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THE JOHNS HOPKINS UNIVERSITY ... 110.201 - LINEAR ALGEBRA. ... Consider the complete matrix B associated to the system.
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Faculty of Arts and Sciences FINAL EXAM - SPRING SESSION 2005 110.201 - LINEAR ALGEBRA.
Examiner: Professor C. Consani Duration: 3 HOURS (9am-12noon), May 12, 2005.
No calculators allowed. Total Marks = 100
Student Name:
TA Name & Session:
Total
1a. [2 marks] Compute the reduced row-echelon form of A. 1b. [2 marks] Determine the rank of A.
1c. [2 marks] Determine a basis of the column space of A. 1d. [2 marks] Determine a basis of the nullspace of A. 1e. [2 marks] For what value(s) of r ∈ R is the following system solvable
Ax =
r
Sol. [1a.]
rref(A) =
[1b.] It follows from the description of rref(A) that rk(A) = 2.
[1c.] A basis of the column space of A is given by the vectors
(^) and
(^) corre-
sponding to the pivot columns in A. [1d.] There are two non-trivial relations among the columns of A. Let denote by v 1 ,... , v 4 the first,...,fourth column of A. We have v 1 − 2 v 2 = 0, 2 v 1 − 2 v 3 − v 4 = 0.
Hence, a basis of the nullspace of A is given by the vectors
and
[1e.] Consider the complete matrix B associated to the system
4 2 3 2 | r
0 0 3 − 6 | r − 4
0 0 0 0 | r − 7
The system is solvable if and only if r = 7.
3a. [4 marks] Give a 3 × 3-matrix A with the following properties: i. AT^ = A−^1. ii. det(A) = 1. (A is not allowed to be a diagonal matrix) 3b. [4 marks] Give a 3 × 3-matrix with the following properties:
i. AT^ = A. ii. A^2 = A. iii. rk(A) = 1. (A is not allowed to be a diagonal matrix)
3c. [4 marks] Suppose A is a 5 × 3-matrix with orthonormal columns. Evaluate the following determinants: i det(AT^ A) ii det(AAT^ ) iii det(A(AT^ A)−^1 AT^ ). 3d. [3 marks] Which value(s) of α ∈ R give det(A) = 0, if
α 2 3 −α α 0 3 2 5
Sol.
[3a.] We consider, for example an orthogonal matrix A, with det(A) = 1. Permutation
matrices such as
(^) or
(^) have this property.
[3b.] We may consider a projection matrix of rank 1: for example (cfr. question 2c.)
A = v(vT^ v)−^1 vT^ , where v =
. Namely the matrix A = 141
[3c.] det(AT^ A) = det(I) = 1 det(AAT^ ) = 0 det(A(AT^ A)−^1 AT^ ) = 0 AAT^ must have dependent columns and determinant zero because A(AT^ x) = 0 for any non-zero vector x in the nullspace of AT^. The 3 × 5-matrix AT^ has 3 linearly independent (orthonormal!) rows and a non-trivial nullspace of dimension 5 − 3 = 2. Notice that det(A(AT^ A)−^1 AT^ ) = det(AAT^ ) = 0 as AT^ A = I. [3d.] det(A) = 5α(α − 1) = 0. Therefore, α = 0 or α = 1.
i. A
ii. A
iii. A is symmetric. The following questions refer to any matrix A with the above properties
4a. [3 marks] Is Ker(A) = { 0 }? Explain your answer. 4b. [3 marks] Is A invertible? Why?
4c. [3 marks] Does A have linearly independent eigenvectors? Explain. 4d. [6 marks] Give a specific example of a matrix A satisfying the above three properties and whose eigenvalues add up to zero.
Sol. [4a.] A has linearly dependent columns: for example, it follows from conditions i.
and ii. that A(
) = 0. This implies that Ker(A) 6 = { 0 }.
[4b.] A is not invertible as Ker(A) 6 = { 0 }. [4c.] Yes, the eigenvectors of a symmetric matrix are linearly independent (and can be chosen to be orthonormal).
[4d.] For example A =
− 3 − 6 a 33
, with a 33 = −5. In fact: A
(^) gives 2 times
the first column of A and A
(^) gives −1 times the second column of A. Then,
from the symmetry condition iii. we get∑ a 13 = a 31 and a 23 = a 32. For a 33 we impose λk = trace(A) = 1 + 4 + a 33 = 0. From this relation we deduce a 33 = −5.
6a. [3 marks] If possible, find an invertible matrix M such that
If it is not possible, state why M cannot exist. 6b. [3 marks] For what real values of c (if any) is
− 1 c 2 c − 4 − 3 2 − 3 4
a symmetric positive definite matrix?
6c. [4 marks] Let A =
. Is the quadratic form q(x, y) associated to A posi- tive definite? Find its principal axes.
Sol.
[6a.] Not possible. The condition means that B =
(^) is similar to A =
. Similar matrices have equal traces and rank. But trace(A) = 3 6 =
trace(B) = 5 and rk(A) = 1 6 = rk(B) = 2. [6b.] Not possible. For symmetric positive-definite matrices all upper-left determi- nants are greater than zero. Note that the 1 by 1 upper-left determinant is −1. [6c.] det(A) < 0, hence A (or equivalently its associated quadratic form q(x, y) = 3 x^2 + 8xy + 3y^2 ) is not positive definite. The eigenvalues of A are: λ 1 = −1 and
λ 2 = 7. The principal axes are the eigenspaces of A, namely E 1 = span{
} and
E 2 = span{
7a. [6 marks] Let A 1 =
. Is A 1 diagonalizable? Why? Is A 1 in-
vertible? Why? Determine the spectral decomposition of A 1 into projection matrices.
7b. [3 marks] Let A 2 =
. Is A 2 invertible? Why? Is A 2 diagonalizable? Why? Determine (if exists) a matrix S and a diagonal matrix D such that S−^1 A 2 S = D. 7c. [6 marks] Describe the linear transformation TA 2 : R^2 → R^2 associated to A 2. Does such A 2 have a decomposition into projection matrices? If yes, give it.
Sol. [7a.]∏ A 1 is symmetric, hence A 1 is diagonalizable. A 1 is invertible as det(A 1 ) = λi = 2 · 1 · (−1) = − 2 6 = 0. The spectral decomposition of A 1 is given by
i=
λixixTi
where xi are eigenvectors associated to the eigenvalues λi. We can choose x 1 =
as eigenvector associated to λ 1 = 2, x 2 =
(^) as eigenvector associated to λ 2 = 1
and x 3 =
(^) as eigenvector associated to λ 3 = −1. It follows that the spectral
decomposition of A 1 is
[7b.] det(A 2 ) = 0 hence A 2 is not invertible. The eigenvalues of A 2 are μ 1 = 0 and μ 2 = −4. They are distinct, hence A 2 is diagonalizable. The columns of the matrix S are made by 2 eigenvectors of A i.e. S =
, whereas D =
is the eigenvalues matrix. [7c.] The linear transformation TA 2 : R^2 → R^2 is determined by a projection P 2 onto
the line spanned by
It follows that A 2 = − 4
is the required decomposition, where P 2 is a projection matrix.
