Dr.

Ahlem Nemer 1

Course : Algebra 3 Year : 2023/2024

Chapter 4 : Vector spaces Department of Computer Science

Solutions

Solution 0.1 .

1) For x ∈ Rn , we find
n
X
< x, x >= x2i < 0 (1)
i=1

and when < x, x >= 0, we get

n
X
x2i = 0, (2)
i=1

then, x1 = x2 = · · · = xn = 0. We deduce that x = 0.

2) For x, y ∈ Rn , we have
n
X
< x, y > = xi yi
i=1
Xn
= yi xi
i=1
= < y, x >

3) For x, y, z ∈ Rn , we have
n
X
< x + y, z > = (xi + yi )zi
i=1
Xn n
X
= xi zi + yi zi
i=1 i=1
= < x, z > + < y, z >

4) For x, y ∈ Rn and a real scalar α, we can get

n
X
< αx, y > = αxi yi
i=1
Xn
= α xi yi
i=1
= α < x, y >

Finally, we can say that < x, y > is an inner product on Rn .

Dr. Ahlem Nemer 2

Solution 0.2 .

1) For f ∈ C[a, b], we have

Z b
< f, f >= w(t)f 2 (t)dt < 0, (3)
a

because w(t)  0 and f 2 (t) < 0 and when

Z b
< f, f >= 0 ⇔ w(t)f 2 (t)dt = 0
a
⇔ w(t)f 2 (t) = 0
⇔ f (t) = 0

because w(t)  0. Then, < f, f >= 0 ⇔ f = 0.

2) For f, g ∈ C[a, b], we have
Z b
< f, g > = w(t)f (t)g(t)dt
a
Z b
= w(t)g(t)f (t)dt
a
= < g, f >

3) For f, g, h ∈ C[a, b], we have

Z b
< f + g, h > = w(t)(f + g)(t)h(t)dt
a
Z b Z b
= w(t)f (t)h(t)dt + w(t)g(t)h(t)dt
a a
= < f, h > + < g, h >

4) For f, g ∈ C[a, b] and a scalar α

Z b
< αf, g > = w(t)(αf )(t)g(t)dt
a
Z b
= α w(t)f (t)g(t)dt
a
= α < f, g >

Finally, we can say that < f, g > is an inner product on C[a, b].
Dr. Ahlem Nemer 3

Solution 0.3
Z π
< f, g > = f (t)g(t)dt
Z0 π
= cos(t) sin(t)dt
0
1 π
Z
= sin(2t)dt
2 0
1 cos(2t) π
= [− ]0
2 2
1
= − (cos(2π) − cos(0))
4
1
= − (1 − 1)
4
= 0

Then, f (t) and g(t) are orthogonal.

Solution 0.4 We have v1 = 1, v2 = x and v3 = x2 . By employing Gram-Schmidt process, we get

u1 = v1 = 1.

< v2 , u1 >
u2 = v2 − u1
k u1 k2
R1
xdx
= x − R−1
1
−1
1dx
1 x2
= x − [ ]1−1
2 2
= x

2
X < v3 , ui >
u3 = v3 − ui
i=1
k ui k2
< v3 , u1 > < v3 , u2 >
= v3 − u1 − u2
k u1 k2 k u2 k2
R1 2 R1 3
−1
x dx x dx
2
= x − R1 − R−1
1 x
−1
1dx −1
x2 dx
1
= x2 −
3
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Solution 0.5
1 1 1
k √ (v1 − v2 ) k2 = < √ (v1 − v2 ), √ (v1 − v2 ) >
2 2 2
1
= (< v1 , v1 > − < v2 , v1 > − < v1 , v2 > + < v2 , v2 >)
2
1
= (< v1 , v1 > + < v2 , v2 >)
2
= 1

1 1 1
k √ (v1 + v2 ) k2 = < √ (v1 + v2 ), √ (v1 + v2 ) >
2 2 2
1
= (< v1 , v1 > + < v2 , v1 > + < v1 , v2 > + < v2 , v2 >)
2
1
= (< v1 , v1 > + < v2 , v2 >)
2
= 1

1 1
< √ (v1 − v2 ), √ (v1 + v2 ) > = 0
2 2
Solution 0.6 .

1) For a scalar α ∈ Y and for x, y 6= 0

0 4 k x − αy k2
= < x − αy, x − αy >
= k x k2 −α < x, y > −α(< y, x > −α k y k2 ).
< y, x >
We take α = . This leads to get
k y k2
< y, x >< x, y >
0 4 k x k2 −
k y k2
| < x, y > |2
= k x k2 − .
k y k2

Then, we have
| < x, y > | 4k x kk y k (4)
2) For x, y ∈ V

k x + y k2 = < x + y, x + y >
= k x k2 + < x, y > + < y, x > + k y k2
4 k x k2 +2 k x kk y k + k y k2
= (k x k + k y k)2 .

Then, we deduce that

k x + y k4k x k + k y k (5)