Let T be a linear transformation of a vector space V into itself. Suppose x ε V is such that Tm(x)=0, Tm-1(x) not equal 0 for some positive integer m. show that x, T(x), …, Tm-1(x) are linearly independent.
In regards to Tm and Tm-1 m and m-1 are upperscript on the T. I don't know how to make it do that on this site.
The Attempt at a Solution
To show linearly independent, show that the only linear combination of transformations that = 0 is the one in which the coefficients are zero. For every xεV and cεR, real numbers, Suppose that c1, c2, ...,cm are scalars s.t. (c1)x + (c2)T(x) + ...(cm)Tm-1(x)=0 then
Σci Tm-1(x) =0 i=1 to m which equals Σci ΣTm-1(x) =0 Therefore all c's =0 hence linearly independent. Tm(x) could not be included since it equals 0 which means c could be nonzero.
This doesn't seem right (too easy)???