How to determine whether the linear operator #T:R^2->R^2# defined by the equations below is one to one?

Question

#w_1=x_1+2x_2#

#w_2=-x_1+x_2#
if so, how to find #T^-1(w_1,w_2)# ?

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Answer 1 · 2018-05-24T17:58:45

Please see the explanation below.

The matrix of the transformation is

#A=((1,2),(-1,1))#

as
#((w_1),(w_2))=A((x_1),(x_2))#

The transformation is one to one if the colums of #A# are linearly independents.

We verify this by writing the augmented matrix and perform a row echelon form.

#((1,2,0),(-1,1,0))#

#<=>#, #R2larr(R2+R1)#, #((1,2,0),(0,3,0))#

#<=>#, #R2larr (R2)/3#, #((1,2,0),(0,1,0))#

#<=>#, #R1larr(R1-2R2)#, #((1,0,0),(0,1,0))#

There is only the trivial solution, the columns are linearly independent and the transformation is one to one.

To calculate the inverse function, calculate

#A^-1=1/3((1,-2),(1,1))#

#x_1=1/3w_1-2/3w_2#

#x_2=1/3w_1+1/3w_2#

Verification :

#A A^-1=((1,2),(-1,1))*((1/3,-2/3),(1/3,1/3))#

#=((1,0),(0,1))#

#=I#