# What is the dimension of the subspace V ⊥ (the orthogonal complement of V )?

Math 254, Spring 2018 Midterm #3, In-Class April 19th,

Tools: Brain/Pen/Pencil/Eraser/Paper. Rules: This is an in-class midterm; see below:

RedID v 2018.04.19.1

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I, , pledge that this exam is completely my own work, and that I did not take, copy, borrow or steal any portions from any other person; furthermore, I did not knowingly let anyone else take, copy, or borrow any portions of my exam. Further, I pledge to abide by the rules set out below. I understand that if I violate this honesty pledge, (i) I will get ZERO POINTS on this exam; (ii) I will get reported to The SDSU Center for Student Rights and Responsibilities; and (iii) I am subject to disciplinary action pursuant to the appropriate sections of the San Diego State University Policies.

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Rules:

• This midterm is closed-book, closed-notes, no phones, no calculators, no phones, no slide-rules, no phones, nor any super-computers allowed. Did I mention NO PHONES?!?

• No communications / internet enabled devices allowed — NO PHONES!

• Write solutions/answers on the attached sheets, and HAND IN the entire packet.

• Note: there should be lots of space to write your solutions, do not feel the need to fill it all…

• Present your solutions using standard notation in an easy-to-read format. It is your job to convince the grader you did the problem correctly, not the grader’s job to decipher cryptic messages scribbled in the margin!

• Your answers MUST logically follow from your calculations in order to be considered! (“Miracle solutions” ⇒ zero points.)

• Perform your computations on the attached pages; draw BOXES around the answers. — THE GRADER WILL NOT GO TREASURE-HUNTING!

• The exam will be graded and returned on or about two weeks after the test date. No grading corrections will be considered once you remove the exam from the lecture hall / professor’s office.

Problem Pts Possible Pts Scored

1 150 2 100 3 (extra credit) 20 4 (extra credit) 30

Total 250

• You MUST stay for at least 20minutes. (Draw an epic unicorn–narwhal battle on the back if you have too much time on your hands!)

• Once ≥5 students have turned in their exams, late-comers MAY NOT be able to take the test.

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1. [§ 5.1–5.2] Let

~v1 =

1 0 1 1 1 1

, ~v2 =

0 1 −1 1 1 1

, ~v3 =

1 1 0 2 2 2

; ~x =

1 1 1 1 1 1

.

(a) i. (10 pts.) What is the dimension of the subspace V = span(~v1, ~v2, ~v3)?

ii. (15 pts.) Why?

(b) (25 pts.) Compute an orthonormal basis for the subspace V .

(c) (25 pts.) What is the QR-factorization of the matrix A = [

~v1 ~v2 ]

?

(d) (25 pts.) Project ~x onto the subspace V .

(e) i. (10 pts.) What is the dimension of the subspace V ⊥ (the orthogonal complement of V )?

ii. (15 pts.) Why?

(f) (25 pts.) Find a basis for V ⊥.

Perform your computations on the attached pages; COLLECT your bases/factorization in ONE place (for each subproblem) and draw BOXES around the answers. — THE GRADER WILL NOT GO TREASURE-HUNTING! Your Boxed Answers MUST logi- cally follow from your calculations in order to be considered!

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2. [§ 6.1] Consider the 7× 7 matrices

A =

1 0 0 2 0 0 0 0 2 0 0 0 0 0 0 0 3 0 0 0 0 4 0 0 4 0 0 0 0 0 0 0 5 0 0 0 0 0 0 0 6 0 0 0 0 0 0 0 7

, B =

0 0 0 0 0 0 7 0 0 0 0 0 6 0 0 0 0 0 5 0 0 4 0 0 4 0 0 0 0 0 3 0 0 0 0 0 2 0 0 0 0 0 1 0 0 2 0 0 0

For each matrix identify all patterns with non-zero products; determine the number of inversions; the signs of each pattern; and combine to form the determinant: (If you have no idea what the combinatorial pattern “method” is about; use your favorite method to compute the determinant in parts a-iv, b-iv.)

(a) Matrix A —

i. (10 pts.) Patterns:

ii. (10 pts.) Number of inversions (for each pattern):

iii. (10 pts.) Sign for each pattern:

iv. (20 pts.) det(A) =

(b) Matrix B —

i. (10 pts.) Patterns:

ii. (10 pts.) Number of inversions (for each pattern):

iii. (10 pts.) Sign for each pattern:

iv. (20 pts.) det(B) =

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3. [§ 6.1–2] For A ∈ R3×3, Sarrus’ rule is a short-cut strategy for computing the determinant:

a11 a12 a13 a11 a12 a21 a22 a23 a21 a22 a31 a32 a33 a31 a32

Products along the right-going “diagonals” contribute with a positive sign, and products along the left-going “diagonals” with a negative sign; i.e.

det(A) = +a11 a22 a33 + a12 a23 a31 + a13 a21 a32 − a13 a22 a31 − a11 a23 a32 − a12 a21 a33.

Now, consider Voldemort’s Rule for A ∈ R4×4:

a11 a12 a13 a14 a11 a12 a13 a21 a22 a23 a24 a21 a22 a23 a31 a32 a33 a34 a31 a32 a33 a41 a42 a43 a44 a41 a42 a43

Voldemort Det(A) = +a11 a22 a33 a44 + a12 a23 a34 a41 + a13 a24 a31 a42 + a14 a21 a32 a43 −a14 a23 a32 a41 − a11 a24 a33 a42 − a12 a21 a34 a43 − a13 a22 a31 a44

(a) (15 pts) Explain why the Voldemort-Determinant is NOT the true determinant — (Do NOT write a 5-page essay!) —

Voldemort Det(A) 6= det(A).

(b) (5 pts) Give a matrix for which

Voldemort Det(A) = 0, det(A) = 1.

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4. Midterm #2 rudely reminded us that we can have at most n linearly indepenent vectors in Rn. Let us consider

~v1 =

1 0 0

, ~v2 =

0 1 0

, ~v3 =

0 0 1

, ~v4 =

1 1 1

,

in R3; clearly ~v4 = ~v1 + ~v2 + ~v3. Now, consider the 4 subsets

S1 = {~v1, ~v2, ~v3}, S2 = {~v1, ~v2, ~v4}, S3 = {~v1, ~v3, ~v4}, S4 = {~v2, ~v3, ~v4}.

Each subset now contains 3 linearly independent vectors. Next, consider

~w1 =

1 0 0

, ~w2 =

0 1 0

, ~w3 =

0 0 1

, ~w4 =