Question

Kernel and image proofl inear algebraI don't understand the solution L(0v)=0w if i put L(v3)=0w is it ok or not why it has to be L(0v)=0w?? is it because one to one 0 has to be mapped to 0???
Step 2 of 3 A linear transformation L : V → W is said to be one to one if L (v.) = L(%) We show L is one to one if and only if ker(L)-o,) Suppose, L is one to one and veker(L) v.-v2 L(v) =0, and L(Q)=0, Since, L is one to one it follows that Therefore ker L={0,} v =0 Comment Step 3 of 3 A Conversely suppose, that ker(L)-(0.) and L(%)=L(%) Hence, v,-v, =0, Therefore, L is one to one Hence L is one to one if and only if ker(L)-0

A linear transformation L: V → W is said to be
one-to-one if L (v1) = L (v2) implies that v1 = v2
(i.e., no two distinct vectors v1, v2 in V get mapped
into the same vector w ∈ W). Show that L is
one-to-one if and only if ker(L) = {0V }

Answer 1

ANSWER:

Given that

A linear transformation L:v->w is said to be one to one

The same vector w $\epsilon$W

Then the L is one to one if and only if ker(L)={$_{0v}$}