(a) Both conditions are satisfied with x = (1, 0) for [tex]\mathbb R^2[/tex] and x = (1, 0, 0) for [tex]\mathbb R^3[/tex]:
||(1, 0)|| = √(1² + 0²) = 1
max{1, 0} = 1
||(1, 0, 0)|| = √(1² + 0² + 0²) = 1
max{1, 0, 0} = 1
(b) This is the well-known triangle inequality. Equality holds if one of x or y is the zero vector, or if x = y. For example, in [tex]\mathbb R^2[/tex], take x = (0, 0) and y = (1, 1). Then
||x + y|| = ||(0, 0) + (1, 1)|| = ||(1, 1)|| = √(1² + 1²) = √2
||x|| + ||y|| = ||(0, 0)|| + ||(1, 1)|| = √(0² + 0²) + √(1² + 1²) = √2
The left side is strictly smaller if both vectors are non-zero and not equal. For example, if x = (1, 0) and y = (0, 1), then
||x + y|| = ||(1, 0) + (0, 1)|| = ||(1, 1)|| = √(1² + 1²) = √2
||x|| + ||y|| = ||(1, 0)|| + ||(0, 1)|| = √(1² + 0²) + √(0² + 1²) = 2
and of course √2 < 2.
Similarly, in [tex]\mathbb R^3[/tex] you can use x = (0, 0, 0) and y = (1, 1, 1) for the equality, and x = (1, 0, 0) and y = (0, 1, 0) for the inequality.
(c) Recall the dot product identity,
x • y = ||x|| ||y|| cos(θ),
where θ is the angle between the vectors x and y. Both sides are scalar, so taking the norm gives
||x • y|| = ||(||x|| ||y|| cos(θ)|| = ||x|| ||y|| |cos(θ)|
Suppose x = (0, 0) and y = (1, 1). Then
||x • y|| = |(0, 0) • (1, 1)| = 0
||x|| • ||y|| = ||(0, 0)|| • ||(1, 1)|| = 0 • √2 = 0
For the inequality, recall that cos(θ) is bounded between -1 and 1, so 0 ≤ |cos(θ)| ≤ 1, with |cos(θ)| = 0 if x and y are perpendicular to one another, and |cos(θ)| = 1 if x and y are (anti-)parallel. You get everything in between for any acute angle θ. So take x = (1, 0) and y = (1, 1). Then
||x • y|| = |(1, 0) • (1, 1)| = |1| = 1
||x|| • ||y|| = ||(1, 0)|| • ||(1, 1)|| = 1 • √2 = √2
In [tex]\mathbb R^3[/tex], you can use the vectors x = (1, 0, 0) and y = (1, 1, 1).
