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Let $T$ be linear transformation from $V$ to $W$. I know how to prove the result that nullity$(T) = 0$ if and only if $T$ is an injective linear transformation. But I still don't intuitively understand why the kernel only containing the zero vector means that $T$ is injective, and vice versa. In contrast, the relation between the image of $T$ and condition of being surjective is easy to see, since in order to map to all of the elements of $W$ the image of $T$ must have the same dimension as $W$. This can intuitively be seen with a diagram of the mapping from $V$ to $W$, for example. I can't really imagine a diagram that plainly shows the injective condition.

In short, what about nullity$(T) = 0$ imples that $T$ is a one-to-one function?

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    $\begingroup$ Take $x,y$ in the domain of $T$ such that $T(x)=T(y)$. Since $T$ is linear, what can you tell about $T(x-y)$? $\endgroup$
    – Git Gud
    Feb 27, 2017 at 20:52
  • $\begingroup$ I understand that that is the direction one would take to prove the result, since then x - y = 0 --> x = y, which implies injectivity. However, I still don't intuitively see what's going on, and how exactly the kernel relates to injectivity $\endgroup$
    – Wesley
    Feb 27, 2017 at 20:54
  • $\begingroup$ I suggest you make it more clear in the question that you're asking about intuition. I sort of got that vibe partially, but ultimately thought you wanted a proof. You can also add the (intuition) tag. $\endgroup$
    – Git Gud
    Feb 27, 2017 at 21:01

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It comes from the fact that $T(x)=T(y)\iff T(x-y)=0$, so that the kernel contains all the information of the injectivity of $T$ : if it is $\{0\}$, then $T(x)=T(y)$ only has solution $x=y$.

In fact, if $T$ is non injective, the dimension of the set of solutions of equation $T(x)=T(y)$ is independent of $y$ : it's the dimension of $\ker(T)$.

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Because if $\ker(T)\ne\{0\}$ then you have several manners to produce the null vector. Hence you cannot be injective.

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  • $\begingroup$ This does not explain the direction $\implies$ of $Ker(T) \implies T$ is injective $\endgroup$
    – Eduardo Sebastian
    Nov 10, 2020 at 23:25
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If you have for example some linear mapping given by the matrix $A$ such that $f(X) = AX$, you can look at the Kernel. If this mapping is injective, the only vector from $X$ that would give me zero would be a trivial vector, otherwise $X$ (it's actually a null space of matrix $A$) contains more vectors that could give me as a result the zero vector (the same result for two different options is a contradiction with the injectivity).

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