Let "T:\\mathbb{R}^2\\to \\mathbb{R}^2" such that "T(x_1,x_2)=(2x_2-2x_1,x_1-x_2)."
We can see that;
"T(0,0)=(2.0-2.0,0-0)\\\\\n\\implies T(0,0)=(0,0)\\\\\n\\text{pre-multiply } T^{-1} \\text{ to both sides}\\\\\n\\implies (0,0)=T^{-1}(0,0)\\\\\n\\implies T^{-1}(0)=0"
Also, we can see that;
"Ker(T):=\\{(x_1,x_2)|x_1=x_2\\}"
And, the basis of "Ker(T)=\\{(1,1)\\}\\implies dim(Ker(T))=1"
Hence, "T" is not one-to-one
Comments
Leave a comment