MATH1231 Assignment 2020 T2
We have a function T:โ3 โ โ2 defined by:
๐ฅ1
โ3๐ฅ
โ 3๐ฅ
๐ฅ1
๐ (๐ฅ2) =
... [Show More] (
๐ฅ3
1
๐ฅ2 โ 5๐ฅ3
3) for all (๐ฅ2) โ โ3.
๐ฅ3
Since it is known that the domain and codomain of T are known vector spaces, to prove T is a linear function it must preserve and hold under vector addition and scalar multiplication. To prove these conditions, we must first form three new vectors:
๐ฅ1
๐ฟ = (๐ฅ2)
๐ฅ3
๐ฆ1
๐ = (๐ฆ2)
๐ฆ3
๐ฅ1 +๐ฆ1
๐ฟ + ๐ = (๐ฅ2 +๐ฆ2).
๐ฅ3 +๐ฆ3
The vector addition condition states:
๐(๐ฃ + ๐ฃ') = ๐(๐ฃ) + ๐(๐ฃ') for all ๐ฃ, ๐ฃ' โ ๐.
This condition can be applied to the given question by calculating ๐(๐ฟ + ๐)
๐ฅ1 +๐ฆ1
โ3(๐ฅ
+ ๐ฆ ) โ 3(๐ฅ
+ ๐ฆ )
๐(๐ฟ + ๐) = ๐ ((๐ฅ2 +๐ฆ2)) = ( 1 1
3 3 )
๐ฅ3 +๐ฆ3
(๐ฅ2 + ๐ฆ2) โ 5(๐ฅ3 + ๐ฆ3)
๐(๐ฟ + ๐) = (โ3(๐ฅ1 + ๐ฅ3) โ 3(๐ฆ1 + ๐ฆ3))
(๐ฅ2 โ 5๐ฅ3) + (๐ฆ2 โ 5๐ฆ3)
Now, we must calculate ๐(๐ฟ) + ๐(๐).
( ) โ3(๐ฅ1 + ๐ฅ3)
๐ ๐ฟ
= ( (๐ฅ
)
โ 5๐ฅ3)
๐(๐) = (โ3(๐ฆ1 + ๐ฆ3))
(๐ฆ2 โ 5๐ฆ3)
๐(๐ฟ) + ๐(๐) = (โ3(๐ฅ1 + ๐ฅ3)) + (โ3(๐ฆ1 + ๐ฆ3))
(๐ฅ2 โ 5๐ฅ3) (๐ฆ2 โ 5๐ฆ)
๐(๐ฟ) + ๐(๐) = (โ3(๐ฅ1 + ๐ฅ3) โ 3(๐ฆ1 + ๐ฆ3))
(๐ฅ2 โ 5๐ฅ3) + (๐ฆ2 โ 5๐ฆ3)
Therefore, we can conclude that:
๐(๐ฟ) + ๐(๐) = (โ3(๐ฅ1 + ๐ฅ3) โ 3(๐ฆ1 + ๐ฆ3))= ๐(๐ฟ + ๐).
(๐ฅ2 โ 5๐ฅ3) + (๐ฆ2 โ 5๐ฆ3)
Thus, it can be confirmed that the function T preserves and is closed under vector addition. Now we must test to see if the function is closed under the condition of scalar multiplication which states:
๐(ฮปv) = ฮป๐ (v) for all ฮป โ F and v โ V.
To prove that the function holds under this condition, we must first take a vector, v โ
โโฏ and a scalar quantity, ฮป โ โ. Using this, we can obtain ๐(ฮปv) and ฮป๐(v).
๐ฅ
v = (๐ฆ).
๐ง
๐ฅ
๐(v)= ๐ ((๐ฆ)). [Show Less]