I Have an N by N Matrix a Whose Columns Span R N When a Invertible

I have an n by n matrix A whose columns span R^n when A invertible.

Can you please explain to me why the columns of the nxn matrix A span R^n when A is invertible? I feel that if matrix A has columns that span R^n, then the inverse of A should likewise share that same characteristic, the spanning. But I'm not sure if that is a sufficient relationship. Can you give an example of matrix A spanning when it is invertible, if possible? How does Theorem 4 relate to this?

Theorem 4 states:

Let A be an mxn matrix. Then the following statements are logically equivalent. That is, for a particular A, either they are true statements or they are all false. a. For each b in R^m, the equation Ax=b has a solution. b. Each b in R^m is a linear combination of the columns of A. c. The columns of A span R^m. d. A has a pivot position in every row.

Solution: Recall that a set S spans the Vector space V, iff

1) S is basis for V 2) S is linearly independent

Furthermore, if each column of an m x n matrix A has a pivot position, then columns of A span Rm

n Let A be an m x n matrix with columns a1,a2,….,an with x  R , such their product represented as Ax is the linear combination of columns of A with the corresponding vector x  Rn :

 x1  x   2 

x3  Ax = [a1,a2,….an]    a1 x1  a2 x2  .....  an xn  :   :    xn 

n Ax =  ai xi i1

So, if every column in the m x n matrix has this unique representation, then column of A span Rm, Hence the set of all such linear combinations => that it is an linearly independent set and forms a basis for Rm. Consider the matrix:

 a11 ,a12 ,a13 ,...... ,a1m  x1  b1  a ,a ,a ,...... ,a x  b   21 22 23 2m  2   2 

a31 ,a32 ,a33 ,...... ,a3m x3  b3         :  :   :   :  :   :       an1 ,an2 ,an3 ,...... ,anm xn  bn 

Which is of the form AX = b, where A is the m x n matrix, X and b are the column matrices.

Now, we have to show that column (A) span Rn.

Without loss of generality, we assume that column(A) are linearly independent. It is enough to show there is a linear combination of col(A) and x and this set is span Rn.

n Taking the first column (A1, say) and x  R such that:

 x1  x   2  n x3  A1X  [a11 ,a 21 ,..a n1 ]    a11 x1  a21 x2  .....  an1 xn   aij xi , j  1  :  i1  :    xn 

This gives a linear combination.

n Taking the second column (A2, say) and x  R such that:

 x1  x   2  n x3  A 2X  [a12 ,a 22 ,..a n2 ]    a12 x1  a22 x2  .....  an2 xn   aij xi , j  2  :  i1  :    xn  This gives another linear combination.

n Likewise for the last column (Am) and x  R such that: We have,

x1  x   2  n x3  A m X  [a1m ,a 2m ,..a nm ]    a1m x1  a2m x2  .....  anm xn   aij xi , j  m  :  i1  :    xn 

Thus we get a linear combination of values:

And we represent the set of linear combinations in a set:

n mn n S = {  aij xi , j  1,2,3,.....,m : aij  R , xi  R } i1

And clearly this set is linearly independent set (as we seen that every column vector b can be represented as linear combination of column(A) and x }

n Ie.,  aij xi  bi ,i  1,2,3,.....,n i1

b1  b   2 

b3  Where b =    :   :    bn 

And hence S span Rn.

NOTE: The above theory is applicable to theorem 4, which is nothing but theorem for any m x n matrix, with properties:

1) Ax = b, x has solution if A is invertible 2) Column(A) are linearly independent in Rn 3) Column(A) forms the basis set Rn 4) Column(A) span Rn