Assuming that a 1, a2, ..., and are linearly related, then
There are numbers k 1, ..., and kn is not all zero, so
k 1a 1+k2a2+...+knan=0
Then only the components of the front part of each vector in the vector group are observed, and obviously there are
k 1b 1+k2b2+...+knbn=0
Because k 1, ..., kn are not 0, according to the definition of linear correlation, we know that
The original vector group b 1, b2, ..., bn are linearly related, and contradictions are obtained!
Therefore, the assumption is not true, and it is still linearly irrelevant after "growth"
(2) Because a 1, a2, ..., and are linearly related, then
There are numbers k 1, ..., and kn is not all zero, so
k 1a 1+k2a2+...+knan=0
Suppose that the vector component is shortened, which is b 1, b2, ..., and bn respectively.
Obviously there is.
k 1b 1+k2b2+...+knbn=0
Because k 1, ..., kn are not 0, according to the definition of linear correlation, we know that
The original vector group b 1, b2, ..., bn is linearly correlated.