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True or False. Please explain your answer. If T: R^n --> R^n where T(X) = Ax, then the reduced echelon form of A is I. Thank you.
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Is A independant or not?
Yes.
if A is a basis for R^n (efficient span); then yes, it row reduces to I (the identity matrix)
but without knowing if the arrow is meant to define a function from R^n ONTO R^n, or from R^n INTO R^n ... i wouldnt know how respond.
spose A = 2 4 1 2 for any x,y in R^2 we have: (2x+4y) (x+2y) which is a vector IN R^2, but since A is not independant, we are not ONTO R^2
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