Injectivity is equivalent to null space equals {0}

Question

Let T : V $\rightarrow$ W. Then T is injective if and only if null T = {0}.

What does the null T = {0} intuitively mean here?

Answer 1

It means that the one and only vector that $T$ sends to $0$ is the $0$ vector itself. As a non-example, think about a projection in $\Bbb{R}^2$ onto the $x$-axis. That projection sends the entire $y$-axis to $0$.

If you want to think of it intuitively, I actually prefer to think about it meaning that $T$ is injective. It means that in the graph of our linear transformation does not have any horizontal bits. If you think about, say, a function $f(x, y) = ax + by$ from $\Bbb{R}^2$ to $\Bbb{R}$ (which defines a plane in $\Bbb{R}^3$), then no matter how you tilt the plane, there's always going to have to be a line contained in the plane that's horizontal. There's no way to prevent this! Thus, the function $f(x, y)$ cannot be injective, regardless of the values of $a$ and $b$, so the nullspace must be larger than $\{(0, 0)\}$.

Answer 2

Though it is guessable from the choice of standard notation you have used, you should have mentioned that $V,W$ are vector spaces and that $T$ is a linear transformation.

As you are asking for intuitive meaning, again I guess you know the formal definition of what null (T) stands for. It is the null space of $T$, consisting of all those vectors $v\in V$ for which $T(v)= 0(\in W)$.

If $T$ is not injective we can find two vectors $u,v\in V, u\ne v$ with $T(u)=T(v)$. Then we will see that the vector $u=v\in 0$ will be in null space of $T$.

If on the other hand if $u,v\in V$ are such that $T(u)=0, T(v)\ne 0$ then by linearity $T(u+v)= T(u)+T(v)= 0+T(v)=T(v)$. That is $T$ takes the same value on $u+v$ and $v$ showing it is non-injective.

Answer 3

In a linear transformation the zero vector is always in the null space. $T(\mathbf 0) = \mathbf 0$

The transformation is injective, $T(u) = T(v) \implies u=v.$ Only the zero vector is in the null space.