Find the dimension of the null space of T.

Find the dimension of the null space of T.

Let E 1, 2, dots , n, where n is an odd positive integer. Let V be the vector space of all functions from E to mathbbR3, where the vector space operations are given by f gk fk gk, for f, g V, k E, fk fk, for f V, mathbbR, k E. Let T colon V V be the map given by Tfk frac12fk fn 1 k, k in E. aFind the dimension of V. My answer is 3. b Find the dimension of the null space of T. I know that rank nullity dimension, but its is of no use here.

The dimension is 3n. Consider the canonical basis e_1,e_2,e_3 of mathbbR3 and, for 1le ile n and 1le jle 3, the map f_i,jcolon EtomathbbR3 defined by f_i,jk begincases e_j ik, 4px 0 ine k endcases Can you prove that f_i,j:1le ile n, 1le jle 3 is a basis for V Now note that assuming n1 Tf_1,jkfrac12f_1,jkf_1,jn1-k begincases frac12e_j k14px 0 1kn 4px frac12e_j kn endcases so Tf_1,jfrac12f_1,jf_n,j More generally, Tf_i,jbegincases dfrac12f_i,jf_n1-i,k ine n12 6px f_i,j in12 endcases Can you finish and compute the rank of T

The space V is isomorphic to the space of real matrices with 3 rows and n columns, the isomorphism Phi being the one that associates f to the matrix with first column given by the components of f1, and so on. So the dimension is 3n. The null space of T is the vector space of functions that get mapped by T to the zero map in V, i.e. the map z in V such that zk 0in Bbb R3. So finmathrmkerT iff fk fn1-k 0, qquad forall kin E. This means that f1-fn, etc. In terms of the matrix isomorphism, since n is odd, this entails that the last fracn-12 columns of the matrix Phif corresponding to f are each the negative of the corresponding column in the first half of the matrix i.e. the n-th column is minus the first, the n-1-th is minus the second, until we arrive at the central column which has no restriction. In other words, f is only determined by its first fracn12 elements. Can you guess what the nullity of T must be