Prove that if a is not square, then either the row vectors of a or the column vectors of a form a linearly dependent set.

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

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 with b the number of columns.

The a row vectors conform a set with more vectors than the dimension of the b dimensional space 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 with a the number of rows.

The b column vectors conform a set with more vectors that the dimension of the a dimensional space 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.

4

step-by-step explanation: