Suppose V is a vector space of finite dimension k and T : V - V is a linear transformation. Suppose T has one single eigenvalue 1 E R, and call n its geometric multiplicity. Which value(s) of n (if any) will result in [T]& being similar to a diagonal matrix, for all bases a of V? Your "values" of n should be in terms of k. If no such value of n exists, explain why not. Clarification 1: In general there are two meanings of multiplicity that apply to eigenvalues. "Multiplicity of an eigenvalue" could mean "number of times (x - 1) appears in the characteristic polynomial x(x) ("Algebraic Multiplicity") or the dimension of the eigenspace Ex ("Geometric Multiplicity"). In this question, we are considering the \\textbf{geometric multiplicity} n of 2. Clarification 2: Saying "There exists basis a of V for which [T'] is diagonal" is equivalent to saying that " [T]& is similar to a diagonal matrix for all bases a of V". Hint: Your solution can consider k > n, k = n, and k