Respuesta :
Answer:
See the proof below.
Step-by-step explanation:
We are assuming that A is not square. So let's assume that A is an axb matrix that is not square because [tex] a \neq b[/tex]
Since is a not square matrix then A needs to has more rows than columns case 1 (a>b) or more columns than rows case 2 (n>m)
Case 1 (a>b)
We need to remember that the dimension for the row space is [tex] R^b[/tex] with b the number of columns.
The a row vectors conform a set with more vectors than the dimension of the b dimensional space [tex] R^b[/tex] where the vectors exists. So then the vectors are linearly dependent.
Case 2(b>a)
We need to remember that the dimension for the colum space is [tex] R^a[/tex] with a the number of rows.
The b column vectors conform a set with more vectors that the dimension of the a dimensional space [tex] R^a[/tex] where the vectors exists. So then the vectors are linearly dependent.
So for both cases when we don't have a square matrix then either the row vectors of A or the column vectors of A form a linearly dependent combination.
