Find an example for each of vectors x, y ∈ V in R. n
for n = 2 or n = 3 such that
applies:
A
||x|| = max{|x1|, ..., |xn|}
||x|| =√n · max{|x1|, ..., |xn|}
(B)
||x + y|| = ||x|| + ||y||
||x + y|| < ||x|| + ||y||
(C)
||x · y|| = ||x|| · ||y||
||x · y|| < ||x|| · ||y||

(a) Both conditions are satisfied with x = (1, 0) for and x = (1, 0, 0) for :

||(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 , 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 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 , you can use the vectors x = (1, 0, 0) and y = (1, 1, 1).

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step-by-step explanation:

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