Problem 1.
Apply the Gram–Schmidt algorithm to these vectors to obtain an orthonormal basis f1, f2, f3.
Problem 2 and 3. Consider the two dimensional vector space V = span(v1, v2), where v1, v2 are given in Problem 1 above. Consider the projection of vectors in R4 onto this space V . Use two different methods to find the matrix P of this projection.
Problem 2.
Method I) Follow the procedure described in Question 2 (3) of the in-class questions on Oct 20 to find the projection matrix P .
Link: https://rutgers.instructure.com/courses/74179/files/12323143/download?wrap=1
Problem 3.
Method II) Utilize an orthonormal basis to find the projection matrix P .
In Problem 1 above, you’ve found an orhthonormal basis f1, f2 for the space V = span(v1, v2) = span(f1, f2). Given a generic y R4, let yJ = Py denote the projected vector. Express yJ
as a linear combination as
Py = yJ = b1f1 + b2f2 (*)
(1) Use the fact that f1, f2 are orthonormal to find b1 b2 in terms of f1, f2, and y. Instruction: operator with symbols f1, f2, y. Do not insert the values of f1, f2 yet.
(2) Insert your answer in (1) back into (*). Identity the matrix P from your result. Express your answer for P in terms of f1, f2.
Instruction: operator with symbols f1, f2. Do not insert the values of f1, f2 yet.
(3) Now insert in the numerical values of f1, f2 (from Problem 1 above) into your answer for P in (2), and check that the result agrees with what you found in Problem 2.
Problem 4. Given a vector space V in Rn. Consider the projection of vectors in Rn onto V , and let P denote the projection matrix. Follow the steps below to prove that P 2 = P and P T = P .
(1) Let f1, . . . , fk be an orthonormal basis for V . (Such an orthonormal basis always exists. We won’t worry about proving its existence here.) Express the projection matrix P in term of f1, . . . , fk.
Hint: generalize the procedure in Problem 3 above.
Instruction: explain all your derivation. Simply showing the answer does not account for a full solution.
(2) Use your answer in (1) to verify P 2 = P and P T = P .
Remark: In Problem 4 of Homework 6, you’ve shown that if P 2 = P and P T = P then P is a projection matrix. Problem 3 here gives the converse statement. Combining these two statements gives
“A square matrix P is a projection matrix if and only if P 2 = P and P T = P .”
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