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Solution: APPM 3310: Matrix Methods — Exam #1 — Summer 2012
Problem 1: (30 pts) Consider the matrix A =
(a) (6 pts) Find a decomposition A = LR where L is special lower triangular and R is in row echelon form (Write down both L and R.) What is the rank of the matrix A?
(b) (6 pts) Is the set
a basis for the range of A? Why or why not?
(c) (6 pts) Find a basis for ker(A).
(d) (6 pts) Now given that b = [4, − 1 , 0]T^ find the general solution of the system Ax = b given above.
(e) (6 pts) Consider the block matrix defined as B =
A A A
A A A
A A A
(^) where A is the same matrix given
above. Write the block matrix B in REF (do not write out the 9 × 12 matrix, instead write it in block form using the fact that R is the REF of A.) What is the rank of B? Justify your answer.
Solution:
(a) We have A =
L=
L=
= R
The REF of A is R =
(^) and so the rank of A is 3.
(b) Yes, they form a basis since we have 3 linearly independent vectors in a space of dimension 3. Note
that det
(^6) = 0, and dim(range(A)) = rank(A) = 3, thus we have a basis by a dimension
argument. (c) Note that z is free, and so
[ A | ~ 0 ] =
x y z w
z
so a basis for kernel(A) is
{[
]T }
(d) Note that b = [4, − 1 , 0]T^ is the sum of the third column and the last column of the A matrix so
A
=^ b^ and so the general solutions is
~x =
t,^ ∀t^ ∈^ R.
(Note that the choice of xp = [0, 0 , 1 , 1]T^ is not unique.)
(e) The rank is 3 since we can reduce B =
A A A
A A A
A A A
→ U =
R R R
O O O
O O O
(^) where R is the row
echelon form of A and so U will have rank 3. Problem 2: (35 pts) Let V be the set of vector valued functions in R^2 with components from the set of polynomials of degree less than or equal to 2, that is
V ≡
{[
f (t) g(t)
] ∣∣
∣∣f (t), g(t) ∈ P(2)
(a) (5 pts) Show that the elements of V are closed under standard vector addition. Justify your answer.
(b) (5 pts) Show that the elements of V are closed under standard scalar multiplication of vectors. Justify your answer.
(c) (5 pts) Is R^2 a subset of V? Justify your answer.
(d) (10 pts) Assuming V is a vector space with real scalars, write down a basis for V, justify your answer. What is the dimension of V?
(e) (5 pts) Write down the coordinates of v =
[
t 6 + 9t
]
(f) (5 pts) Is v =
[
t 6 + 9t
]
in the span of the vectors v 1 =
[
t
]
and v 2 =
[
t π
]
? Justify your answer.
Solution: (a) Note that
[
f 1 (t) g 1 (t)
]
[
f 2 (t) g 2 (t)
]
[
f 1 (t) + f 2 (t) g 1 (t) + g 2 (t)
]
. Now note that f 1 (t) + f 2 (t) and g 1 (t) + g 2 (t)
are in P(2) since P(2) is a vector space, and so
[
f 1 (t) + f 2 (t) g 1 (t) + g 2 (t)
]
∈ V.
(b) Note that c
[
f (t) g(t)
]
[
cf (t) cg(t)
]
and note that cf (t) and cg(t) are in P(2) for any c ∈ R since P(2)
is a vector space and so
[
cf (t) cg(t)
]
∈ V.
(c) Yes, since any real number is an element of P(2) we have that
[
a b
]
∈ V for any real numbers a and
b, thus R^2 ⊂ V. (d) The following set forms a basis for V {[ 1 0
]
[
t 0
]
[
t^2 0
]
[
]
[
t
]
[
t^2
]
since this set spans V and is a linearly independent set of elements in V. Note that dim(V) = 6.
(e) Using the ordered basis given in part (d), we have v =
[
]T
(f) No, using coordinates we wish to solve
0 π 6 1 0 9 0 0 0
which is (clearly) inconsistent, so v is not in the span of v 1 and v 2. Problem 3: (35 pts) Short answer. Each of the questions below are unrelated. Justify your answers.
(a) Define a norm in R^2 by ‖x‖ ≡
xT^ x, show that if xT^ y = 0 then ‖x + y‖^2 = ‖x‖^2 + ‖y‖^2. (b) Does 〈u, v〉 ≡ u 1 v 1 + 2u 2 v 3 + 3u 3 v 2 define a norm in R^3? Why or why not? (c) Always True or False: For any n × n matrix A, det(AT^ A) = det(A^2 ) (Justify your answer.) (d) Always True or False: For an n × m matrix A, if rank(A) = n then kernel(A) = {~ 0 } (Justify)
