Show with a similar argument that lo is, in fact, the smallest index such that Co E Klo. HELPFUL INFORMATION To analyze the conjugate gradient method, we start by introducing the so-called Krylov subspace, Km := Km(A, ro) := span {ro, Aro, Arro, . .., Am-1ro}. It is clear that the dimension of Km is at most m. Sometimes, I'm has dimension less than m, for example, if ro was an eigenvector, then dim(Km)=1, for m = 1, 2, ..., since Aro = Aro. A proof of the next two statements is left as a homework exercise: Let lo be the minimal index for which Aloro E Klo but Alo-Bro & Klo-1. Then there are coefficients, co, . .., Clo-1 satisfying Coro + Ci Aro + . . . Clo-1 Alo-dro = Aloro. (8 If co = 0, multiplying (8) by A- would imply that Alo-'ro E Klo-1. As we have assumed that this is not the case, we must have co # 0 and so eo = Aro = Col(A-ro - Ciro - C2Aro . . . - Clo-1A-2To) E Klo. Please answer in detail. Thank you so much