Answer to Question #286443 in Linear Algebra for kevin

Question

Define the kernel and the image of a linear transformation K : M → N and hence Show that the Kernel and image of K are vector subspaces of M and N respectively

Answer 1

$\text{We represent the kernel of K by ker K, and the image by Im K}\\ ker K = \{m \in M: K(m)=0_N\}\\ \text{\bf{\underline{PROOF}}}\\ f(0) = 0_N \implies 0_m \in ker K \neq \empty\\ \text{Let m, n $\in ker K \implies$ K(m) = $0_N$, K(n) = $0_N$ }.\\ \text{Then $K(m) = K(n) = 0_N \implies K(m) - K(n) = 0_N$}\\ \implies K(m-n) = 0_N(\text{Since K is a homomorphism)}\\ \implies m-n \in kerK\\ \text{Also for r $\in$ R, a $\in$ kerK, then $K(a) = 0_N$. By the definition of R-module }\\ \text{homomorphism}\\ K(ra) = rK(a)=r0_N=0_N \\ i.e\quad K(ra) = 0_N \text{. So $ra \in ker K $. And kerK is a module }$

$Im K = \{K(x):x \in M\}\\ \text{Let $n_1, n_2 \in K(m)$. Then there exists $m_1, m_2 \in M$ such that $K(m_1)=n_1, K(m_2) = n_2$}\\ n_1 +n_2 = K(m_1)+K(m_2)=K(m_1+m_2)\text{(Since K is a homomorphism)}\\ \text{Therefore $n_1+ n_2 \in ImK$}\\ \text{Now let r $\in$ and $n \in Im K$. Then there exists m such that K(m) = n and rn = rK(m) }\\ =f(rm).\\ \text{But $rn\in N$, since N is an R-module. Therefore $K(rm) \in ImK$}$