Applied Linear Algebra Exercises Solution 3 is the solution of 6.1 problem. Please solve the 6.2 Questions (a) to (h) to use Solution 3. Thank you 6. Revie

Applied Linear Algebra Exercises Solution 3 is the solution of 6.1 problem. Please solve the 6.2 Questions (a) to (h) to use Solution 3. Thank you 6. Review Exercise: Please write a review to one of the four solutions to the subsequent
problem (it is problem 5 from homework 6) using the list of questions given in 6.2. (Note that
you cannot write a review for your own solution.) Do not forget to write the number of the
solution you are reviewing.
We recommend to read all four solutions before writing your review.
6.1 Problem
Let {41, 42, …, Un} be an orthonormal basis for a subspace s of an inner product space V,
and let p be the projection of v € V onto S.
(a) Show that p= = {v, 44) 2.
(b) Show that P = ||0||
6.2 Questions
(a) Are the necessary definitions identified correctly? (If not, suggest the definitions that
need to be recalled.)
(b) Are the statements that need to be proven identified correctly? (If not, suggest the
correct versions.)
(c) Are all the given statements identified correctly? (If not, suggest the statements that
are missing.)
(d) Is the idea of the proof clear? (If not, suggest a better explanation.)
(e) Is the problem split into simple enough steps? (If not, suggest a better partition.)
(f) Is each step explained in a clear manner? (If not, suggest a better explanation.)
(g) Does the proof use the statement it intends to prove? (If so, show the place where it is
used.)
(h) Are all the cases covered? (If not, suggest missed cases.)
Solution 3
(a) Each (v, u:) gives the component of v that is on vector u. Therefore if you add the
components (v, u1u1 + (v, u2)u2 + … + (v, Ure) Un you obtain a projection p of v
on S. >
pi(0, 2) mi.
(b) We know ||P|| = “=1(0, ). So we need to get v-p + (v-p,p) = {v,p) – ||P||2 →
Si=1(v, (0, 1)) – D=(v, u:)2 = 0 so plv-p. Using Pythagorean theorem,
||0 + w||2 = || 0 || + || 12 || 2 = ||0 – p+p|l2 = ||0 – p|l2 + ||P|12||0||2|P||2
P|