Linear Algebra The Gram Schmidt Orthonormalization Process-2, guaranteed and verified 100% Pass

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The Gram-Schmidt Orthonormalization Process


Def. Let 𝑉 be an inner product space. A subset of 𝑉 is an orthonormal basis for 𝑉 if it
is an ordered basis that is orthonormal.


Ex. The standard basis in ℝ𝑛 is an orthonormal basis.


3 4 4 3
Ex. The set {< , >, < , − >} is an orthonormal basis for ℝ2 since
5 5 5 5

3 4 3 4 2 2 4 3
4 −3 2 2
‖< , >‖ = √( ) + ( ) = 1, ‖< , − >‖ = √( ) + ( ) = 1
5 5 5 5 5 5 5 5

3 4 4 3 12 12
and (< , >) ∙ (< , − >) = − = 0.
5 5 5 5 25 25




Writing vectors in terms of an orthonormal basis can greatly simplify calculations.
Ex. Suppose {𝑣1 , 𝑣2 , 𝑣3 , 𝑣4 } is an orthonormal basis for an inner product space 𝑉
and 𝑣 = 2𝑣1 − 3𝑣2 + 𝑣3 + 4𝑣4 . Find ‖𝑣‖.


‖𝑣 ‖ = √< 𝑣, 𝑣 >= √< (2𝑣1 − 3𝑣2 + 𝑣3 + 4𝑣4 ), (2𝑣1 − 3𝑣2 + 𝑣3 + 4𝑣4 ) >
But since {𝑣1 , 𝑣2 , 𝑣3 , 𝑣4 } is orthonormal we have that < 𝑣𝑖 , 𝑣𝑗 > = 𝛿𝑖𝑗 .


Thus we have:

‖𝑣 ‖ = √22 + (−3)2 + 12 + 42 = √30.

, 2


Theorem: Let 𝑉 be an inner product space and 𝑆 = {𝑣1 , … , 𝑣𝑘 } be an orthogonal
<𝑣,𝑣𝑖 >
subset of nonzero vectors. If 𝑣 ∈ 𝑠𝑝𝑎𝑛(𝑆) then 𝑣 = ∑𝑘𝑖=1 𝑣 . In particular, if
‖𝑣𝑖 ‖2 𝑖
𝑆 is an orthonormal set then 𝑣 = ∑𝑘𝑖=1 < 𝑣, 𝑣𝑖 > 𝑣𝑖 .
The set {< 𝑣, 𝑣𝑖 >} are called the Fourier coefficients of 𝑣.


Proof: 𝑣 ∈ 𝑠𝑝𝑎𝑛(𝑆) ⟹ 𝑣 = ∑𝑘𝑖=1 𝑎𝑖 𝑣𝑖 , 𝑎1 , … , 𝑎𝑘 ∈ ℝ.


< 𝑣, 𝑣𝑗 > =< ∑𝑘𝑖=1 𝑎𝑖 𝑣𝑖 , 𝑣𝑗 >


=< 𝑎𝑗 𝑣𝑗 , 𝑣𝑗 >


2
= 𝑎𝑗 ‖𝑣𝑗 ‖ .


<𝑣,𝑣𝑗 >
⟹ 2 = 𝑎𝑗 .
‖𝑣𝑗 ‖



Corollary: Let 𝑉 be an inner product space and 𝑆 = {𝑣1 , … , 𝑣𝑘 } be an orthogonal
subset of nonzero vectors. Then 𝑆 is linearly independent.


Proof: Suppose that 𝑎1 𝑣1 + ⋯ + 𝑎𝑘 𝑣𝑘 = 0. Let’s show that 𝑎1 = ⋯ = 𝑎𝑘 = 0.
By the previous theorem
<𝑎1 𝑣1 +⋯+𝑎𝑘 𝑣𝑘 , 𝑣𝑗 >
𝑎𝑗 = 2 = 0; 1 ≤ 𝑗 ≤ 𝑘.
‖𝑣𝑗 ‖

Thus 𝑆 = {𝑣1 , … , 𝑣𝑘 } is linearly independent.