<p>I have an n by n matrix A whose columns span R^n when A invertible. </p><p>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? </p><p>Theorem 4 states: </p><p>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.</p><p>Solution: Recall that a set S spans the Vector space V, iff</p><p>1) S is basis for V 2) S is linearly independent</p><p>Furthermore, if each column of an m x n matrix A has a pivot position, then columns of A span Rm</p><p> 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 : </p><p> x1  x   2 </p><p>x3  Ax = [a1,a2,….an]    a1 x1  a2 x2  .....  an xn  :   :    xn </p><p> n Ax =  ai xi i1</p><p>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:</p><p> a11 ,a12 ,a13 ,...... ,a1m  x1  b1  a ,a ,a ,...... ,a x  b   21 22 23 2m  2   2 </p><p>a31 ,a32 ,a33 ,...... ,a3m x3  b3         :  :   :   :  :   :       an1 ,an2 ,an3 ,...... ,anm xn  bn </p><p>Which is of the form AX = b, where A is the m x n matrix, X and b are the column matrices.</p><p>Now, we have to show that column (A) span Rn.</p><p>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.</p><p> n Taking the first column (A1, say) and x  R such that:</p><p> x1  x   2  n x3  A1X  [a11 ,a 21 ,..a n1 ]    a11 x1  a21 x2  .....  an1 xn   aij xi , j  1  :  i1  :    xn </p><p>This gives a linear combination.</p><p> n Taking the second column (A2, say) and x  R such that:</p><p> 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.</p><p> n Likewise for the last column (Am) and x  R such that: We have,</p><p>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 </p><p>Thus we get a linear combination of values:</p><p>And we represent the set of linear combinations in a set:</p><p> n mn n S = {  aij xi , j  1,2,3,.....,m : aij  R , xi  R } i1</p><p>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 }</p><p> n Ie.,  aij xi  bi ,i  1,2,3,.....,n i1</p><p>b1  b   2 </p><p>b3  Where b =    :   :    bn </p><p>And hence S span Rn.</p><p>NOTE: The above theory is applicable to theorem 4, which is nothing but theorem for any m x n matrix, with properties:</p><p>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</p>