This is a question from a past linear algebra final exam. I am trying to find the range and kernel of a linear transformation $T:\mathbb{R}^5 \rightarrow \mathbb{R}^2$ represented by $T(x)=Ax$, where $A$ is the matrix $$ \begin{bmatrix} 1 & 0 & 0 & 1 & 2\\ 2 & 1 & 1 & 2 & 4\\ 0 & 1 & 1 & 0 & 0\\ \end{bmatrix} $$
I can find the range by finding the column space of $A$, which are just the vectors $(1,2,0)^T$ and $(0,1,1)^T$. I can find the kernel by solving the equation $Ax=0$ which has only the trivial solution. Therefore, T is one-to-one. For $T$ to be onto, the rank of $T$ must equal the dimension of $\mathbb{R}^2$, and since the rank is 2, $T$ is also onto. Since it is both onto and one-to-one, $T$ is also an isomorphism.
However, I don't understand how a transformation from $\mathbb{R}^5$ to $\mathbb{R}^2$ can be represented by a 3x5 matrix. Applying the transformation on each column of the identity matrix should produce an image in $\mathbb{R}^2$, but how can this be if each column in this matrix has 3 elements? I see that each column has a zero, but this still doesn't make sense to me since the second and third columns have a z element. It is possible this was a typo in the question if no one has an answer.
For the sake of completeness, this answer summarizes the answers to the exam question and your questions and the comments. I assume that $T:\Bbb R^5 \rightarrow \Bbb R^3.$
The range of $T$ is the subspace of $\Bbb R^3$ spanned by $$\left\{ \begin{bmatrix} 1\\ 2\\ 0\end{bmatrix}, \begin{bmatrix} 0\\ 1\\ 1\end{bmatrix} \right\}.$$ $T$ is not onto because $\Bbb R^3$ has three dimensions, so the image $T$ would require three linearly-independent spanning vectors.
You know how to find the kernel of $T,$ so I will just give an answer: It is the subspace of $\Bbb R^5$ spanned by $$\left\{ \begin{bmatrix} -2\\ 0\\ 0\\ 0\\ 1\end{bmatrix}, \begin{bmatrix} -1\\ 0\\ 0\\ 1\\ 0\end{bmatrix}, \begin{bmatrix} 0\\ -1\\ 1\\ 0\\ 0\end{bmatrix} \right\}.$$ As a partial check on the answer, we see that the dimensions of the range and kernel sum to the dimension of $V.$
A linear transformation $T:\Bbb R^n \rightarrow \Bbb R^m$ with $n > m$ is not one-to-one and may or may not be onto. If $n < m,$ $T$ is not onto and may or may not be one-to-one. In neither case is $T$ one-to-one.