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Solution: APPM 3310: Matrix Methods — Exam #2 — Summer 2012
Problem 1. (30 points) The following problems are unrelated:
(a) What is the definition of a positive definite matrix?
(b) If F > 0 and if B > 0 then is it true that F + B > 0? Why or why not?
(c) Let q(x) = xT^ M x for any x ∈ Rn^ where M > 0, show that the expression
〈x, y〉 ≡
[q(x + y) − q(x) − q(y)]
is the same as the inner product defined by M.
Solution: (a) An n × n matrix K is positive definite if it is symmetric and if xT^ Kx > 0 for all x 6 = 0.
(b) It is true since xT^ (F + B)x = xT^ F x + xT^ Bx and xT^ F x > 0 for all x 6 = 0 and xT^ Bx > 0 for all x 6 = 0 which implies xT^ F x + xT^ Bx > 0 (since this is a sum of positive terms) for all x 6 = 0 and also note that (F + B)T^ = F T^ + BT^ = F + B, so F + B is positive definite.
(c) Note that, 〈x, y〉 =
[q(x + y) − q(x) − q(y)]
=
[
(x + y)T^ M (x + y) − xT^ M x − yT^ M y
]
[
(xT^ + yT^ )(M x + M y) − xT^ M x − yT^ M y
]
[
(xT^ M x + xT^ M y + yT^ M x + yT^ M y) − xT^ M x − yT^ M y
]
[
xT^ M y + yT^ M x
]
and note that xT^ M y is a scalar so it is symmetric thus xT^ M y = (xT^ M y)T^ = yT^ M T^ x = yT^ M x (note M > 0 implies M T^ = M ) and so
〈x, y〉 =
[
xT^ M y + yT^ M x
]
[
2 xT^ M y
]
= xT^ M y
therefore the expression 〈x, y〉 ≡ 12 [q(x + y) − q(x) − q(y)] is the same as the inner product defined by M.
Problem 2. (30 points) The following problems are unrelated:
(a) Find a basis for all the four fundamental subspaces of the matrix F =
1+i 2 i − 1 0 1 1 − 2 − 2 − 2 i 1 −i
(b) Show that every positive definite m × m matrix B is a Gram matrix (you must specify an inner product and a set of vectors and show that B is the associated Gram matrix.)
(c) Let ‖ · ‖F and ‖ · ‖B be two norms, does ‖v‖ ≡
(‖v‖F + ‖v‖B ) define a norm? (Either explicitly show that this satisfies all the properties of a norm or give an explicit example of when it fails to satisfy one of the properties.)
Solution: (a) Note that by doing the row operation R∗ 3 = (1 − i)R 1 + R 3 to the augmented form [F |b] yields, (^)
1+i 2 i − 1 b 1 0 1 1 b 2 − 2 − 2 − 2 i 1 −i b 3
1+i 2 i − 1 b 1 0 1 1 b 2 0 0 0 (1 − i)b 1 + b 3
and so, range(F ) = span
1 + i 0 − 2
2 i 1 − 2 − 2 i
and corange(F ) = span
1 + i 2 i − 1
and the consistency condition (1 − i)b 1 + b 3 = 0 implies (b 1 b 2 b 3 )
1 −i 0 1
(^) = 0, and the fact that the
range is orthogonal to the cokernel implies coker(F ) = span
1 −i 0 1
. To find the kernel, we let b = 0
which yields,
1+i 2 i − 1 0 0 1 1 0 0 0 0 0
(1+i)x + 2iy − z = 0 y + z = 0 z = z
x y z
2 i+ 1+i − 1 1
(^) z =
3+i 2 − 1 1
(^) t, ∀t ∈ R
so ker (F ) = span
3+i 2 − 1 1
(b) Define 〈x, y〉 ≡ xT^ By (since B > 0 this defines an inner product) and consider the e 1 , e 2 ,... , em then B = (〈ei, ej 〉)ij =
eTi Bej
ij =^ bij^ , so^ B^ is the associated Gram matrix to^ 〈x,^ y〉 ≡^ x
T (^) By and
e 1 , e 2 ,... , em.
(c) Yes, this defines a norm:
Positivity: Note that since ‖v‖F ≥ 0 and ‖v‖B ≥ 0 for any vector v since they are norms, this implies
that ‖v‖ =
(‖v‖F + ‖v‖B ) ≥ 0 for any vector v. Now if ‖v‖ =
(‖v‖F + ‖v‖B ) = 0 this implies
that ‖v‖F = 0 and ‖v‖B = 0 which in both cases implies that v = 0 since they are norms.
Homogeneity: Note that
‖cv‖ =
(‖cv‖F + ‖cv‖B ) =
(|c|‖v‖F + |c|‖v‖B ) = |c|
(‖v‖F + ‖v‖B )
= |c|‖v‖
by the homogeneity of the F -norm and B-norm.
Triangle Inequality: Note that, ‖v + w‖ =
(‖v + w‖F + ‖v + w‖B )
≤
(‖v‖F + ‖w‖F + ‖v‖B + ‖w‖B )
=
(‖v‖F + ‖v‖B ) +
(‖w‖F + ‖w‖B ) = ‖v‖ + ‖w‖
by the fact that the F -norm and B-norm satisfy the Triangle Inequality and so we have a norm.
Problem 3. (40 points) The following problems are unrelated:
(a) Find the closest vector in the space S spanned by (1, 1 , 0 , 0)T^ and (0, 0 , 1 , 1)T^ to b = (3, 1 , 2 , 1)T^. What is the distance of b from the space S? Show all work.
