Proof that these statements about injectivity are equivalent? Let T : $R^n$ → $R^m$ be a linear transformation, and let x,y be vectors in $R^n$,$R^m$ respectively. Prove that the following statements are equivalent: A: T is not injective.
B: the set {$x$ ∈ $R^n$: T($x$) = 0} is infinite (i.e., has infinitely many elements).
C: for every y ∈ image(T), the set {x ∈ $R^n$: T(x) = y} is infinite.
I'm stumped. Can anyone help?
A: Hints (to show $(1) \iff (2))$: 
(1) Prove that $\ker T := \{v \in \mathbb{R}^n : Tv = 0\}$ is a subspace of $\mathbb{R}^n$
(2) Prove that a linear map $T: \mathbb{R}^n \to \mathbb{R}^m$ is injective if and only if $\ker T = \{0\}$.

Thus, if $T$ is non-injective, then $\ker T \neq \{0\}$. Since this is an $\mathbb{R}$-vector space, what do you know about the amount of elements in $\ker T$?
Conversely, if $\ker T$ contains infinitely many elements, $T$ can't be injective.

To show $(1)\iff (3)$, use dual statements to the ones I provided above.
A: Recall that a map $T$ is injective iff $\ker(T) = 0$. We prove this now.
Assume that $T$ is injective, then we have that $T(x) = T(y) \implies x= y$. If $z \in \ker(T)$, then we have $T(z) = 0$, but $T$ being a linear map means that $T(0) = 0$, so we have $T(0) = T(z) \implies z = 0$. So the only element in $\ker(T)$ is $0$; i.e., $\ker(T) = 0$.
Assume that $\ker(T) = 0$. We wish to show that $T$ is injective. Let $T(x) = T(y)$, then we have $T(x) - T(y) = 0$. By linearity, this gives $T(x-y) = 0$, and since $\ker(T)$ is trivial we have $x -y = 0$, or $ x = y$. So $T$ is injective.
Now, we show that $\ker(T)$ is a subspace. Recall that a subset is a subspace if it is closed under scalar multiplication and addition. Let $a,b \in \ker(T)$, we need to first show that $a +b \in \ker(T)$. But this is clear, since $T(a+b) = T(a) + T(b) = 0 + 0  =0$. Next, if $r \in \mathbb{R}$, $a \in \ker(T)$, we want to show that $ra \in \ker(T)$. This follows from the fact that $T(ra) = rT(a) = r \cdot 0 = 0$. So it is a vector subspace.
$(1) \implies (2)$: If $T$ is not injective, then we have $\ker(T) \neq 0$, so there is some $z \in \ker(T)$. Furthermore, $rz \in \ker(T)$ for all $r \in \mathbb{R}$, so we have that $\ker(T)$ has infinitely many elements.
$(2) \implies (3)$: If $\ker(T)$ is infinite, we have infinitely many $z \in \ker(T)$. We want to show that, for $y \in \text{Im}(T)$, the set $\{x \in \mathbb{R}^n : T(x) = y\}$ is infinite. This is easy; we have that there is some $x$ which maps to $y$, since it's in the image, and for every $z \in \ker(T)$ we have that $T(x+z) = T(x) + T(z) = T(x) = y$, so this set is infinite.
$(3) \implies (1)$: This also follows from definitions. Since the set is infinite, we have that there are at least two so that $T(x) = T(y)$ but $x \neq y$, so $T$ fails to be injective.
