Considering the set of linear transformations from V to W, given each are not injective does this mean that The set doesn’t form a Subspace? I would have thought this would be the case since the 0 vector from V could be mapped to another vector as well as the 0 vector in W? Or would this not be the case since it was given they are a set of LINEAR transformations? To be a Subspace would need the transformation to the 0 vector in W, is this prevented given they are Not injective?
Set of all non- injective linear transformations a Subspace
linear algebralinear-transformationsvector-spaces
Related Solutions
We need $U \subset V$ to say that $U$ is linear independent. Otherwise we wouldn't be able to form linear combinations on $U$. Also we need $U \subset V$ because our map $L$ maps from $V$ to $W$ and we want to prove something about $L(U)$, so we need to be able to apply $L$ to $U$.
Here is a proof of the claim:
Since $U$ is linear independent we have for vectors $u_i \in U$ for all $c_i \in \mathbb{F}$ with $$\sum_i c_i u_i = 0$$ that $c_i = 0 $ for every $i$.
Now we need to show that the set $L(U)$ is linear independent, so we look at elements $L(u_i)$. We need to show that for all $c_i \in \mathbb{F}$ with $$\sum_i c_i L(u_i) = 0$$ $c_i = 0$ for every $i$ holds. Since $L$ is linear we see that $$0 =\sum_i c_i L(u_i) = L \left(\sum_i c_i u_i \right) $$ holds. Now, since $L$ is injective we know that $L$ only sends $0$ to $0$, so $\sum_i c_i u_i = 0$ follows. And because of the linear independence of $U$ we conclude $c_i =0$ for every $i$ and thus $L(U)$ is linear independent.
Your iff-statement is wrong. Only one direction holds. There is an easy counterexample: Every subset $\{v\} \subset V$ with $v \neq 0$ is linear independent since it consists of only one element which is not zero. Now every linear map $L: V\to W$ with $L(v) \neq 0$ sends $\{v\}$ to a linear independent subset of $W$ but these maps don't have to be injective in general.
However if your subset $U \subset V$ is a basis of $V$ and gets send to a linear independent subset $L(U) \subset W$ of $W$, then $L$ is injective. We need to show $\ker(L) = \{0\}$. So suppose $L(v) = 0$. Since $U$ is a Basis we can write $ v = \sum_i c_i u_i$ for some $c_i \in \mathbb{F}$. This means $$ 0 = L(v) = L\left(\sum_i c_i u_i\right) = \sum_i c_i L(u_i) $$ and because $L(U)$ is linear independent it follows that $c_i = 0$ for every $i$ and as such $$v = \sum_i c_i u_i = \sum_i 0 u_i = 0$$ so $\ker(L) = \{0\}$ and we conclude that $L$ is injective.
The most basic example of a non-injective linear map is $z : \mathbb{R} \to \mathbb{R}$ defined by $z(x) = 0$.
The theory of when a linear map is injective is studied extensively in linear algebra. There are many theorems about when such a map is injective when the vector spaces are finite-dimensional.
Best Answer
This is not a subspace. For example, if you are in $\mathbb{R}^2$, consider $$\varphi(x,y)=(x,0) \quad \text{and} \quad \psi(x,y)=(0,y)$$
Both are non-injective, but $\varphi+\psi = \mathrm{id}$ is injective.