... the second slot aswell as the first slot. Proof:u,v + w=v +w,u=v,u+w,u=v,u+w,u=u,v+u,w;here u,v, w ∈ V.In an inner-product space, we have conjugate homogeneity in the second ... get an operator on W. Specifically,define S ∈L(W) bySw = PW,U(Tw)for w ∈ W. By our induction hypothesis, S has an eigenvalue λ.Wewill show that this λ is also an eigenvalue for T.Let w ∈ W ... an operator whose matrix with respect toThese two exercisesshow that 5.16 failswithout the hypothesisthat an upper-triangular matrix isunder consideration.some basis contains only 0’s on...