Find the range of a linear map $T\colon\mathbb{R}^2 \to \mathbb{R}^4$.

Question

Given the matrix $$A=\begin{pmatrix} 1 & 1 \\ 2 & 1 \\ 2 & -3 \\ 1 & -3 \end{pmatrix},$$ compute the range of $$T\colon\Bbb R^2\to\Bbb R^4,\quad x\mapsto Ax.$$

I'm not sure how to proceed. Do you take the transpose of $A$ which equals the coordinates given by $(x,y)$ in $R^4$ and then RREF to determine the vector bases and that computes the range?

Answer 1

Hint: The image of the linear transformation represented by a matrix is always the span of the columns.

In this case, $T\begin {pmatrix}x\\y\end{pmatrix}=x\begin {pmatrix}1\\2\\2\\1\end{pmatrix}+y\begin {pmatrix}1\\1\\-3\\-3\end{pmatrix}\,,\forall \begin {pmatrix}x\\y\end{pmatrix}\in\Bbb R^2$.

Answer 2

Hint: Calculate $T(e_1),T(e_2)$. These two vectors span the image of $T$. If two vectors are independent, then they form a basis of $Im(T)$. If not, then find linearly independent vectors from $T(e_1),T(e_2)$.