Consider the map T: R^m → R^m such that T(x_1, x_2, ..., x_m) = (x_1, x_2 - x_1, ..., x_m - x_{m-1}). Then:
a) T is a one-to-one linear operator.
b) T is an onto linear operator.
c) T is both one-to-one and onto linear operator.
d) T is neither one-to-one nor onto linear operator.
Answer (1)
To determine the properties of the map T : R m → R m defined as T ( x 1 , x 2 , ... , x m ) = ( x 1 , x 2 − x 1 , ... , x m − x m − 1 ) , let's analyze its linearity, injectivity (one-to-one), and surjectivity (onto):
Linearity : For T to be a linear transformation, it must satisfy the following properties for all vectors u , v ∈ R m and scalars c :
Additivity : T ( u + v ) = T ( u ) + T ( v )
Homogeneity : T ( c u ) = c T ( u )
Considering the way T is defined, it is constructed using basic vector operations (addition and subtraction of components), which is consistent with linearity. Therefore, T is a linear operator.
Injectivity (One-to-one) : A function is one-to-one if different inputs lead to different outputs. Analytically, T is injective if T ( x ) = T ( y ) ⇒ x = y .
Let's consider T ( x ) = 0 , where 0 = ( 0 , 0 , ... , 0 ) . This means:
x 1 = 0
x 2 − x 1 = 0 ⇒ x 2 = x 1 = 0
Continue in this manner...
x m − x m − 1 = 0 ⇒ x m = x m − 1 = 0
Hence, the only solution to T ( x ) = 0 is x = 0 . This indicates that T is injective.
Surjectivity (Onto) : For T to be onto, each y ∈ R m should have a corresponding x ∈ R m such that T ( x ) = y .
Take y = ( y 1 , y 2 , ... , y m ) . To find x , solve:
x 1 = y 1
x 2 = y 1 + y 2
...
x m = y m − 1 + y m
This construction shows that for any y ∈ R m , we can find an x such that T ( x ) = y , proving T is surjective.
Combining these properties, we conclude the answer is c) T is both one-to-one and onto linear operator.
