It is stated that TU = I, the identity matrix. This implies that T is invertible (non-singular) and the inverse of T is U. If T is non-singular, then the column vectors form a basis for the column-space of T (that is, all of the column vectors are linearly independent.) This can be shown by the fact that Tx = 0^n has only the trivial solution: x = 0^n from which follows from T being a non-singular matrix. The rank of T is defined to be the dimension of the column-space of T (or row-space, as the dimension of each are always identical.) Lastly, the dimension is defined to be the number of vectors forming a basis - in this case, n. Thus, the rank of T is n, or: rank(T) = n