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Whenever I try to learn about a relationship, I try to reason intuitively why a theorem or lemma should make sense. I know often times this is increasingly difficult to achieve. However, I have the following:

Let $ L: \mathbb{V} \to \mathbb{W} $ be a linear mapping. $L$ is one-to-one if and only if Ker($L$) = $\{\vec{0}\}$

I can intuitively explain the forward direction, that if $L$ is one-to-one, then only one value can possibly map to $\vec{0}$ and that must be $\vec0$.

When I consider the other direction, it is hard for me to intuitively explain that the property of a linear mapping being one-to-one follows naturally from the fact that its kernel contains only $\vec0$. I know that the algebraic proof of this is trivial, but I feel like without a more intuitive understanding, I am misunderstanding something about linear mappings and hence not fully appreciating the importance of this relationship.

Thanks in advance.

EDIT:

The question is, can I explain the logic behind this principle without implicitly referring to the proof?

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4 Answers 4

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I try to see it like this:

The kernel $K$ of a linear map $L:V \to W$ is a kind of universal fiber. The mathematical version of this is that the fiber $L^{-1}(w)$ of $w$ is $v + K$, for any $v$ with $L(v)=w$. So the fiber is essentially a translated version of the kernel.

To represent this pictorially, I imagine $K$ moving around in $V$, and at any given point, all the points in this moving $K$ are sent to the same point in $W$.

Now, an injective map is one with all fibers of cardinality one. So using the above, it is quite intuitive to see that if the kernel is $\{0\}$, all the fibers are singletons.

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  • $\begingroup$ This is very clever @Nitrogen ! +1 $\endgroup$
    – user171326
    May 6, 2017 at 10:34
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If $f$ is not injective, there are $v\ne w$ such that $f(v)=f(w)$. By linearity, then, $f(v-w)=0$, so the kernel contains $v-w$. But this is nonzero because $v\ne w$.

I find this pretty intuitive. The key underlying fact is that a linear function applies in the same way to differences between vectors as to vectors themselves, so if there's a nonzero difference that becomes no difference, there must also be a nonzero vector that becomes zero.

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Intuitively, you can think of it as follows. Because the map is linear, the only way two values get mapped to the same thing is if they differ by an element of the kernel. This is because $A(x) = A(y)$ implies $A(x) - A(y) = A(x-y) = 0$ so that $x-y$ is in the kernel.

If the kernel only contains $0$, we're fine because the only way $x-y$ is in the kernel is if $x-y=0$, or that $x=y$, and the map is 1-1. But if the kernel contains other elements, say there exists $0 \neq k \in Ker(A)$, then for any $x$ in your space, $A(x + k) = A(x) + A(k) = A(x)$ so that $x+k$ and $x$ get mapped to the same value, which violates being 1-1.

Essentially, because the map is linear, you can "shift" elements around by elements of the kernel without affecting the result of the map. This will cause issues being 1-1 unless the only element you can "shift" by is $0$.

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We know that the kernel is a subvectorspace. Hence if $x_1, x_2$ are (two different elements, nonzero) in the kernel, we have that $x_1 - x_2$ is in the kernel, so we find that $$0 = L(x_1 - x_2) = L(x_1) - L(x_2)$$ by linearity. This would mean that $L$ is not one-to-one, since we have two different points with the same image.

I think this is just such a basic concept, that any 'non algebraic argument' can be translated to an algebraic argument en vice versa. I could translate my algebraic explanation into words as follows: If we suppose that the map is not injective, then there are different elements having the same image, hence by linearity their difference is mapped onto zero. Therefore the kernel is not trivial. But this is essentially what I wrote above.

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  • $\begingroup$ I feel like you did not answer the OP's question... quoting them: "I know that the algebraic proof of this is trivial, but I feel like without a more intuitive understanding, I am misunderstanding something about linear mappings and hence not fully appreciating the importance of this relationship.". $\endgroup$
    – C. Falcon
    Jan 29, 2017 at 20:40
  • $\begingroup$ you are right, I overlooked this in the OP's question, however, in my defence, I find this quite intuitive... $\endgroup$
    – Student
    Jan 29, 2017 at 20:43

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