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The answers key for the linear algebra midterm examination 2 held at uc san diego in 2003. The exam covers various topics related to linear algebra, including definitions, subspaces, rank, and matrix representations of linear transformations. Students are required to demonstrate their understanding by providing answers and justifications.
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March 5, 2003
Instructor: Sam Buss, UC San Diego
Write your name or initials on every page before beginning the exam.
You have 50 minutes. There are seven problems. You may not use calculators, notes, textbooks, or other materials during this exam. You must show your work in order to get credit. Good luck!
Student ID:
Tuesday section time:
Total
(a) โA is a symmetric matrix.โ ANSWER: A = AT^.
(b) โA is an skew-symmetric matrix.โ ANSWER: A = โAT^.
(c) โv 1 ,... , vn is a basis for V .โ ANSWER: v 1 ,... , vn are linearly independent and span V.
(d) โU and V are orthogonal (U โฅ V ).โ ANSWER: For every u โ U and every v โ V , uT^ v = 0.
(e) โU is the orthogonal complement of V (U = V โฅ^ ).โ ANSWER: U = {u : for all v โ V , uT^ v = 0}
(a) What is the dimension of R(A)? ANSWER: r
(b) What is the dimension of N (A)? ANSWER: n โ r
(c) What is the dimension of R(AT^ )? ANSWER: r
(d) What is the dimension of N (AT^ )? ANSWER: m โ r
. Let x = (2, 2 , 2 , 0)T^. Express x in the form x = p + q where p โ U and q โ U โฅ^. What are the vectors p and q?
ANSWER: By inspection, the two vectors that span U are orthogonal; however, the first one is not a unit vector. To get two orthonomal vectors that span U , we convert the first vector to a unit vector:
u 1 = (0, 0 , 1 /
2)T^ and u 2 = (1, 0 , 0 , 0)T^.
Now, p = ใu 1 , xใu 1 + ใu 2 , xใu 2. We compute: ใu 1 , xใ =
2 and ใu 2 , xใ = 2. Thus, p =
2 u 1 + 2~u 2 = (2, 0 , 1 , 1)T^.
To finish up, q = x โ p = (0, 2 , 1 , โ1)T^.
1 2 ,^ โ^
โ 3 2
. Define f : R^2 โ R^2 by
f (x) = the vector projection of x onto u.
(a) What is the value of f (e 1 )?
ANSWER: (1/ 4 , โ
(b) Give the matrix that represents f.
ANSWER: We also find that f (e 2 ) = (โ
3 / 4 , 3 /4)T^. Therefore the matrix that represents f is ( 1 / 4 โ
x 0 1 3 y 4 1 2
Find the best least squares fit by a linear function. In other words, find the linear function, f (x) = c 0 + c 1 x, which best approximates these data values in the least squares sense.
ANSWER: Let A be the matrix
Then
AT^ A =
and AT
We need to solve the matrix equation ( 3 4 4 10
c 0 c 1
When we do this (work omitted), we get c 0 = 3 and c 1 = โ 1 / 2.
Thus, f (x) = 3 โ 12 x is the best least squares fit.
