How can I find the image of a linear application?

Question

Given the real vector spaces $\mathbb{R}^2$ and $\mathbb{R}^4$, and the application:

$$f: (a, b, c, d) \in \mathbb{R}^4 \mapsto (a + b, c - 2d) \in \mathbb{R}^2$$

How can I find the image of $f$ and its dimension?

I know the dimension of the image is equal to the rank of the matrix given by $a+b$ and $c-2d$, so it's $2$ I guess, but how can I find the image of $f$?

Answer 1

$f$ is linear. Therefore it’s image is a linear subspace of $\mathbb R^2$. Also $f(1,0,0,0) =(1,0)$ and $f(0,0,1,0)= (0,1)$. As $\{(1,0),(0,1)\}$ spans $\mathbb R^2$, the image of $f$ is all $\mathbb R^2$ and the dimension of the image of $f$ is equal to $2$.

Answer 2

Try to think of the question as asking that you have to find all the points in $(x,y)\in \mathbb{R}^2$ which can be expressed in the form $(a+b,c-2d)$ where $a,b,c,d \in \mathbb{R}$. Clearly, choosing $(a,b,c,d)=(x,0,y,0)$ gives us the required point in $\mathbb{R}^2$ and we find that all the points in $\mathbb{R}^2$ can be written in the required form, so that the image is the whole set $\mathbb{R}^2$.

The same setup can be written in the matrix notation in the form:

$$\begin{pmatrix} 1 & 1 & 0 & 0\\ 0 &0& 1 &-2 \end{pmatrix} \times \begin{pmatrix} a\\ b\\ c\\ d\\ \end{pmatrix}= \begin{pmatrix} a+b\\ c-2d \end{pmatrix} $$

We see that the rank of the matrix, which is the dimension of the linear map can be at most the minimum of the number of rows or columns. Here we can span the whole of $\mathbb{R}^2$ by the map, so the dimension must be at least $2$ and the above argument tells us that it is exactly $2$.

Answer 3

You said it your self that the matrix corresponding to this linear map is $$\begin{bmatrix} 1&1&0&0\\0&0&1&-2\end{bmatrix}$$ If you see this matrix, there are two linearly independent columns which means rank of this map is $2$ in this case it is full rank. Thus, image is full space.