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MAT3701 Assignment 2 Complete Solutions UNISA 2024 Due date 16 August 2024 LINEAR ALGEBRA NEW
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MAT3701ASSIGNMENT 2
FULL SOLUTIONS
Due date: 16 August 2024, Friday.
Complete Solutions
LINEAR ALGEBRA
UNISA
2024
,SOLUTION:
We are given ||T(x)|| = ||x||, for all x.
We know <x,x> = 0 if and only if x = 0 and <T(x), T(x)> = 0 if and only if T(x) = 0.
Therefore, T(x) = 0 if and only if x = 0.
So N(T) = {0} and therefore, T is one to one.
SOLUTION
To prove that ∣∣⋅∣∣ defined by ∣∣(a,b)∣∣=max{∣a∣,∣b∣} is a norm on R2, we need to verify that
it satisfies the three properties of a norm:
1. Non-negativity and definiteness:
∣∣(a,b)∣ ∣≥0 and ∣∣(a,b)∣∣=0 ⟺ (a,b)=(0,0)
2. Homogeneity (or absolute scalability):
∣∣c⋅(a,b)∣∣=∣c∣⋅∣∣(a,b)∣∣for allc∈R
3. Triangle inequality:
, ∣∣(a1,b1)+(a2,b2)∣∣ ≤ ∣∣(a1,b1)∣∣ + ∣∣(a2,b2)∣∣
Let's verify each property step by step.
1. Non-negativity and Definiteness
For any (a,b)∈R2:
• Non-negativity: ∣∣(a,b)∣∣=max{∣a∣,∣b∣} ≥ 0 because the absolute values of real
numbers are always non-negative.
• Definiteness:
o If (a,b)=(0,0), then ∣∣(0,0)∣∣=max{∣0∣,∣0∣} = 0.
o Conversely, if ∣∣(a,b)∣∣=0, then max{∣a∣,∣b∣}=0 implies that
both ∣a∣=0 and ∣b∣=0, hence a=0 and b=0. Thus, (a,b)=(0,0).
2. Homogeneity
For any (a,b)∈R2 and c∈R:
∣∣c⋅(a,b)∣∣=∣∣(ca,cb)∣∣=max{∣ca∣,∣cb∣}=∣c∣⋅max{∣a∣,∣b∣}=∣c∣⋅∣∣(a,b)∣∣
3. Triangle Inequality
For any (a1,b1) and (a2,b2)∈R2:
∣∣(a1,b1)+(a2,b2)∣∣=∣∣(a1+a2,b1+b2)∣∣=max{∣a1+a2∣,∣b1+b2∣}
We need to show that:
max{∣a1+a2∣,∣b1+b2∣}≤max{∣a1∣,∣b1∣}+max{∣a2∣,∣b2∣}
Consider the two cases for the maximum of ∣a1+a2∣ and ∣b1+b2∣:
1. If ∣a1+a2∣≤max{∣a1∣,∣a2∣}:
∣a1+a2∣≤∣a1∣+∣a2∣≤max{∣a1∣,∣b1∣}+max{∣a2∣,∣b2∣}
2. If ∣b1+b2∣≤max{∣b1∣,∣b2∣}:
∣b1+b2∣≤∣b1∣+∣b2∣≤max{∣a1∣,∣b1∣}+max{∣a2∣,∣b2∣}
Thus, in both cases, we have:
max{∣a1+a2∣,∣b1+b2∣}≤max{∣a1∣,∣b1∣}+max{∣a2∣,∣b2∣}
Therefore, the triangle inequality holds.
Since all three properties are satisfied, ∣∣⋅∣∣=max{∣a∣,∣b∣} is indeed a norm on R2.
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