MATRICES - TRANSFORMS
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Linear algebra here. So I would like to understand this 1-1 and onto of a matrix. Let's say we have a mapping of T: R3-->R2 where T(x1,x2,x3)=(x1+x3,x2-2*x3)<---R2 this not 1-1 since DimNul(A)=1. Which makes the Ker(T)={1}. T is onto simply because there is a solution. Or because rank(A)=2. (No zero rows) but if we had a mapping o R3-->R3 then it won't be 1-1 nor onto, right? Can a matrix be onto but not 1-1 and the other way around?

GIVING BELOW THE GENERAL TEST PROCEDURE FOR YOUR PROBLEM
HOW TO CHECK A L.T.MATRIX FOR ONE TO ONE & ON TO
T[U]=M*U=V
T[X1,X2,X3]=M*[X1,X2,X3]=[X1+X3,X2-2X3]…SO WE GET
M=				U=
1	0	1	*	X1	=	X1+X3
0	1	-2		X2		X2-2X3
				X3
ONE TO ONE ….		CHECK LINEAR INDEPENDENCE OF COLUMN VECTORS..
WE HAVE 3 COLUMN VECTORS IN R2 OF DIMENSION 2
SO THEY ARE LINEARLY DEPENDENT .
SO THE TRANSFORM CAN NOT BE ONE TO ONE ….
ON TO …	CHECK WHETHER THE COLUMN VECTORS SPAN THE RANGE SPACE
NR3=R3+4R1
1	0	1
0	1	-2
WE FIND THAT THERE ARE 2 L.I. CONUMNS .
WE NEED 2 L.I. VECTORS IN R2 TO SPAN R2 .
SO THEY CAN SPAN R2
SO THE TRANSFORM IS ON TO .