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In my linear algebra classes, we often just assume that a kernel is a subspace. However, how do we prove this?

My current idea is to apply subspace theorem (closed under addition, multiplication, and contains the zero vector). However when I try working it out, I can not seem to get it.

Does someone know how to actually prove this?

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5 Answers 5

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Suppose $T$ is a linear transformation $T:V \rightarrow W$ To show $Ker(T)$ is a subspace, you need to show three things:

1) Show it is closed under addition.

2) Show it is closed under scalar multiplication.

3) Show that the vector $0_v$ is in the kernel.

To show 1, suppose $x,y\in Ker(T)$. Then $T(x+y)=T(x)+T(y)=0_w+0_w=0_w$ Hence $x+y$ also exists in the kernel and so the kernel is closed on addition.

To show 2, Assume $\lambda \in F,x\in Ker(T)$ so it follows that $T(\lambda x)=\lambda T(x)=\lambda 0_w=0_w$ So again $\lambda v\in Ker(T)$ so it is closed on scalar multiplication.

Finally it is simple to see 3, $\forall v \in V,T(0_v)=T(v+(-v))=T(v)+T(-v)=T(v)-T(v)=0_w$ so indeed $0_v$ is in the kernel.

So you have shown that $Ker(T)$ is a subspace of $V$.

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  • $\begingroup$ Aren't conditons (1) and (2) alone enought to say that some nonempty subset M of a vector space is a subspace ? We can get (3) from (1) and (2) the following way, take v from M (it is possible because M is nonempty). Then (-1)v is in M by (2), and 0=v+(-v)=v+(-1)v is in M by (1). $\endgroup$ Feb 12, 2021 at 14:50
  • $\begingroup$ @ЮрійЯрош What you say is true only if M is nonempty as you said. However, some definitions of vector spaces do not exclude the possibility that M is $\phi$. See here: math.stackexchange.com/questions/627940/… $\endgroup$ Feb 12, 2021 at 15:05
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This is a simple applicaiton of properties of linear transformations. As a sketch of the proof:

(1) Contains zero: $$f(0)=f(0_F 0_V)=0_Ff(0_V)=0$$ (2)Scalar multiplication: $$f(cv)=cf(v)=c0_V=0$$ (3) Addition $$f(v_1+v_2)=f(v_1)+f(v_2)=0+0=0$$

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Let $V$ and $W$ linear spaces and $f:V \to W$ linear. Let $K:=ker(f)$

From $f(0)=0$ we get $ 0 \in K$.

Now suppose that $x,y \in K$ and $\alpha, \beta$ are scalars. Use linearity to derive

$f(\alpha x+ \beta y)= \alpha f(x) + \beta f(y)$. Do you now see that $\alpha x+ \beta y \in K$ ?

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Let $f : E\rightarrow F$ be a linear mapping. we have :

  • $0_E \in Kerf$
  • $\forall x,y\in Kerf,\ \forall \alpha \in \mathbb{K}, f(\alpha x +y)=\alpha f(x)+f(y)=0_F \Rightarrow \alpha x +y\in Kerf$ because $f(x)=f(y)=O_F$. Then $Kerf$ is a subspace of $E$.
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The kernel of F is a subspace of V. Proof Since F(O) = 0, we see that 0 is in the kernel. Let v, w be in the kernel. Then F(v + w) = F(v) + F(w) = 0 + 0 = 0, so that v + w is in the kernel. If c is a number, then F(cv) = cF(v) = 0 so that cv is also in the kernel. Hence the kernel is a subspace.

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