MS-C1342 Linear algebra, V/2023 Noferini / Puska

Linear algebra
Exercise sheet 3 / Model solutions

1. Let f : R2 → R be such that

 

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T 2 1
f (x) = x Ax where A= .
1 2
The matrix A can be decomposed as A = U T ΛU , where
" #
√1 − √1
 
2 2 1 0
U = √1 √1
and Λ = .
2 2
0 3

(a) Show that y T Λy ≥ y T y for all y ∈ R2 and xT Ax ≥ xT x for all x ∈ R2 . (Hint:

Write A = U T ΛU and set y = U x.)

1/2 T
1/2
(b) Show that xT Ay ≤ xT Ax y Ay . (Hint: Write xT Ay = (Λ1/2 U x)T (Λ1/2 U y)
and use the Cauchy–Schwarz inequality, |x y| ≤ kxk2 kyk2 for all x, y ∈ R2 .)
T

(c) Show that f satisfies the triangle inequality : f (x + y) ≤ f (x) + f (y). Is f a norm?
Solution.
(a) By direct calculation,
y T Λy = y12 + 3y22 ≥ y12 + y22 = y T y. (1)
Using the decomposition of A,
xT Ax = xT U T ΛU x = (U x)T Λ(U x).
Denoting y = U x and using (1),
(U x)T Λ(U x) = y T Λy ≥ y T y = xT U T U x.
Noticing that U T U = I completes the proof.
(b) Using the decomposition and hint gives
xT Ay = xT U T ΛU y = (Λ1/2 U x)T (Λ1/2 U y).
Application of the Cauchy–Schwarz inequality to the RHS gives:
|(Λ1/2 U x)T (Λ1/2 U y)| ≤ kΛ1/2 U xk2 kΛ1/2 U yk2

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The Euclidian norm is defined as kxk2 = xT x so that

1/2 T T
1/2
xT Ay ≤ kΛ1/2 U xk2 kΛ1/2 U yk2 = xT U T ΛU x y U ΛU y ,
which completes the proof.

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(c) By direct calculation and by symmetry of A,

(x + y)T A(x + y) = xT Ax + 2xT Ay + y T Ay.

Using (b) we estimate this from above as

(x + y)T A(x + y) ≤ xT Ax + 2(xT Ax)1/2 (y T Ay)1/2 + y T Ay,

which can be written as

2
(x + y)T A(x + y) ≤ (xT Ax)1/2 + (y T Ay)1/2 .

Taking a square root from both sides completes the proof that f (x + y) ≤ f (x) +
f (y). Lastly, f is indeed a norm, as it also
√ satisfies the other two requirements because
f (x) ≥ kxk2 by item (a) and f (αx) = α2 xT Ax = |α|f (x).

2. Let q 1 = [0, 1, 1]T , q 2 = [1, 1, −1]T and q 3 = [2, −1, 1]T .

(a) Show that q 1 , q 2 and q 3 are mutually orthogonal with respect to the Euclidian inner
product.
(b) Represent the vectors v = [1, 2, 3]T and w = [4, −2, 1]T as a linear combination of the
vectors q 1 , q 2 and q 3 . (Hint: Write v = α1 q 1 + α2 q 2 + α3 q 3 . Use the orthogonality
to find coefficients α1 , α2 , α3 .)

Solution.

(a) By a direct computation,

q 1 · q 2 = 0 + 1 − 1 = 0, q 2 · q 3 = 2 − 1 − 1 = 0, q 1 · q 3 = 0 − 1 + 1 = 0.

(b) For each coefficient αi of v in the orthogonal basis {q i }i=1,...,3 we can use the formula
qi · v
αi = ,
qi · qi
from which we get that

α1 = 5/2, α2 = 0, α3 = 1/2.

With a similar technique one can show that w = (−1/2)q 1 + (1/3)q 2 + (11/6)q 3 .

3. Let q 1 = [0, 3, 2]T , q 2 = [5, 1, −1]T and q 3 = [2, −2, 2]T and define the inner product

hx, yi = x1 y1 + 2x2 y2 + 3x3 y3 . (2)

(a) Show that q 1 , q 2 and q 3 are mutually orthogonal with respect to the inner product
defined in (2).

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(b) Represent the vectors v = [1, 2, 3]T and w = [4, −2, 1]T as a linear combination of the
vectors q 1 , q 2 and q 3 .

Solution.

(a) By direct calculation,

hq 1 , q 2 i = 0 × 5 + 2 × 3 × 1 + 3 × 2 × (−1) = 6 − 6 = 0,
hq 1 , q 3 i = 0 × 2 + 2 × 3 × (−2) + 3 × 2 × (2) = −12 + 12 = 0,
hq 2 , q 3 i = 5 × 2 + 2 × 1 × (−2) + 3 × (−1) × 2 = 10 − 4 − 6 = 0.

(b) Similarly to the previous exercise, we may write

3 3
X hv, q j i X hv, q j i
v= qj = 2 qj .
i=1
hq j , q j i i=1
kq j k
| {z }
norm induced
by the inn.prod.

Let us then compute the coefficients in the sum:

hv, q 1 i 0 + 12 + 18
= = 1,
hq 1 , q 1 i 0 + 18 + 12
hv, q 2 i 5+4−9
= = 0,
hq 2 , q 2 i 25 + 2 + 3
hv, q 3 i 2 − 8 + 18 1
= = ,
hq 3 , q 3 i 4 + 8 + 12 2

so that
1
v = q1 + q3.
2
Similarly for the vector w we get
hw, q 1 i 0 − 12 + 6 1
= =− ,
hq 1 , q 1 i 30 5
hw, q 2 i 20 − 4 − 3 13
= = ,
hq 2 , q 2 i 30 30
hw, q 3 i 8+8+6 11
= = ,
hq 3 , q 3 i 24 12

so that
1 13 11
w = − q1 + q2 + q3.
5 30 12
4. Let x, y ∈ Rn . Show that:

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(a) kxk∞ ≤ kxk2 ≤ kxk1 ,

√
(b) kxk1 ≤ nkxk2 ≤ nkxk∞ ,
(c) |x · y| ≤ kxk∞ kyk1 .

(Hint: (b) Write kxk1 as an inner product between x and a vector containing elements −1
and 1. Apply the Cauchy–Schwarz inequality.)
Solution.

(a) First of all

n
!1/2
 1/2 X
kxk∞ = max |xj | = max |xj |2 ≤ |xj |2 = kxk2 .
j=1,...,n j=1,...,n
j=1

On the other hand

n
!2 n n X
n n
X X X X
kxk21 = |xj | = |xj |2 + |xj ||xk | ≥ |xj |2 = kxk22
j=1 j=1 j=0 k=1 j=1
j6=k
| {z }
≥0

(b) Define the vector x̂ = [sgn(x1 ), sgn(x2 ), . . . , sgn(xn )]T , where

(
1, if t ≥ 0,
sgn(t) =
−1 if t < 0.

Clearly
n
X n
X
x̂ · x = sgn(xj )xj = |xj |.
j=1 j=1

Now the Cauchy–Schwarz inequality gives

n
!1/2 n
!1/2
X X √
kxk1 = x̂·x ≤ kx̂k2 kxk2 = sgn(xj )2 kxk2 = 1 kxk2 = nkxk2 .
j=1 j=1

On the other hand

n
!1/2 n
!1/2 n
!1/2
X X X
kxk2 = |xj | ≤ max |xi |2 = kxk2∞
i=1,...,n
j=1 j=1 j=1

1/2 √
= nkxk2∞ = nkxk∞ ,

that is, equivalently, √

nkxk2 ≤ nkxk∞ .

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(c) We have
n
X n
X
|x · y| = xj y j ≤ |xj yj |
j=1 j=1
n
X Xn
= |xj ||yj | ≤ ( max |xi |)|yj |
i=1,...,n
j=1 j=1 | {z }
=kxk∞
n
X
= kxk∞ |yj | = kxk∞ kyk1 .
j=1

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