18.06SC Unit 2 Exam Solutions

18.06SC Unit 2 Exam Solutions 1 (24 pts.) Suppose q1 , q2 , q3 are orthonormal vectors in R3 . Find all possible values for these 3 by 3 determinants and explain your thinking in 1 sentence each. � � (a) det q1 q2 q3 = � � (b) det q1 + q2 q2 + q3 q3 + q1 = � � � � (c) det q1 q2 q3 times det q2 q3 q1 = Solution. (a) The determinant of any square matrix with orthonormal columns (“orthogonal matrix”) is ±1. (b) Here are two ways you could do this: (1) The determinant is linear in each column: � � � � � � det q1 + q2 q2 + q3 q3 + q1 = det q1 q2 + q3 q3 + q1 + det q2 q2 + q3 q3 + q1 � � � � = det q1 q2 + q3 q3 + det q2 q3 q3 + q1 � � � � = det q1 q2 q3 + det q2 q3 q1 Both of these determinants are equal (see (c)), so the total determinant is ±2. (2) You could also use row reduction. Here’s what happens: � � � � det q1 + q2 q2 + q3 q3 + q1 = det q1 + q2 −q1 + q3 q3 + q1 � � = det q1 + q2 −q1 + q3 2q3 � � = 2 det q1 + q2 −q1 + q3 q3 � � = 2 det q1 + q2 −q1 q3 � � = 2 det q2 −q1 q3 � � = 2 det q1 q2 q3 � Again, whatever det q1 q2 q3 � is, this determinant will be twice that, or ±2. (c) The second matrix is an even permutation of the columns of the first matrix (swap q1 /q2 then swap q2 /q3 ), so it has the same determinant as the first matrix. Whether the first matrix has determinant +1 or −1, the product will be +1. 2 (24 pts.) Suppose we take measurements at the 21 equally spaced times t = −10, −9, . . . , 9, 10. All measurements are bi = 0 except that b11 = 1 at the middle time t = 0. � and D � to fit those 21 points (a) Using least squares, what are the best C by a straight line C + Dt ? (b) You are projecting the vector b onto what subspace ? (Give a basis.) Find a nonzero vector perpendicular to that subspace. Solution. (a) If the line went exactly through the 21 points, then the 21 equations     1 −10 0          1 −9  0      ..   ..   .. . . .  C   =        1 1 0  D      ..   .. ..  . . .      1 10 0 would be exactly solvable. Since we can’t solve this equation Ax = b exactly, we look for a least-squares solution AT Ax̂ = AT b.      � 21 0 C 1    =   � 0 770 D 0 �= So the line of best fit is the horizontal line C 1 , 21 � = 0. D (b) We are projecting b onto the column space of A above (basis: � �T � 1 . . . 1 , −10 . . . There are lots of vectors perpendicular to this subspace; one is the error vector e = � �T 1 b − PA b = 21 (ten −1’s) 20 (ten −1’s) . 10 �T ). 3 (9 + 12 + 9 pts.) The Gram-Schmidt method produces orthonormal vectors q1 , q2 , q3 from independent vectors a1 , a2 , a3 in R5 . Put those vectors into the columns of 5 by 3 matrices Q and A. (a) Give formulas using Q and A for the projection matrices PQ and PA onto the column spaces of Q and A. (b) Is PQ = PA and why ? What is PQ times Q ? What is det PQ ? (c) Suppose a4 is a new vector and a1 , a2 , a3 , a4 are independent. Which of these (if any) is the new Gram-Schmidt vector q4 ? (PA and PQ from above) PQ a4 1. kPQ a4 k 2. a4 − aT 4 a1 a1 aT 1 a1 − aT 4 a2 a2 aT 2 a2 − aT 4 a3 a3 aT 3 a3 k norm of that vector k 3. a4 − PA a4 ka4 − PA a4 k Solution. (a) PA = A(AT A)−1 AT and PQ = Q(QT Q)−1 QT = QQT . (b) PA = PQ because both projections project onto the same subspace. (Some people did this the hard way, by substituting A = QR into the projection formula and simplifying. That also works.) The determinant is zero, because PQ is singular (like all non-identity projections): all vectors orthogonal to the column space of Q are projected to 0. (c) Answer: choice 3. (Choice 2 is tempting, and would be correct if the ai were replaced by the qi . But the ai are not orthogonal!) 4 (22 pts.) Suppose a 4 by 4 matrix has the same entry × throughout its first row and column. The other 9 numbers could be anything like 1, 5, 7, 2, 3, 99, π, e, 4.   × × × ×      × any numbers   A=    × any numbers    × any numbers (a) The determinant of A is a polynomial in ×. What is the largest possible degree of that polynomial ? Explain your answer. (b) If those 9 numbers give the identity matrix I, what is det A ? Which values of × give det A = 0 ?  × × × ×   × A=  ×  × 1 0 0 1 0 0    0   0  1 Solution. (a) Every term in the big formula for det(A) takes one entry from each row and column, so we can choose at most two ×’s and the determinant has degree 2. (b) You can find this by cofactor expansion; here’s another way:    1 × × × 1 − 3× × ×        0 1 1 0 0  1 0    = × det det(A) = × det     0 1 0 1 0  0 1    1 0 0 1 0 0 0 = ×(1 − 3×). This is zero when × = 0 or × = 13 .     1 0 0    0   = ×(1−3×) det  0 1 0      0  0 0 1 1 ×