im}(T) = \text{span}\{T(e_{1}), T(e_{2}), T(e_{3}), T(e_{4})\} = \text{span}\{1\} = \textbf{R}
\end{align*}
The last relation holds because $\dim\textbf{R} = 1$.
Finally, it can be claimed that $\dim\text{Im}(T) = 1$ and $\dim\ker(T) = 4 - 1 = 3$ due to the Rank-Nullity theorem.
Hopefully this helps!
A: Given  that $T: \mathbf{R}^4 \to \mathbf{R}$ is defined by
$$
T(\mathbf{x}) = x_1 + x_2 + x_3 + x_4
$$
Note that we find
$$
T(\mathbf{e}_1) = 1, T(\mathbf{e}_2) = 1, T(\mathbf{e}_3) = 1,T(\mathbf{e}_4) = 1
$$
Hence, it follows that
$$
\mbox{Image}(T) = \mbox{Range}(T) = \mbox{span}\{ T \mathbf{e}_1,
T \mathbf{e}_2, T \mathbf{e}_3, T \mathbf{e}_4 \} = \mbox{span} \{ 1 \}
= \mathbf{R}
$$
Thus, $\mbox{rank}(T) = \mbox{dim (Image)}(T) = 1$.
By the Rank-Nullity Theorem, it follows that
$$
\mbox{rank}(T) + \mbox{nullity}(T) = \mbox{dimension of the domain of} \ T = 4
$$
Hence, we deduce that
$$
\mbox{nullity}(T) = 4 - 1 = 3
$$
Alternatively, we can compute the nullity of $T$ by a direct calculation.
The null space of $T$ is given as the subspace of $\mathbf{R}^4$ consisting
of the solutions of the equation
$$
x_1 + x_2 + x_3 + x_4 = 0
$$
Thus, the null space of $T$ consists of the vectors
$$
\mathbf{x} = \left[ \begin{array}{c}
 - x_2 - x_3 - x_4 \\
 x_2 \\
x_3 \\
x_4 \\
\end{array} \right]
$$
Hence, the null space of $T$ is spanned by the basis
$$
\mathcal{B} = \left\{ \left[ \begin{array}{c}
 -1 \\
1 \\
 0 \\
0 \\ \end{array} \right], \left[ \begin{array}{c}
 -1 \\
0 \\
 1 \\
0 \\ \end{array} \right],  \left[ \begin{array}{c}
 -1 \\
0 \\
 0 \\
1 \\ \end{array}  \right] \right\}
$$
This also shows that $\mbox{nullity}(T) = 3$.
