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I am struggling with an algebra problem here is what we got :

Assume we have :

$$L : E_0 to E_1 quadtextandquad K : E_1 to E_2 $$

We have to show that :

$$dim(ker(K circ L)) leq dim(ker(L)) + dim(ker(K)).$$

I define $H$ as :

$$H : ker(K circ L) to E_2, qquad H(v) = L(v) quad textfor quad v ∈ ker(K circ L)$$

Now I have to prove that :

$$ ker(L) = ker(H) $$

But how can I show that :

$$Rightarrow qquad ker(L) subseteq ker(K circ L) quad textand quad ker(K circ L) subseteq ker(L) quad ?$$

I got an idea for this one : $L(v) = 0$ implies $K(L(v)) = 0$ ... But I stuck with this one.

$$ Rightarrow qquad operatornameim(H) subseteq ker(K) quad ? $$

How can I prove? I only know and prove that $operatornameim(Kcirc L) subseteq operatornameim(K)$.

Thanks.

Generally you cannot show that $ker (mathcal KL) = ker mathcal L$: if $mathcal Kv =0$ for all $v in E_1$, where $E_1 neq 0$, and $mathcal L$ surjective, then $ker(mathcal KL) = E_0$ but $ker mathcal Lsubsetneq E_0$ [otherwise $mathcal L$ would not be surjective: some $v in E_1 setminus 0$ cannot be mapped by $mathcal L$].

Assume $dim(E_0) = n <+infty$. By the Rank-Nullity Theorem, $ dim(ker mathcal L)+ dim(mathrm im, mathcal L) = n$, also $
dim(ker (mathcal KL)) + dim(mathrm im (mathcal KL)) =n
$.

Then
beginalign*
dim(ker (mathcal KL)) &= n - dim (mathrm im (mathcal KL))\
&= dim (ker mathcal L) + dim(mathrm im (mathcal L )) \
&quad - dim (mathrm im(mathcal KL)).
endalign*

Note that the mapping $mathcal Kvert _mathrm im,mathcal L colon mathrm im, mathcal L to E_2$ is a linear mapping, and we could know that $mathrm im(mathcal Kvert _mathrm im, mathcal L) = mathrm im (mathcal KL)$: $z in mathrm im (mathcal Kvert _mathrm im ,mathcal L) iff exists y in mathrm im, mathcal L, mathcal Ky = z iff exists x in E_0, y = mathcal Lx, z = mathcal Ky iff exists x in E_0, z = mathcal KLx$. So again by the Rank-Nullity Theorem,
$$
dim (mathrm im, mathcal L) = dim(ker (mathcal Kvert _mathrm im,mathcal L)) + dim (mathrm im(mathcal KL)).
$$
Thus
$$
dim (mathrm im, mathcal L) - dim (mathrm im(mathcal KL)) =dim(ker (mathcal Kvert _mathrm im,mathcal L)) leqslant dim (ker mathcal K),
$$
the $leqslant$ holds because $mathrm im ,mathcal L subset E_1$, and possibly there is some $y in (E_1setminus mathrm im, mathcal L)$ that $mathcal Ky = 0$.

Hence the inequality we want to prove.