Question 15310
 Let x1<...< xm . For I = 1,…,m, define linear operators
			Ti : Pn   --> R1
By Tip = p(xi). If m<=n, show that (T1,...Tm) is linearly independent. What happens if m>n? Explain
 
 Since you did not explan everything clear. No any of your work was shown.
 And you posted in wrong place. I don't quite feel like to answer it.

 Still .. for my weak mind

 Your {{{P[n]}}} supposed to be the n -dim space of polynomials over R with degree <= n-1. Here, I use y as the variable in {{{P[n]}}}.

  Sol: Let {{{ f[i](y) = y^(i-1) }}}, i =1,2,..,n. We know that these n  
  polynomials form a basis of {{{P[n]}}}.
 
  If {{{x[i] | 1<= i<=m }}} are m distinct real numbers.

  To show (T1,...Tm) is linearly independent. We have to show that
 
 {{{ SIGMA }}} {{{c[i]T[i](p)}}} = 0 for any p in {{{P[n]}}} ...(*)
 implies {{{c[i]}}} = 0 for all real number  {{{c[i]}}} i=1,..,m
 When m <= n,
 Chooseing p to be one of the n polynomials of the basis 
 {{{f[j] |j=1,,..n}}}, say {{{f[j] }}} for some 1<= j<= n
 We have {{{ SIGMA }}} {{{c[i]T[i](f[j])}}} = 0
 Since {{{f[j] (y) = y^(j-1) }}}} and {{{ T[i](y^(j-1)) = (x[i]^(j-1)) }}}
 for all i=1,2,..,m and fixed j.

 We get a system of n equations of m variables {{{c[i]}}} from (*)
 [ Note: i summation over 1 to m ]

 {{{ SIGMA }}} {{{c[i] }}} = 0  (j=1) 
 {{{ SIGMA }}} {{{c[i] x[i]}}} = 0 (j=2)
 {{{ SIGMA }}} {{{c[i] x[i]^2}}} = 0 (j=3)
 ...
 {{{ SIGMA }}} {{{c[i] x[i]^(n-1)}}} = 0 (j=n)

 This system is equivalent to the matrix equation

 AX = 0, where A = ({{{x[j]^(i-1)}}}) i=1,2,..,n, j=1,2,..,m.
 X= ({{{c[1]}}},{{{c[2]}}},...,{{{c[m]}}}) : mx1 column vector and 0 is nx1 matrix 
  
 Now consider the rank of the nxm matrix A, we know that
 rank A <= min(m,n). By now A = ({{{x[j]^(i-1)}}}), since all {{{x[i]}}}
 are distinct. Let k = min(m,n), the square kxk submatrix of A has
 determinant = II (x[i] -x[j]) (i > j) , which is non-zero , so we have
 rank A = min(m,n).
 (e.g.
      ({{{ 1 }}} {{{ 1 }}} {{{ 1 }}})
      ({{{x[1]}}} {{{x[2]}}} {{{x[3] }}})
     ({{{x[1]^2}}} {{{x[2]^2}}} {{{x[3]^2 }}} )
 with det = (x[1] - x[2]) (x[2] - x[3])(x[3] - x[1]) <> 0 )
      
 If m <= n, then rank(A) = m, 
 The linear transformation
 A: {{{R^m}}}--> {{{R^n}}}, m = nullity (A) + rank(A) = nullity (A) + m.
 So, we have nullity(A) = 0, this means null(A) = {0}.
 That is, the equation A X = 0 has unique solution of zero vector
  Hence, all {{{c[i]}}} = 0 ,i=1,2,..m .
 This proves that (T1,...Tm) is linearly independent.
 
 If m > n, then rank(A) = n. By m = nullity (A) + rank(A) = nullity (A) + n.
 Hence,  nullity (A) = m-n > 0. This means AX = 0 has nonzero solution.
 Therfore, not all {{{c[i]}}} = 0 ,i=1,2,..m .
 This shows that (T1,...Tm) is linearly dependent if m > n.

 This proof is not quite easy for beginners. Try to read carefully to
 understand it.

 Kenny