Gram-Schmidt process linear algbera.pptx

4/11/2023 Presented by Md. Al-Amin 1
Our Topic:
Gram-Schmidt Process
Presented By: Md. Al-Amin
Student Id:181202
Mathematics Discipline,
Khulna University,Khulna
Bangladesh

Welcome To Our Presentation
We are Presentation Group 1.
Student Id:181201,181202,181203,181204,181205,181206,181207,181209,181211
Overview-what you’ll learn?
Orthogonal And
Orthonormal
set and basis
Projection Thoery
Geometrical Interpretation of this theory
Calculating Projection
Real life example of Projection Theory
And This Calculation
Application Of Gram-
Schmidt Process
(QR-Decomposition)
And Calculating this
Legendre Polynomial
And how can calculate

Let , V= 𝑣1, 𝑣2, … , 𝑣𝑟 𝑖𝑠 𝑎 𝑠𝑒𝑡 𝑖𝑛 𝑎 𝑟𝑒𝑎𝑙 𝑖𝑛𝑛𝑒𝑟 𝑝𝑟𝑜𝑑𝑢𝑐𝑟 𝑠𝑝𝑎𝑐𝑒. V will be said to be
if all pairs of distinct vectors in the set are orthogonal.
Orthogonal
OrthoNormal Orthogonal + each vector has norm 1
Suppose U = 𝒖𝟏, 𝒖𝟐 , 𝑢3 𝑤ℎ𝑟𝑒𝑒 𝒖𝟏 = 1,0,0 , 𝒖𝟐 = 0,1,0 , 𝒖𝟑 = (0,0,1)
X
Y
0
𝑢1 = 1,0,0
𝑢
2
=
0,1,0
Z
𝐻𝑒𝑟𝑒, 𝒖𝟏, 𝒖𝟐 = 1 ∙ 0 + 0 ∙ 1 + 0 ∙ 0. = 0,
∴ 𝑇ℎ𝑒 𝑎𝑛𝑔𝑙𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝒖𝟏, 𝒖𝟐 𝑖𝑠 90° 𝑜𝑟 𝑜𝑡ℎ𝑜𝑔𝑜𝑛𝑎𝑙
𝐻𝑒𝑟𝑒, 𝒖𝟐, 𝒖𝟑 = 0 ∙ 0 + 1 ∙ 0 + 0 ∙ 1 = 0,
∴ 𝑇ℎ𝑒 𝑎𝑛𝑔𝑙𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝒖𝟐, 𝒖𝟑 𝑖𝑠 90° 𝑜𝑟 𝑜𝑡ℎ𝑜𝑔𝑜𝑛𝑎𝑙
So ,The set U is called to be orthogonal.
𝐻𝑒𝑟𝑒, 𝒖𝟑, 𝒖𝟏 = 0 ∙ 1 + 0 ∙ 0 + 1 ∙ 0 = 0,
∴ 𝑇ℎ𝑒 𝑎𝑛𝑔𝑙𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝒖𝟑, 𝒖𝟏 𝑖𝑠 90° 𝑜𝑟 𝑜𝑡ℎ𝑜𝑔𝑜𝑛𝑎𝑙
𝑁𝑜𝑤 , , 𝒖𝟏 = 12 + 02 + 02 = 1,
𝒖𝟐 = 02 + 12 + 02 = 1 𝑎𝑛𝑑
𝒖𝟑 = 02 + 02 + 12 = 1 ∴ 𝐮𝟏, 𝐮𝟐, 𝐮𝟑 ℎ𝑎𝑣𝑒 𝑛𝑜𝑟𝑚 𝟏 𝑖𝑛𝑑𝑖𝑣𝑖𝑑𝑢𝑎𝑙𝑙𝑦.
So , U is also orthonormal.

Construction of an orthonormal set
Let,
𝒗𝟏 = 0,1,0 , 𝒗𝟐 = 1,0,1 , 𝒗𝟑 = 1,0, −1 and assume that 𝑅3 has the Euclidean inner product .It follows the set of
vectors 𝑆 = 𝑣1, 𝑣2, 𝑣3 is orthogonal.
since 𝒗𝟏, 𝒗𝟐 = 𝒗𝟏, 𝒗𝟑 = 𝒗𝟐, 𝒗𝟑 = 0
The Euclidean norms of the vectors are 𝒗𝟏 =1, 𝒗𝟐 = 2, 𝒗𝟑 = 2
𝒖𝟏 =
v1
𝒗𝟏
=, 0,1,0 , 𝐮2 =
𝒗𝟐
𝒗𝟐
=
1
2
, 0,
1
2
, 𝐮3 =
𝑣3
𝒗𝟑
=
𝟏
𝟐
, 𝟎, −
𝟏
𝟐
S = 𝐮𝟏, 𝒖𝟐, 𝒖𝟑 is orthonormal as 𝐮1, 𝒖𝟑 = 𝐮𝟏, 𝒖𝟑 = 𝒖𝟐, 𝒖𝟑 = 𝟎 𝒂𝒏𝒅
𝒖𝟏 = 𝒖𝟐 = 𝒖𝟑 = 1
Theorem:::
𝐼𝑓 𝑆 = 𝒗𝟏, 𝒗𝟐, … , 𝒗𝒏 𝑖𝑠 𝑎𝑛 𝑜𝑟𝑡ℎ𝑜𝑔𝑜𝑛𝑎𝑙 𝑠𝑒𝑡 𝑜𝑓 𝑛𝑜𝑛𝑧𝑒𝑟𝑜 𝑣𝑒𝑐𝑡𝑜𝑟𝑠 𝑖𝑛 𝑎𝑛 𝑖𝑛𝑛𝑒𝑟 𝑝𝑟𝑜𝑑𝑢𝑐𝑡 𝑠𝑝𝑎𝑐𝑒, 𝑡ℎ𝑒𝑛 𝑆 𝑖𝑠 𝑙𝑖𝑛𝑒𝑎𝑟𝑙𝑦 𝑖𝑛𝑑𝑒𝑝𝑒𝑛𝑑𝑒𝑛𝑡.

Proof: 𝐿𝑒𝑡 , 𝑘1𝒗𝟏 + 𝑘2𝒗𝟐 + ⋯ + 𝑘𝑛𝒗𝒏 = 0 … (1)
For this proof ,we must show that 𝑘1 = 𝑘2 = ⋯ = 𝑘𝑛 = 0
𝐹𝑜𝑟 𝑒𝑎𝑐ℎ 𝑣𝑖 𝑖𝑛 𝑆 , it follows from (1) that,
𝑘1𝒗𝟏 + 𝑘2𝒗𝟐 + ⋯ + 𝑘𝑛𝒗𝒏 , 𝒗𝒊 = 0, 𝒗𝒊 = 0
⇒ 𝑘1 𝒗𝟏, 𝒗𝒊 + 𝑘2 𝒗𝟐, 𝒗𝒊 + ⋯ + 𝑘𝑛 𝒗𝒏, 𝒗𝒊 = 0
⇒ 𝑘1 𝒗𝟏, 𝒗𝟏 + 𝑘1 𝒗𝟏, 𝒗𝟐 + ⋯ = 0
𝑘𝑖 𝒗𝒊, 𝒗𝒊 + 𝑘𝑖/𝑗 𝒗𝒊, 𝒗𝒋 = 0
But, 𝒗𝒋, 𝒗𝒊 = 0 𝑤ℎ𝑒𝑛 𝑗 ≠ 𝑖 ∵ 𝐿𝑒𝑡, 𝒗𝟏, 𝒗𝟐 ℎ𝑎𝑠 𝑎𝑛𝑔𝑙𝑒 90° 𝑎𝑠 𝑣1, 𝑣2 𝑎𝑟𝑒 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 𝑜𝑓 𝑜𝑟𝑡ℎ𝑜𝑔𝑜𝑛𝑎𝑙 𝑠𝑒𝑡 𝑜𝑓 𝑑𝑖𝑠𝑡𝑖𝑛𝑐𝑡 𝑒𝑙𝑒𝑚𝑒𝑛𝑡
𝑘𝑖 𝒗𝒊, 𝒗𝒊 = 0 𝑏𝑢𝑡 𝒗𝒊, 𝒗𝒊 ≠ 0 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 𝑆 𝑖𝑠 𝑛𝑜𝑛 𝑍𝑒𝑟𝑜 𝑠𝑒𝑡 , 𝑠𝑜 𝑣𝑖 𝑐𝑎𝑛 𝑛𝑜𝑡 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑧𝑒𝑟𝑜.
∴ 𝑘𝑖 = 0 → 𝑘1 + 𝑘2 + ⋯ + 𝑘𝑛 = 0
So , S is linearly independent.
An Orthonormal Basis :
In an inner product space , a basis consisting of orthonormal vectors is called an Orthonormal basis, and a basis consist
of orthogonal vectors is called an orthogonal basis.

An Orthonormal Basis
Check that whether the following vectors form an orthogonal basis or orthonormal basis
𝒖𝟏 = 𝟎, 𝟏, 𝟎 , 𝒖𝟐 =
𝟏
𝟐
, 𝟎,
𝟏
𝟐
, 𝒖𝟑 =
𝟏
𝟐
, 𝟎, −
𝟏
𝟐
For checking orthogonal basis or orthonormal basis we have to follow some rules
If S = {𝑢1, 𝑢2, . . . , 𝑢𝑛} is an orthogonal set of nonzero vectors in an inner product space,then S is linearly
independent. That means for orthogonal basis the vectors will be linearly independent.
As, 𝑢1, 𝑢2 = 𝑢1, 𝑢3 = 𝑢2, 𝑢3 =0 ;we can say that it is orthogonal basis for 𝑅3
More over, An orthogonal set in which each vector has norm 1 is said to be orthonormal.
For this,as, 𝑢1 = 𝑢2 = 𝑢3 =1;Now, we can say that it is orthonormal.
So, vectors form an orthonormal basis
Theorem::
(a) If S = {𝒗𝟏, 𝒗𝟐, … , 𝒗𝒏} is an orthogonal basis for an inner product space V, and if 𝑢 is any vector in V, then
𝒖 =
𝒖,𝒗𝟏
𝒗𝟏
𝟐 ⋅ 𝒗𝟏 +
𝒖,𝒗𝟐
𝒗𝟐
𝟐 ⋅ 𝒗𝟐 + ⋯ +
𝒖,𝒗𝒏
𝒗𝒏
𝟐 ⋅ 𝒗𝒏
(b) If S = {𝒗𝟏, 𝒗𝟐, … , 𝒗𝒏} is an orthonormal basis for an inner product space V, and if 𝑢 is any vector in V, then
𝑢= 𝑢, 𝑣1 𝑣1 + 𝑢, 𝑣2 𝑣𝟐 + ⋯ + 𝑢, 𝑣𝑛 𝑣𝒏

Proof:
Since 𝑆 = 𝒗𝟏, 𝒗𝟐, … , 𝒗𝒏 is a basis for V , every vector u in V can be expressed in the form
𝐮 = 𝑐1𝒗𝟏 + 𝑐2𝒗𝟐 + ⋯ + 𝑐𝑛𝒗𝒏
We will complete the proof by showing that ci =
𝒖,𝒗𝒊
𝒗𝒊
2
for i = 1, 2, . . . , n. To do this, observe first that
𝑢, 𝑣𝑖 = 𝑐1𝒗𝟏 + 𝑐2𝒗𝟐 + ⋯ + 𝑐𝑛(𝒗𝒏, 𝒗𝒊
=c1 𝑣1, 𝑣𝑖 +c2 𝑣2, 𝑣𝑖 + ⋯ + cn 𝑣𝑛, 𝑣𝑖
Since S is an orthogonal set, all of the inner products in the last equality are zero except the i-th, so we have
𝑢, 𝑣𝑖 =ci 𝑣𝑖, 𝑣𝑖 = ci 𝑣𝑖
2
Or, ci =
𝑢,𝑣𝑖
𝑣𝑖
2
𝑢 s=
𝒖,𝒗𝟏
𝒗𝟏
2 ,
𝒖,𝒗𝟐
𝒗𝟐
2 , … ,
𝒖,𝒗𝒏
𝒗𝒏
2
We also know that, An orthogonal set in which each vector has norm 1 is said to be orthonormal.
That means, 𝑣1 = 𝑣2 = 𝑣3 =… = 𝑣𝑛 =1
𝒖 s= 𝒖, 𝒗𝟏 , 𝒖, 𝒗𝟐 , … , 𝒖, 𝒗𝒏

An Orthonormal Basis from an Orthogonal Basis
Problem (A): Show that the vectors
𝒘𝟏= (0,2,0), 𝒘𝟐=(3,0,3), 𝒘𝟑 = −4,0,4
form an orthogonal basis and use that basis to find an orthonormal basis by normalizing each vector.
Solution: The given vectors form an orthogonal set since
𝒘𝟏, 𝒘𝟐 =0, 𝒘𝟏, 𝒘𝟑 =0, 𝒘𝟐, 𝒘𝟑 =0
𝒗𝟏=
𝒘𝟏
𝒘𝟏
= (0,1,0), 𝒗𝟐=
𝒘𝟐
𝒘𝟐
= (
1
2
,0,
1
2
) 𝒗𝟑=
𝒘𝟑
𝒘𝟑
= (-
1
2
,0,
1
2
)

Problem (B): Express the vector u=(1,2,4) as a linear combination of the orthogonal basis vectors obtained in
part (A)
Solution: It follows from formula
𝒖 = 𝒖, 𝒗𝟏 𝒗𝟏 + 𝒖, 𝒗𝟐 𝒗𝟐 + 𝒖, 𝒗𝟑 𝒗𝟑
Now
𝒖, 𝒗𝟏 =(1,2,4).(0,1,0)=2
𝒖, 𝒗𝟐 =(1,2,4).(
1
2
,0,
1
2
)=
5
2
𝒖, 𝒗𝟑 =(1,2,4).(-
1
2
,0,
1
2
)=
3
2
Hence that
(1,2,4)= 2(0,1,0)+
5
2
1
2
, 0,
1
2
+
3
2
−
1
2
, 0,
1
2

W
u
0
𝒘𝟏 = projw𝒖
𝒘𝟐 = projw⊥𝒖
𝑊⊥
𝒗𝟏, 𝒗𝟐, … , 𝒗𝒓 𝑜𝑟𝑡ℎ𝑜𝑔𝑜𝑛𝑎𝑙 𝑏𝑎𝑠𝑖𝑠 𝑓𝑜𝑟 𝑊
If W is a finite-dimensional subspace of an
inner product space V , then every vector u
in V can be expressed in exactly one way as
Projection Theory
u=w1+w2
where w1 is in W and w2 is in
Here , w1= projwu = the orthogonal
projection of u on W and
𝒘𝟐 = 𝑝𝑟𝑜𝑗𝒘⊥𝒖
= 𝑡ℎ𝑒 𝑜𝑟𝑡ℎ𝑜𝑔𝑜𝑛𝑎𝑙 𝑝𝑟𝑜𝑗𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝒖 𝒐𝒏 𝑾⊥
𝒖 = 𝑝𝑟𝑜𝑗𝑤𝒖 + 𝑝𝑟𝑜𝑗𝑤⊥𝒖
⇒ 𝑝𝑟𝑜𝑗𝑤⊥ 𝒖 = 𝒖 − 𝑝𝑟𝑜𝑗𝑤𝒖
∴ 𝒖 = 𝑝𝑟𝑜𝑗𝑤𝒖 + (𝒖 − 𝑝𝑟𝑜𝑗𝑤𝒖)
𝑝𝑟𝑜𝑗𝑤𝒖 + (𝒖 − 𝑝𝑟𝑜𝑗𝑤𝒖)

v
According to projection theory,
y = u +v where
𝐮 = 𝑝𝑟𝑜𝑗𝑢𝐲 𝐯 = 𝑝𝑟𝑜𝑗𝑢⊥𝐲

Calculating
Projections
𝑷𝒓𝒐𝒃𝒍𝒆𝒎: 𝐿𝑒𝑡 , 𝑅3
ℎ𝑎𝑣𝑒 𝑡ℎ𝑒 𝑒𝑢𝑐𝑙𝑖𝑑𝑒𝑎𝑛 𝑖𝑛𝑛𝑒𝑟 𝑝𝑟𝑜𝑑𝑢𝑐𝑡, 𝑎𝑛𝑑 𝑙𝑒𝑡 𝑊 𝑏𝑒 𝑡ℎ𝑒 𝑠𝑢𝑏𝑠𝑝𝑎𝑐𝑒 𝑠𝑝𝑎𝑛𝑛𝑒𝑑 𝑏𝑦 𝑡ℎ𝑒
𝑜𝑟𝑡ℎ𝑜𝑛𝑜𝑟𝑚𝑎𝑙 𝑣𝑒𝑐𝑡𝑜𝑟𝑠 𝒗𝟏 = 0,1,0 𝑎𝑛𝑑 𝒗𝟐 = (−
4
5
,0,
3
5
) .
𝐹𝑖𝑛𝑑 𝑜𝑟𝑡ℎ𝑜𝑔𝑜𝑛𝑎𝑙 𝑝𝑟𝑜𝑗𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝒖 = 1,1,1 𝑜𝑛 𝑊 𝑎𝑛𝑑 𝑐𝑜𝑚𝑝𝑜𝑛𝑡 𝑜𝑓 𝒖 𝑜𝑟𝑡ℎ𝑜𝑔𝑜𝑛𝑎𝑙 𝑡𝑜 𝑊.
Solution:::
We know,
𝐼𝑓 𝑣1, 𝑣2, 𝑣3 𝑖𝑠 𝑎𝑛 𝑜𝑟𝑡ℎ𝑜𝑛𝑜𝑟𝑚𝑎𝑙 𝑏𝑎𝑠𝑖𝑠 𝑓𝑜𝑟 𝑊 𝑎𝑛𝑑 𝑢 𝑖𝑠 𝑎𝑛𝑦 𝑣𝑒𝑐𝑡𝑜𝑟 𝑖𝑛 𝑉 , 𝑡ℎ𝑒𝑛 𝑜𝑟𝑡ℎ𝑜𝑔𝑜𝑛𝑎𝑙 𝑝𝑟𝑜𝑗𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑢 𝑜𝑛 𝑊,
𝑝𝑟𝑜𝑗𝑊 𝑢 = 𝑢, 𝑣1 𝑣1 + 𝑢, 𝑣2 𝑣2 + ⋯ + 𝑢, 𝑣𝑟 𝑣𝑟
𝑇ℎ𝑒 𝑜𝑟𝑡ℎ𝑜𝑔𝑜𝑛𝑎𝑙 𝑝𝑟𝑜𝑗𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑢 = 1,1,1 𝑜𝑛 𝑊 𝑖𝑠 𝑝𝑟𝑜𝑗𝑤𝑢 = 𝑢, 𝑣1 𝑣1 + 𝑢, 𝑣2 𝑣2
= 1 0,1,0 + −
1
5
−
4
5
, 0,
3
5
=
4
25
, 1, −
3
25
𝑇ℎ𝑒 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑜𝑓 𝑢 𝑜𝑟𝑡ℎ𝑜𝑔𝑜𝑛𝑎𝑙 𝑡𝑜 𝑊 𝑖𝑠 𝑝𝑟𝑜𝑗𝑊⊥𝑢 = 𝑢 − 𝑝𝑟𝑜𝑗𝑊𝑢
= 1,1,1 −
4
25
, 1, −
3
25
=
21
25
, 0,
28
25
Now we see that, 𝑝𝑟𝑜𝑗𝑤𝑢 is orthogonal to both 𝒗𝟏 = 0,1,0 𝑎𝑛𝑑 𝒗𝟐 = (−
4
5
,0,
3
5
) .
𝐵𝑒𝑐𝑎𝑢𝑠𝑒 𝑝𝑟𝑜𝑗𝑊⊥𝑢, 𝒗𝟏 =
21
25
⋅ 0 + 0 ∙ 1 +
28
25
∙ 0 = 0 , 𝑝𝑟𝑜𝑗𝑤𝑢, 𝒗𝟐 =
21
25
⋅ −
4
5
+ 0 ∙ 0 +
28
25
∙
3
5
= −
84
125
+
84
125
=0
So , this vector is orthogonal to each vector in the space W spanned by 𝒗𝟏 = 0,1,0 𝑎𝑛𝑑 𝒗𝟐 = (−
4
5
,0,
3
5
) as it should be.

Geometrical Interpretation of Orthogonal Projection
X
Y
0 a
u
𝒘𝟏 = 𝑘𝒂
𝒘𝟐
𝐴𝑐𝑐𝑜𝑟𝑑𝑖𝑛𝑔 𝑡𝑜 𝑝𝑟𝑜𝑗𝑒𝑐𝑡𝑖𝑜𝑛 𝑡ℎ𝑒𝑜𝑟𝑦, 𝒖 = 𝒘𝟏 + 𝒘𝟐
𝒖 = 𝑘𝒂 + 𝒘𝟐 ; 𝑎 = 𝑏𝑎𝑠𝑖𝑠 𝑣𝑒𝑐𝑡𝑜𝑟 𝑓𝑜𝑟 𝑥 𝑎𝑥𝑖𝑠
∴ 𝒖, 𝒂 = 𝑘𝒂 + 𝒘𝟐 ∙ 𝒂 = 𝑘 𝒂 𝟐 + 𝒘𝟐, 𝒂
⇒ 𝒖, 𝒂 = 𝑘 𝒂 𝟐 𝑤2, 𝑎 has angle 90°
∵ cos 90 = 0, 𝑤2, 𝑎 = 0
∴ 𝑘 =
𝒖,𝒂
𝒂 𝟐
𝑇ℎ𝑒 𝑝𝑟𝑜𝑗𝑒𝑐𝑡𝑖𝑜𝑛 𝑤1 = 𝑘𝒂 =
𝒖, 𝒂
𝒂 𝟐
⋅ 𝒂
= 𝑃𝑟𝑜𝑗𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑢 𝑜𝑛 𝑋 − 𝑎𝑥𝑖𝑠
𝑆𝑖𝑚𝑖𝑙𝑎𝑟𝑙𝑦, 𝑖𝑓 𝑏 𝑖𝑠 𝑎 𝑏𝑎𝑠𝑖𝑠 𝑣𝑒𝑐𝑡𝑜𝑟 𝑎𝑙𝑜𝑛𝑔 𝑌 − 𝑎𝑥𝑖𝑠 𝑡ℎ𝑒𝑛 𝑃𝑟𝑜𝑗𝑒𝑐𝑡𝑜𝑟 𝑜𝑓 𝑢 𝑜𝑛 𝑌 𝑎𝑥𝑖𝑠 =
𝒖,𝒃
𝒃 𝟐 ⋅ 𝒃
𝑖𝑓 𝑐 𝑖𝑠 𝑎 𝑏𝑎𝑠𝑖𝑠 𝑣𝑒𝑐𝑡𝑜𝑟 𝑎𝑙𝑜𝑛𝑔 𝑍 − 𝑎𝑥𝑖𝑠 𝑡ℎ𝑒𝑛 𝑃𝑟𝑜𝑗𝑒𝑐𝑡𝑜𝑟 𝑜𝑓 𝑢 𝑜𝑛 𝑍 𝑎𝑥𝑖𝑠 =
𝑢,𝑐
𝑐 2 ⋅ 𝑐
We know,
𝐼𝑓 𝑣1, 𝑣2, 𝑣3 𝑖𝑠 𝑎𝑛 𝑜𝑟𝑡ℎ𝑜𝑔𝑜𝑛𝑎𝑙 𝑏𝑎𝑠𝑖𝑠 𝑓𝑜𝑟 𝑊 𝑎𝑛𝑑 𝑢 𝑖𝑠 𝑎𝑛𝑦 𝑣𝑒𝑐𝑡𝑜𝑟 𝑖𝑛 𝑉 , 𝑡ℎ𝑒𝑛 𝑜𝑟𝑡ℎ𝑜𝑔𝑜𝑛𝑎𝑙 𝑝𝑟𝑜𝑗𝑒𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑢 𝑜𝑛 𝑊,
𝑝𝑟𝑜𝑗𝑊 𝑢 =
𝑢,𝑣1
𝑣1
2 ⋅ 𝑣1 +
𝑢,𝑣2
𝑣2
2 ⋅ 𝑣2 + ⋯ +
𝑢,𝑣𝑟
𝑣𝑟
2 ⋅ 𝑣𝑟
= the sum of orthogonal projection on “axes” determined by the basis vectors for the subspace W.

Every nonzero finite-dimensional inner product space has an orthonormal basis.
Let , W be nonzero finite-dimensional inner product space and 𝒖𝟏, 𝒖𝟐, … , 𝒖𝒓 𝑖𝑠 𝑎𝑛𝑦 𝑏𝑎𝑠𝑖𝑠 𝑓𝑜𝑟 𝑊.
𝑇ℎ𝑒 𝑓𝑜𝑙𝑙𝑜𝑤𝑖𝑛𝑔 𝑠𝑒𝑞𝑢𝑒𝑛𝑐𝑖𝑎𝑙 𝑠𝑡𝑒𝑝𝑠 𝑤𝑖𝑙𝑙 𝑝𝑟𝑜𝑑𝑢𝑐𝑒 𝑎𝑛 𝑜𝑟𝑡ℎ𝑜𝑔𝑜𝑛𝑎𝑙 𝑏𝑎𝑠𝑖𝑠 𝒗𝟏, 𝒗𝟐, … , 𝒗𝒓 for W.
Step 1 : Let ,𝒗𝟏 = 𝒖𝟏
Step 2 : 𝒗𝟐= 𝒖𝟐 − 𝑝𝑟𝑜𝑗𝑊1
𝒖𝟐
∴ 𝒗𝟐 = 𝒖𝟐 −
𝑢2,𝑣1
𝑣1
2 ⋅ 𝑣1
𝒖𝟏 = 𝒗𝟏
𝑢2
𝑝𝑟𝑜𝑗𝑊1
𝒖𝟐
𝒗𝟐 = 𝒖𝟐 − 𝑝𝑟𝑜𝑗𝑊1
𝑢2
𝒗𝟐 = 𝒖𝟐 − 𝑝𝑟𝑜𝑗𝑊1
𝑢2
𝒗𝟐
𝑢3
Step 3: 𝒗𝟑= 𝒖𝟑 − 𝑝𝑟𝑜𝑗𝑤2
𝒖3
∴ 𝒗𝟑 = 𝒖𝟑 −
𝑢3,𝑣1
𝑣1
2 ⋅ 𝑣1−
𝑢3,𝑣2
𝑣2
2 ⋅ 𝑣2
𝑊1 𝑠𝑝𝑎𝑛𝑛𝑒𝑑 𝑏𝑦 𝒗𝟏
𝑊2 𝑠𝑝𝑎𝑛𝑛𝑒𝑑 𝑏𝑦 𝒗𝟏 , 𝒗𝟐
Similarly, 𝒗𝒓 = 𝒖𝒓 −
𝑢𝑟,𝑣1
𝑣1
2 ⋅ 𝑣1−
𝑢𝑟,𝑣2
𝑣2
2 ⋅ 𝑣2 − ⋯ −
𝑢𝑟,𝑣𝑟−1
𝑣𝑟−1
2 ⋅ 𝑣𝑟−1
In this Way we can calculate ,Gram-Schmidt Process.
𝒗𝟑

Calculating
Problem : Assume that the vector space 𝑅3 has the Euclidean inner product . Apply the
Gram-Schmidt Process to transfer the basis vectors
𝒖𝟏= 1,1,1 , 𝒖𝟐 = 0,1,1 , 𝒖𝟑 = 0,0,1 into an orthogonal basis 𝒗𝟏, 𝒗𝟐, 𝒗𝟑
and normalize that vectors into 𝑞1, 𝑞2, 𝑞3
Step 1: Let , 𝒗𝟏 = 𝒖𝟏 = 1,1,1
Step 2: 𝒗𝟐 = 𝒖𝟐 − 𝑝𝑟𝑜𝑗𝑊1
𝒖𝟐 = 𝒖𝟐 −
𝒖𝟐,𝒗𝟏
𝒗𝟏
𝟐 ⋅ 𝒗𝟏 = 0,1,1 −
2
3
1,1,1 = −
2
3
,
1
3
,
1
3
Step 3 : 𝒗𝟑= 𝒖𝟑 − 𝑝𝑟𝑜𝑗𝑤2
𝒖𝟑= 𝒖𝟑 −
𝒖𝟑,𝒗𝟏
𝒗𝟏
𝟐 ⋅ 𝒗𝟏 −
𝒖𝟑,𝒗𝟐
𝒗𝟐
𝟐 ⋅ 𝒗𝟐
= 0,0,1 −
1
3
(1,1,1)−
0+0+1 3
4
9
+
1
9
+
1
9
−
2
3
,
1
3
,
1
3
= 0,0,1 −
1
3
(1,1,1) −
1
3
2
3
−
2
3
,
1
3
,
1
3
= (0, −
1
2
,
1
2
)
Thus , 𝒗𝟏 = 𝟏, 𝟏, 𝟏 𝒗𝟐 = −
𝟐
𝟑
,
𝟏
𝟑
,
𝟏
𝟑
𝒗𝟑= (𝟎, −
𝟏
𝟐
,
𝟏
𝟐
)
𝑨𝒏 𝒐𝒓𝒕𝒉𝒐𝒏𝒐𝒓𝒎𝒂𝒍 𝒃𝒂𝒔𝒊𝒔 𝒇𝒐𝒓 𝑹𝟑
𝒊𝒔
𝒒𝟏 =
𝒗𝟏
𝒗𝟏
=
1
3
,
1
3
,
1
3
, 𝒒𝟐 =
𝒗𝟐
𝒗𝟐
=
−2
6
,
1
6
,
1
6
, 𝒒𝟑 =
𝒗𝟑
𝒗𝟑
= 0, −
1
2
,
1
2

Legendre Polynomial
𝑝𝑛 𝑥 =
1
2𝑛𝑛!
𝑑𝑛
𝑑𝑥𝑛 (𝑥2 − 1)𝑛
Apply of Gram-Schmidt with Legendre polynomials
Problem:
Let the vector space 𝑝2 have the inner product 𝑝, 𝑞 = −1
1
𝑝 𝑥 𝑞 𝑥 𝑑𝑥
Apply the Gram-Schmidt process to transform the standard basis 1, 𝑥, 𝑥2
for 𝑝2 into an orthogonal basis
∅1 𝑥 , ∅2 𝑥 , ∅3(𝑥) .
Solution:
Take 𝒖𝟏=1,𝒖𝟐= x and 𝒖𝟑=𝑥2
Step 1:𝒗𝟏=𝒖𝟏=1
Step 2:We have
𝒖𝟐, 𝒗𝟏 = −1
1
𝑥𝑑𝑥 = 0 So 𝒗𝟐 = 𝒖𝟐−
𝒖𝟐,𝒗𝟏 𝒗𝟏
𝒗𝟏
𝟐 = 𝒖𝟐=x

Step 3: We have
𝒖𝟑, 𝒗𝟏 = −1
1
𝑥2
𝑑𝑥 =
𝑥3
3 −1
1
=
2
3
𝒖𝟑, 𝒗𝟐 = −1
1
𝑥3
𝑑𝑥 =
𝑥4
4 −1
1
= 0
𝒗𝟏
𝟐= 𝒗𝟏, 𝒗𝟏 = −1
1
1𝑑𝑥 = 𝑥 −1
1
= 2
So, 𝒗𝟑=𝒖𝟑-
𝒖𝟑,𝒗𝟏
𝒗𝟏
𝟐 𝒗𝟏-
𝒖𝟑 ,𝒗𝟐
𝒗𝟐
𝟐 𝒗𝟐=𝑥2-
1
3
Thus, we have obtained the orthogonal basis ∅1 𝑥 , ∅2 𝑥 , ∅3(𝑥) in which
∅1(x)=1, ∅2(x)=x, ∅3(x)=𝑥2
-
1
3

QR-
decompositon
/QR-
factorization
Matrix version of Gram-
Schimdt orthogonalization
process
Developed in 1950s by
John G.F Franchis and
Vera N. Kublanovskaya
Here,A=m×n matrix
where,A=QR.
Q=m × n matrix,R=n × n
upper triangular invertible
matrix
A has linearly independent
columns,Q forms an
orthonormal basis for A.

𝐓𝐡𝐞𝐨𝐫𝐞𝐦:
𝐈𝐟 𝐀 𝐢𝐬 𝐚𝐧 𝐦 × 𝐧 𝐦𝐚𝐭𝐫𝐢𝐱 𝐰𝐢𝐭𝐡 𝐥𝐢𝐧𝐞𝐚𝐫𝐥𝐲 𝐢𝐧𝐝𝐞𝐩𝐞𝐧𝐝𝐞𝐧𝐭𝐜𝐨𝐥𝐮𝐦𝐧 𝐯𝐞𝐜𝐭𝐨𝐫, 𝐭𝐡𝐞𝐧 𝐀 𝐜𝐚𝐧 𝐛𝐞
𝐟𝐚𝐜𝐭𝐨𝐫𝐞𝐝 𝐚𝐬 𝐀 = 𝐐𝐑 ,
𝐰𝐡𝐞𝐫𝐞 𝐐 = 𝐦 × 𝐧 𝐦𝐚𝐭𝐢𝐱 𝐰𝐢𝐭𝐡 𝐨𝐫𝐭𝐡𝐨𝐧𝐨𝐫𝐦𝐚𝐥 𝐜𝐨𝐥𝐮𝐦𝐧 𝐯𝐞𝐜𝐭𝐨𝐫
𝐑 = 𝐧 × 𝐧 𝐢𝐧𝐯𝐞𝐫𝐭𝐢𝐛𝐥𝐞 𝐮𝐩𝐩𝐞𝐫 𝐭𝐫𝐢𝐚𝐧𝐠𝐮𝐥𝐚𝐫 𝐦𝐚𝐭𝐫𝐢𝐱
Solution: To solve this problem, suppose that the column vectors of A are 𝑢1,𝑢2,…,𝑢𝑛 and that Q has orthonormal
column vectors 𝑞1, 𝑞2, … , 𝑞𝑛.Thus A and Q can be written in partitioned form as
A=[𝑢1|𝑢2|….|𝑢𝑛] and Q=[𝑞1|,𝑞2|……𝑞𝑛]
𝑢1,𝑢2,…, 𝑢𝑛 are expressible in terms of 𝑞1, 𝑞2,…𝑞𝑛 as
𝑢1=<𝑢1,𝑞1>𝑞1+<𝑢1, 𝑞2>𝑞2+….<𝑢1, 𝑞𝑛 >𝑞𝑛
𝑢2=<𝑢2,𝑞1>𝑞1+<𝑢2, 𝑞2>𝑞2+….<𝑢2, 𝑞𝑛 > 𝑞𝑛
. . . .
. . . .
. . . .
𝑢𝑛=<𝑢𝑛,𝑞1>𝑞1+ <𝑢𝑛,𝑞2>𝑞2+….<𝑢𝑛,𝑞𝑛>𝑞𝑛

This relationship can be expressed in matrix form using Gram-Schimdt Process as:
[𝑢1|𝑢2|…|𝑢𝑛] =[𝑞1|𝑞2|…|𝑞𝑛]
< 𝑢1, 𝑞1 > < 𝑢2, 𝑞1 > …
< 𝑢1, 𝑞2 >
.
.
< 𝑢2, 𝑞2 >
.
.
…
.
.
< 𝑢1, 𝑞𝑛 > < 𝑢2, 𝑞𝑛 > …
< 𝑢𝑛, 𝑞1 >
< 𝑢𝑛, 𝑞2 >
.
.
< 𝑢𝑛, 𝑞𝑛 >
Or more briefly as A=QR …(1)
Here, R is the upper triangular invertible matrix .Thus,all entries below the main diagonal of R
are zero, and R has the form
R=
< 𝑢1, 𝑞1 > < 𝑢2, 𝑞1 > …
0
.
.
< 𝑢2, 𝑞2 >
.
.
…
.
.
0 0 …
< 𝑢𝑛, 𝑞1 >
< 𝑢𝑛, 𝑞2 >
.
.
< 𝑢𝑛, 𝑞𝑛 >
This equation (1) is a factorization of A into the product of A matrix Q with orthonormal
column vectors and an invertible upper triangular matrix R. We call equation (1) is a QR-
decomposition of A.

Find a QR-decomposition of A=
𝟏 𝟎 𝟎
𝟏 𝟏 𝟎
𝟏 𝟏 𝟏
Solution: The column vectors of A are
𝑢1 =
1
1
1
, 𝑢2 =
0
1
1
𝑎𝑛𝑑 𝑢3=
0
0
1
Applying the Gram-Schimdt process with normalization to these column vectors yields the orthonormal
vectors( 𝐭𝐡𝐚𝐭 𝐩𝐫𝐨𝐜𝐞𝐬𝐬 𝐢𝐬 𝐬𝐡𝐨𝐰𝐧 𝐩𝐫𝐞𝐯𝐢𝐨𝐮𝐬)
𝑞1 =
1
3
1
3
1
3
, 𝑞2 =
−2
6
1
6
1
6
𝑎𝑛𝑑 𝑞3= −
0
1
2
1
2
We know, R=
< 𝑢1, 𝑞1 > < 𝑢2, 𝑞1 > < 𝑢3, 𝑞1 >
0 < 𝑢2, 𝑞2 > < 𝑢3, 𝑞2 >
0 0 < 𝑢3, 𝑞3 >
=
3
3
2
3
1
3
0
2
6
1
6
0 0
1
2
It follows that a QR-decomposition of A is
1 0 0
1 1 0
1 1 1
=
1
3
−2
6
0
1
3
1
6
−1
2
1
3
1
6
1
2
3
3
2
3
1
3
0
2
6
1
6
0 0
1
2
A = Q R

That’s it.
Thank You all for seeing our presentation.

Gram-Schmidt process linear algbera.pptx

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