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Math 111 Final Exam
May 21, 2004
Your Name
Student Number
Section Number
1.
For more space, write on the opposite side.
2.
If your answer is too complicated, you must have made a mistake.
3.
Show all your work. Cross o. (instead of erase) the undesired parts.
4.
Provide all the details. Your reason counts most of the points.
Number Score
1
2
3
4
5
6
Total
1
(1) (20 points) Let
A =
. ..
001
010
. . .
2) (5 points) Let Q(.x)= .xT A.x be the quadratic form associated with the real symmetric matrix A.
..
100
1) (15 points) Diagonalize A;
212223
Find the maximum value and minimum value of Q(.x) subject to condition ||.x|| =
= 1.
+ x
+ x
x
Solution. 1) Skipped. 2) There are the maximum and minimum eigenvalues of A, i.e., 1 and .1 respectively.
(2) (20 points) Let M22 be the collection of all 22 real matrices. With matrix addition and scalar multiplication by real numbers, M22 becomes a real vector space of dimension 4.
1) (5 points) The trace de.nes a map T : M22 R1 . By de.nition, T (A)=[a11 + a22] for any 2 2 matrix A =[aij]. Show that T a linear map. (here we view R1 as the collection of all 1 1 matrices.)
2) (15 points) Find a basis for the null space of T .
Solution. 1) Obvious because the dependence on aij is linear explicitly. 2) Many ways to .nd a basis way. A systematic way of doing it is to turn it into a problem in R4 because M22 can be naturally identi.ed with R4 . Clearly, T is onto, so the null space of T is 3-dimensional, so you can solve the problem even by inspection. One possible answer would be
.. 01 .. 00 .. 10 ..
,, .
00 10 0 .1
(3) (15 points, 5 points each)
Let (.v1,.v2, .v3) be an orthonormal basis of R3 . Let A = .v1.v1 T + .v2.v2 T . (Note that if .v is a column matrix, then .vT , being the transpose of .v, is a row matrix. So .v.vT , being the matrix product of .v with .vT , is a square matrix.)
1) Show that .v1, .v2 and .v3 are eigenvectors of A.
2) Show that A is diagonalizable.
3) What is the rank of A?
Solution. 1) Simple computations show that .v1, .v2 are eigenvectors of A with eigenvalue 1 and .v3
is an eigenvector of A with eigenvalue 0.
2) Since (.v1,.v2, .v3) is a basis made of eigenvectors, A is diagonalizable.
3) Then rank of A is 2 because (.v1,.v2) is a basis for the column space of A.
(4) (15 points)
Let P be a real symmetric nn-matrix. Suppose that P 2 = P . Show that the linear map sending .x to P.x is the orthogonal projection onto the column space of P .
Solution. Let I be the n n-identity matrix. For any .x in Rn, we have
.x = P.x +(I . P ).x.
Since P.x is in the column space of P , all we need to show is that (I . P ).x is in the orthogonal complement of the column space of P , but that is clear from this computation: for any .y,(I . P ).x P.y = P (I . P ).x .y (P is symmetric) = (P . P 2).x .y = .0 .y (P = P 2) =0.
(5) (10 poi