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 3×5 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.
Best 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.