Answer:
is linearly dependent set.
Step-by-step explanation:
Given:
is a linearly dependent set in set of real numbers R
To show: the set
is linearly dependent.
Solution:
If
is a set of linearly dependent vectors then there exists atleast one
such that 
Consider 
A linear transformation T: U→V satisfies the following properties:
1. 
2. 
Here,
∈ U
As T is a linear transformation,

As
is a linearly dependent set,
for some 
So, for some 

Therefore, set
is linearly dependent.
Answer:
CDF is scaled down by a factor of 1/2 of ABC
Step-by-step explanation: