MTH2021 Linear Algebra with Applications
Sample Exam – Handwritten Questions
School of Mathematics, Monash University
14. Endow P2 with the inner product hp, qi =
R 1
1 p(x)q(x) dx, and consider the subspace W = span(B) where B = {p1, p2} is the
linearly independent set with elements
p1(x) = x
2 1, p2(x) = x 1.
(a) Use the Gram-Schmidt algorithm to construct an orthonormal basis forW .
We first apply the Gram-Schmidt algorithm to construct an orthogonal basis B˜ = {q1, q2} from B. We begin by setting
q1 = p1. To find q2, we compute
q2 = p2 hp2, q1ihq1, q1i q1.
But
hq1, p2i = hp1, p2i =
Z 1
1
p1(x)p2(x)dx =
Z 1
1
(x2 1)(x 1) dx =
Z 1
1
(x3 x2 x+ 1) dx = 4
3
hq1, q1i = hp1, p1i =
Z 1
1
p1(x)p1(x)dx =
Z 1
1
(x2 1)2 dx =
Z 1
1
(x4 2x2 + 1) dx = 16
15
so that
p2(x) hp2, q1ihq1, q1i q1(x) = (x 1)
4/3
16/15
(x2 1) = 5
4
x2 + x+
1
4
.
Since we are going to normalize B˜ at the end of the orthogonalization procedure, we are free to choose the scale of q2 as we
find convenient. Set q2(x) = 5x2 4x 1.
The set B˜ is orthogonal. We can now obtain an orthonormal set by choosing r1 = 1kq1kq1 and r2 =
1
kq2kq2, so that
r1(x) =
1
kq1kq1(x) =
p
15
4
q1(x) =
p
15
4
x2
p
15
4
r2(x) =
1
kq2kq2(x) =
1
4
q2(x) =
5
4
x2 x 1
4
.
(b) Let q(x) = x2 x. Express q as linear combination of the elements of the orthonormal basis found in Part (a).
q = hq, r1ir1 + hq, r2ir2,
with
hq, r1i =
Z 1
1
q(x)r1(x) dx =
Z 1
1
p
15
4
(x4 x3 x2 + x) dx = 1p
15
hq, r2i =
Z 1
1
q(x)r2(x) dx =
Z 1
1
1
4
(5x4 9x3 + 3x2 + x) dx = 1.
[6+2=8 marks]
15. Let A =
246 2 02 6 0
0 0 0
35.
(a) Explain why A is orthogonally diagonalizable.
A is symmetric, therefore it is orthogonally diagonalizable.
(b) The vectors v1 = (1, 1, 0), v2 = (1, 1, 0), and v3 = (0, 0, 1) are eigenvectors of A. What are the eigenvalues of A?
Av1 = 8v1, so 1 = 8 is an eigenvalue of A.
Av2 = 4v2, so 2 = 4 is an eigenvalue of A.
Av3 = 0, so 3 = 0 is an eigenvalue of A.
Since A is a 3 ⇥ 3 matrix, it can have at most 3 eigenvalues. Therefore 1 = 8, 2 = 4, 3 = 0 comprise the full set of
eigenvalues of A.
(c) Find a matrix Q which orthogonally diagonalizes A.
An orthonormal basis for the eigenspace of 1 = 8 is { 1p2 (1, 1, 0)}.
An orthonormal basis for the eigenspace of 2 = 4 is { 1p2 (1, 1, 0)}.
An orthonormal basis for the eigenspace of 3 = 0 is (0, 0, 1)}.
Therefore the matrix
Q =
266664
1p
2
1p
2
0
1p
2
1p
2
0
0 0 1
377775
orthogonally diagonalizes A.
(d) Let B =
242 2 01 1 0
1 1 0
35, and note that BTB = A. What are the singular values of B?
The singular values of B are the square roots of the eigenvalues of BTB. Since BTB = A, the singular values of B are
1 = 2
p
2, 2 = 2 and 3 = 0.
(e) Use the fact that null(BT ) = span{(0, 1,1)} to find a singular value decomposition for B.
Using the singular values found above, we set
⌃ =
242p2 0 00 2 0
0 0 0
35 .
Part (c) showed that
Q =
266664
1p
2
1p
2
0
1p
2
1p
2
0
0 0 1
377775
is an orthogonal matrix, whose columns are eigenvectors of A = BTB, corresponding respectively to the singular values
1,2,3. Therefore, set V = Q. To construct the matrix U , begin by setting
u1 =
1
1
B
0@ 1p21p
2
0
1A =
2410
0
35
u2 =
1
2
B
0@ 1p21p
2
0
1A =
24 0 1p
2
1p
2
35
Then, since
⇣
0, 1p
2
, 1p
2
⌘
is an orthonormal basis for null(BT ), set
u3 =
24 01p
2
1p
2
35 .
Therefore, setting
U =
⇥
u1 u2 u3
⇤
=
241 0 00 1p
2
1p
2
0 1p
2
1p
2
35
we have the singular value decomposition B = U⌃V T .
[1 + 3 + 2 + 2 + 4 = 12 marks]
16. Let V be a vector space and let U andW be subspaces of V , such that U is not a subset ofW , andW is not a subset of U .
(a) Is U \W closed under addition? Is U \W closed under scalar multiplication? Does U \W form a subspace of V ?
Let v1,v2 2 U \W . Since U is a subspace, v1+v2 2 U . SineW is a subspace, v1+v2 2W . Therefore v1+v2 2 U \W ,
and so U \W is closed under addition.
Let k be a scalar. Since U is a subspace, kv1 2 U . SinceW is a subspace, kv1 2W . Therefore kv1 2 U \W , and so U \W
is closed under scalar multiplication.
Since 0 2 U and 0 2W , we have U \W 6= ;. Since U \W is nonempty and closed under addition and scalar multiplication,
it is a subspace of V .
(b) Is U [W closed under addition? Is U [W closed under scalar multiplication? Does U [W form a subspace of V ?
Since U 6✓W andW 6✓ U , there exists u 2 U \W andw 2W \U . If u+w belonged to U , then so would (u+w)u = w.
But w 62 U by assumption, and so u+w 62 U . Similarly, u+w 62 W and so u+w 62 U [W . But u,w 2 U [W and so
U [W is not closed under addition.
Let k be a scalar and let v 2 U [W . If v 2 U then kv 2 U since U is a subspace, and hence kv 2 U [W . Similarly, if
v 2W then kv 2W sinceW is a subspace, and hence kv 2 U [W . It follows that if v 2 U [W then ku 2 U [W and so
U [W is closed under scalar multiplication.
Since U [W is not closed under addition, it is not a subspace of V .
[4 + 4 = 8 marks]
17. Let V be a vector space with basis BV = {v1, . . . ,vn}, and letW be a vector space with basis BW = {w1, . . . ,wn}. Let T : V !W
be defined so that for all x1, . . . , xn 2 R
T

nX
i=1
xivi
!
=
nX
i=1
xiwi
(a) Show that T is linear
Let x,y 2 V . Then there exist x1, . . . , xn, y1, . . . , yn 2 R such that
x =
nX
i=1
xivi and y =
nX
i=1
yivi
If k 2 R we have
T (x+ ky) = T

nX
i=1
xivi + k
nX
i=1
yivi
!
= T

nX
i=1
(xi + k yi)vi
!
=
nX
i=1
(xi + k yi)wi
=
nX
i=1
xiwi + k
nX
i=1
yiwi
= T

nX
i=1
xivi
!
+ k T

nX
i=1
yivi
!
= T (x) + k T (y)
which shows that T is linear.
(b) Show that T is bijective.
Let x 2 V . Then there exist x1, . . . , xn 2 R such that
x =
nX
i=1
xivi.
If x 2 ker(T ) then
0 = T (x) = T

nX
i=1
xivi
!
=
nX
i=1
xiwi
But since BW is linearly independent, it follows that x1 = . . . xn = 0 and so x = 0. Therefore, ker(T ) = {0} and so T is
injective.
Since nullity(T ) = 0 from the previous part, the rank-nullity theorem implies that rank(T ) = dim(V ). But dim(V ) =
dim(W ) by assumption, and therefore dim(ran(T )) = dim(W ). The range of T is therefore an n-dimensional subspace of
the n-dimensional codomainW , and so ran(T ) = W , which shows that T is surjective.
[5 + 5=10 marks]
18. A square matrix A is lower triangular if (A)ij = 0 for all i < j. Show that the product of two lower triangular matrices is lower
triangular.
Let A and B be lower triangular n⇥ n matrices, and let i < j.
(AB)ij =
nX
k=1
(A)ik(B)kj
=
iX
k=1
(A)ik(B)kj +
nX
k=i+1
(A)ik(B)kj .
Since (A)ik = 0 for all k > i, the second sum vanishes. Likewise, (B)kj = 0 for all k  i when i < j, implying the first sum
vanishes. Therefore (AB)ij = 0 for all i < j, which implies that AB is lower triangular.
[4 marks]
19. Let S be a non-empty set of vectors in a vector space V . Prove that if v 2 S can be expressed as a linear combination of the other vectors
in S, then span(S \ {v}) = span(S).
(Hint: to show that span(S \ {v}) = span(S), it suffices to show that span(S \ {v}) ✓ span(S) and span(S) ✓ span(S \ {v}).)
Suppose S = {v1,v2, . . . ,vm,v}. If w 2 span(S \ {v}) then w 2 span(S) since w = w + 0v. So span(S \ {v}) ✓ span(S).
Suppose v =
Pm
i=1 civi. If w 2 span(S) then
w =
mX
i=1
aivi + am+1v
=
mX
i=1
aivi + am+1
mX
i=1
civi
=
mX
i=1
(ai + am+1ci)vi 2 span(S \ {v})
which implies that span(S) ✓ span(S \ {v}). But span(S \ {v}) ✓ span(S) so in fact span(S \ {v}) = span(S).
[4 marks]
20. Let V be an inner product space. Prove that for all u,v 2 V
ku+ vk  kuk+ kvk
(Hint: You may use the fact that the square root function f : [0,1) ! [0,1), f(x) = px, is an increasing function, meaning that
f(x) > f(y) whenever x > y. You also may find the Cauchy-Schwarz inequality |hu,vi|  kuk kvk useful.)
ku+ vk2 = hu+ v,u+ vi
= hu,ui+ 2hu,vi+ hv,vi
 kuk2 + kvk2 + 2|hu,vi|
 kuk2 + kvk2 + 2kukkvk (via Cauchy Schwarz)
= (kuk+ kvk)2.
We have therefore showed that
ku+ vk2  (kuk+ kvk)2.
Since the square root is an increasing function, taking the square root of both sides of the above inequality implies
ku+ vk  kuk+ kvk
as required.
[4 marks]