matricies-----Linear algebra

[Sarahisme]

Can i just check if i have done this question right?

Question:............

"Suppose that a linear transformation T is defined by a 4×5 matrix. Can T to be one to one? Can T be onto? Give reasons for your answers."

 

 

 

This is my answer:.............

 

Let the 4x5 matrix that defines transformation T be:

A = [# * * * *]

[* # * * *]

[* * # * *]

[* * * # *]

 

#'s indeicate some possible pivot positions

 

for the transformation T to be one to one, the columns oof matrix A have to be lineraly independent (by a particular therom). That is T is one to one iff Ax=0 has only the trvial solution. However, because there will always be at least one free variable in the reduced forms of A, therefore the columns of A are not linearly independent. Thus the transformation T can not be one-to-one.

 

 

The transformation T is onto (map R^n -> R^m) iff the columns of A span R^m (By a Theorem). Also by another Theorem, the columns of A span R^m iff A has a pivot position in every row. So because A is a 4x5 matrix, it is defintely possible for A to have a pivot position in every row. Thus T can be onto so long as there is a pivot position in each row when the matrix has been reduced.

 

phew!

how'd i go?

 

Cheers

Sarah

[Sarahisme]

??????????? hello? lol i have no friends

[matt grime]

5 column vectors in R^4? Can they be independent?

[Sarahisme]

no they are not

[matt grime]

then there was no need to do

 

 

"for the transformation T to be one to one, the columns oof matrix A have to be lineraly independent (by a particular therom). That is T is one to one iff Ax=0 has only the trvial solution. However, because there will always be at least one free variable in the reduced forms of A, therefore the columns of A are not linearly independent. Thus the transformation T can not be one-to-one."

[Sarahisme]

well couldn't hurt could it