Linear-Algebra-The-Matrix-Representation-of-a-Linear-Transformation-3, guaranteed and verified 100% Pass

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The Matrix Representation of a Linear Transformation


Def. Let 𝑉 be a finite dimensional vector space. An ordered basis for 𝑽 is a basis for 𝑉
with a specific order.


Ex. In ℝ3 let 𝐵 = {𝑒1 , 𝑒2 , 𝑒3 } where 𝑒1 =< 1,0,0 >, 𝑒2 =< 0,1,0 >,
𝑒3 =< 0,0,1 >.
𝐵 is called the standard ordered basis for ℝ3 .
𝐶 = {𝑒2 , 𝑒1 , 𝑒3 } is a different ordered basis for ℝ3 .
Even though 𝐵 and 𝐶 contain the same basis vectors, they appear in different
orders in each set.


As we will see shortly, when we express vectors in terms of a basis, the order of the
basis matters.
Just as 𝑒1 , 𝑒2 , … , 𝑒𝑛 is the standard ordered basis for ℝ𝑛 , {1, 𝑥, 𝑥 2 , … , 𝑥 𝑛 } is the
standard ordered basis for 𝑃𝑛 (ℝ).


Def. Let 𝐵 = {𝑣1 , 𝑣2 , … , 𝑣𝑛 } be an ordered basis for a finite dimensional vector space
𝑉. For 𝑣 ∈ 𝑉 , let 𝑎1 , … , 𝑎𝑛 be the unique real numbers such that
𝑣 = 𝑎1 𝑣1 + ⋯ + 𝑎𝑛 𝑣𝑛 .
we define the coordinate vector of 𝒗 relative to 𝐵 by
𝑎1
[𝑣]𝐵 = [ ⋮ ].
𝑎𝑛

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Ex. 𝐵 = {𝑣1 , 𝑣2 , 𝑣3 } = {𝑒1 , 𝑒2 , 𝑒3 } and 𝐵′ = {𝑤1 , 𝑤2 , 𝑤3 } = {𝑒2 , 𝑒1 , 𝑒3 } are ordered bases
for ℝ3 . The vector 𝑣 =< 5, −3,2 > is given by:
< 5, −3,2 >= 5𝑒1 − 3𝑒2 + 2𝑒3
= 5𝑣1 − 3𝑣2 + 2𝑣3 .

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Thus we have: [𝑣 ]𝐵 = [−3 ].
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On the other hand:
< 5, −3,2 >= 5𝑒1 − 3𝑒2 + 2𝑒3
= −3𝑤1 + 5𝑤2 + 2𝑤3 .

−3
Which gives us: [𝑣 ]𝐵 ′ = [ 5].
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Ex. Let 𝑉 = 𝑃2 (ℝ) and 𝐵 = {𝑣1 , 𝑣2 , 𝑣3 } = {1, 𝑥, 𝑥 2 }, 𝐵′ = {𝑤1 , 𝑤2 , 𝑤3 } = {𝑥 2 , 𝑥, 1}
ordered bases for 𝑉. Then
𝑓 (𝑥 ) = 3 − 4𝑥 + 5𝑥 2 is represented by:
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𝑓 (𝑥 ) = 3 − 4𝑥 + 5𝑥 = 3𝑣1 − 4𝑣2 + 5𝑣3 ⟹ [𝑓 ]𝐵 = [−4 ] .
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𝑓 (𝑥 ) = 3 − 4𝑥 + 5𝑥 2 = 5𝑤1 − 4𝑤2 + 3𝑤3 ⟹ [𝑓 ]𝐵′ = [−4 ] .
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