18.06 (Spring 14) Problem Set 6
This problem set is due Thursday, April 3, 2014 by 4pm in E17-131. The problems are out
of the 4th edition of the textbook. This homework has 8 questions worth 80 points in total.
Please WRITE NEATLY. You may discuss with others (and your TA), but you must turn
in your own writing.
3 1
2
1. Suppose every vector in R is multiplied by the matrix A =
.
2 4
(a) Describe the shape that comes from multiplying every vector in the square of
side 2 centered at (0, 0) by A.
(b) What is the area of that shape ?
Solution:
(a) We check where the corners of the squares are being sent by A:
1
4
−1
−2
2
1
−4
−1
A
=
,A
=
,A
=
=
,A
−2
−1
−6
−1
1
6
1
2
Every point in the interior of the square is a convex combination of the corners, so
(since we only apply a linear transformations) all the points
ofthe interior
in the
2
−4
−2
4
.
,
,
,
square will be sent to convex combinations of the vectors
−2
−6
2
6
So the shape we obtain is the parallelogram with the corners given above.
(b) det(A) · area of the square = 10 · 4 = 40
2. Suppose you have any shape with area 1 in the xy plane. If you multiply every vector
in that shape by the same A as above, then the area is multiplied by the determinant
of A. Why is this ?
Solution:
RR
The area of the initial shape S in the plane xy is
S(xy) dA. The area of the
RR
shape after applying the transformation is
glance we see
AS(xy) dA. After a quick
RR
that the Jacobian matrix J = A. Hence after a change of variable
AS(xy) dA =
RR
RR
S(xy) det(J)dA = det(A) ·
S(xy) dA.
A simpler, yet not rigorous solution is to argue that the statement holds for a parallelogram, then extend it to arbitrary shapes by partitioning them into infinitesimally small parallelograms. Indeed, given a parallelogram determined by vectors
u, v, its area is det([u v]). Then, after applying A, the area becomes det([Au Av]) =
det(A[u v]) = det(A) det([u v]), so the area gets multiplied by A.
3. Find the determinant of the 3 × 3 matrix A whose (i, j) entry is 3|i−j| . Find the
inverse of A using the cofactor matrix and the determinant (A−1 = C T / det A).
1
Solution:


1 3 9
A = 3 1 3 , det(A) = 13 + 32 · 9 + 32 · 9 − 92 − 32 − 32 = 64
9 3 1


−8 24
0
C =  24 −80 24 
0
24 −8


−1
3
0
1
1
C T =  3 −10 3 
A−1 =
det(A)
8
0
3
−1
4. (p. 252, problem 11) Suppose that CD = −DC and find the flaw in this reasoning:
Taking determinants gives |C| |D| = − |D| |C|. Therefore |C| = 0 or |D| = 0. One or
both of the matrices must be singular.
Solution:
CD = −DC ⇒ det(CD) = (−1)n det(DC), and not − det(DC). If n is even, CD
can be invertible.
5. Prove that every orthogonal matrix Q has determinant +1 or −1.
Solution:
Q is orthogonal, hence QT Q = I. Hence det(Q)2 = det(QT ) det(Q) = det(QT Q) =
det(I) = 1. So det(Q) ∈ {±1}.
T
6. Find the determinant
 of I + M , if M is the rank one matrix M = vv , where v is a
a
column vector v =  b .
c
Solution:


1 + a2
ab
ac
1 + b2
bc 
I + M =  ab
ac
bc
1 + c2
ab
ab 1 + b2 1 + b2
bc
bc
− ab det(I + M ) = (1 + a ) ac 1 + c2 + ac ac
bc bc
1 + c2 2
= (1 + a2 )(1 + b2 + c2 ) − ab · ab + ac · (−ac)
= 1 + a2 + b2 + c2
2
Side note: If you know Sylvester’s determinant theorem, the problem becomes incredibly easy. Applied to this particular instance, we get that det(I+vv T ) = det(1+v T v) =
1 + kvk2 .
7. Suppose An is the n × n symmetric tridiagonal matrix with a subdiagonal of 1’s, a
main diagonal of 3’s, and a superdiagonal of 1’s. By cofactors of row 1, connect the
determinant of An to the determinants of An−1 and An−2 .
Solution:
The first row has nonzero elements only on the first two columns.
 The first cofac1 1
0
... 0
 0



tor is C11 = An−1 , the second is C12 =  .
 We can compute
.
 .

An−2
0
det(C12 ) = 1 · det(An−2 ) + 1 · 0 = det(An−2 ) (since when taking the second cofactor
we get only zeroes in the first column). Putting everything together, we finally get
det(An ) = 3 · det(An−1 ) − 1 · det(An−2 ).
8. (a) Find orthonormal
vectors q1 , q2 , q3 such that q1 and q2 span the column space of

1 1
A = 2 2 . Use Gram-Schmidt.
2 −2
(b) Which of the four fundamental subspaces for A will contain q3 ?
 
4
(c) Solve Ax = 3 by least squares.
2
Solution:
(a) Let c1 and c2 be the columns of A. Then q1 =


c1
kc1 k
 
1
c −q ·(cT q )
= 13 2, q2 = c2 −q1 ·(c2T q1 ) =
k2 1 2 1k
2
8



 
(1/9)· 16 

2
−20
1 

 = √
4 . To get q3 , pick an arbitrary vector c3 that is ob
3 5
8 

−5
(1/9)· 16 


−20 viously not in the column space
  of A, and continue doing Gram-Schmidt. For
1
c −q ·(q T c )−q ·(q T c )
example we can pick c3 = 1, then compute q3 = c3 −q1 ·(q1T c3 )−q2 ·(q2T c3 ) =
k3 1 1 3 2 2 3k
0
 
2
√1 −1.
5
0
3
(b) q3 ∈ N (AT ), since N (AT ) ⊥ C(A), and q3 ⊥ C(A).
 
−1 4
9 1
14
9 1
14
T
T


(c) Solve A Ax = A 3 , equivalent to
x=
. Hence x =
=
1 9
6
1 9
6
2
9 −1 14
3/2
1
=
.
80 −1
9
6
1/2
9. MATLAB problems: Please go to lms.mitx.mit.edu to finish this part.
4