Linear-Algebra Composition of Linear Transformations, guaranteed and verified 100% pass

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Composition of Linear Transformations


Theorem: Let 𝑉, 𝑊, and 𝑍 be vector spaces and let 𝑇: 𝑉 → 𝑊 and 𝑈: 𝑊 → 𝑍 be
linear then 𝑈 ∘ 𝑇 = 𝑈𝑇: 𝑉 → 𝑍 is linear.


Proof: Let 𝑢, 𝑣 ∈ 𝑉 and 𝑐 ∈ ℝ then
𝑈𝑇(𝑐𝑢 + 𝑣) = 𝑈(𝑇 (𝑐𝑢 + 𝑣))
= 𝑈(𝑐𝑇(𝑢) + 𝑇(𝑣))
= 𝑈(𝑐𝑇(𝑢)) + 𝑈(𝑇(𝑣))
= 𝑐𝑈(𝑇(𝑢)) + 𝑈(𝑇(𝑣))
= 𝑐𝑈𝑇(𝑢) + 𝑈𝑇(𝑣).
Thus 𝑈𝑇 is linear.


Theorem: Let 𝑉 be a vector space. Let 𝑇, 𝑈1 , 𝑈2 ∈ ℒ(𝑉) then
a. 𝑇(𝑈1 + 𝑈2 ) = 𝑇𝑈1 + 𝑇𝑈2 and (𝑈1 + 𝑈2 )𝑇 = 𝑈1 𝑇 + 𝑈2 𝑇
b. 𝑇(𝑈1 𝑈2 ) = (𝑇𝑈1 )𝑈2
c. 𝑇𝐼 = 𝐼𝑇 = 𝑇.
d. 𝑐 (𝑈1 𝑈2 ) = (𝑐𝑈1 )𝑈2 = 𝑈1 (𝑐𝑈2 ) for any 𝑐 ∈ ℝ.


Proof of a.: Let 𝑣 ∈ 𝑉 then
(𝑇(𝑈1 + 𝑈2 ))(𝑣) = 𝑇(𝑈1 (𝑣 ) + 𝑈2 (𝑣))
= 𝑇𝑈1 (𝑣 ) + 𝑇𝑈2 (𝑣)
= (𝑇𝑈1 + 𝑇𝑈2 )(𝑣).
Similarly for (𝑈1 + 𝑈2 )𝑇 = 𝑈1 𝑇 + 𝑈2 𝑇.

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Let 𝑇: 𝑉 → 𝑊 and 𝑈: 𝑊 → 𝑍 be linear transformations of vector spaces. The key
fact is that if we have matrix representations of 𝑇 and 𝑈 as 𝐴 and 𝐶 respectively,
then the matrix representation of the composition 𝑈𝑇 is 𝐶𝐴. That is the matrix
product of 𝐶 and 𝐴.
We can see this by letting 𝐵1 = {𝑣1 , … , 𝑣𝑛 }, 𝐵2 = {𝑤1 , … 𝑤𝑚 }, and
𝐵3 = {𝑧1 , … , 𝑧𝑝 } be ordered bases for 𝑉, 𝑊, and 𝑍 respectively. Thus for any
basis vector 𝑣𝑗 ∈ 𝑉 we have:

𝑈𝑇(𝑣𝑗 ) = 𝑈 (𝑇(𝑣𝑗 )) = 𝑈(∑𝑚
𝑘=1 𝐴𝑘𝑗 𝑤𝑘 )

= ∑𝑚
𝑘=1 𝐴𝑘𝑗 𝑈(𝑤𝑘 )
𝑝
= ∑𝑚
𝑘=1 𝐴𝑘𝑗 (∑𝑖=1 𝐶𝑖𝑘 𝑧𝑖 )
𝑝
= ∑𝑖=1(∑𝑚
𝑘=1 𝐶𝑖𝑘 𝐴𝑘𝑗 )𝑧𝑖
𝑝
= ∑𝑖=1 𝐷𝑖𝑗 𝑧𝑖
where 𝐷𝑖𝑗 = ∑𝑚
𝑘=1 𝐶𝑖𝑘 𝐴𝑘𝑗 .



Def. Let 𝐶 be an 𝑚 × 𝑛 matrix and 𝐴 an 𝑛 × 𝑝 matrix. We define the product of
𝑪 and 𝑨 to be
(𝐶𝐴)𝑖𝑗 = ∑𝑚
𝑘=1 𝐶𝑖𝑘 𝐴𝑘𝑗 for 1 ≤ 𝑖 ≤ 𝑚, 1 ≤ 𝑗 ≤ 𝑝.


Thus if the matrix representation of 𝑇: 𝑉 → 𝑊 is 𝐴 and the matrix representation
of 𝑈: 𝑊 → 𝑍 is 𝐶 then the matrix representation of 𝑈𝑇: 𝑉 → 𝑍 is 𝐶𝐴.


Ex. Let 𝑇: ℝ2 → ℝ3 and 𝑈: ℝ3 → ℝ2 be linear transformations defined by
𝑇 (< 𝑎1 , 𝑎2 >) =< 2𝑎1 − 𝑎2 , 𝑎2 , 𝑎2 − 3𝑎1 >
𝑈(< 𝑏1 , 𝑏2 , 𝑏3 >) =< 𝑏1 + 𝑏2 + 2𝑏3 , 𝑏2 − 3𝑏3 >
with respect to the standard ordered bases 𝐵1 for ℝ2 and 𝐵2 for ℝ3 . Find a
matrix representation of 𝑈𝑇 with respect to these bases.