Let $L:\mathbb{R^3}\to \mathbb{R^2}$ be a linear transformation defined by $L($x$) = (x_1+x_2,x_2+x_3)^T$ for all $(x_1,x_2,x_3)^T ∈\mathbb{R^3}$
If $S$ is a subspace of $\mathbb{R^3}$ spanned by the standard basis vectors for $\mathbb{R^3}$: e$_1$, e$_3$ then we can find the image of $S$
- if x $∈S$ then x must be of the form $(α, 0, β)^T$
- Hence, $L($x$) = (α, β)^T$
- So, image of $S$ denoted $L(S)=\mathbb{R^2}$
Find $range(L)$
- Since image of $S$ is $\mathbb{R^2}$ then the $range(L)$ is also $\mathbb{R^2}$
What is the relation between the first part and the second part?
I suppose if I were to calculate part 2. without considering part 1. I would do this:
- if v $∈V$ then v must be of the form $(α, θ, β)^T$
- Hence $L($v$) = (α + θ, θ + β)^T$
- With this, I would assume that the image of $V$ or the $range(L) = \mathbb{R^2}$ because the addition of different scalars would probably still include all the values of $\mathbb{R^2}$?
But I am still curious about the relation between the first and second parts
$S$ is a subset of your domain, so if you know the image of $S$ is $\mathbb R^2$ then your range contains at least $\mathbb R^2$. But by definition of $L$ being a function from $\mathbb R^3$ to $\mathbb R^2$, it can't be more than $\mathbb R^2$ so it must in fact be $\mathbb R^2$.
As for your method of calculating part 2, to formalise the last bullet point given any $(\alpha, \beta)^T \in \mathbb R^2$ we know $L(\alpha, 0, \beta)^T = (\alpha, \beta)^T$ so $(\alpha, \beta)$ is in the range of $L$. This is essentially the same argument as the one involving the image of $S$, just phrased differently.