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NPTEL – Physics – Mathematical Physics - 1
Lecture 12
Gram-Schmidt Orthogonalization (GSO) [continued]
Example 1.
Two vectors u and v where 𝑒, 𝑣 ∈ 𝑉 are said to be orthogonal if < 𝑒|𝑣 > = 0. Similarly a
real matrix P is said to be orthogonal if P is invertible and if 𝑃𝑇 = π‘ƒβˆ’1 or
𝑃𝑃𝑇 = 𝑃𝑇𝑃 = 1. Besides the rows and columns of P form an orthonormal set of vectors.
Also det |𝑃| = Β±1.
Suppose 𝑀 β‰  0, let v be any vector in V, it needs to be shown that,
𝑐 = <𝑣|𝑀>
= <𝑣|𝑀>
<𝑀|𝑀> ‖𝑀‖2
is the unique scalar such that
𝑉 = 𝑣 βˆ’ 𝑐𝑀 is orthogonal to w.
The proof can proceed as follows. In order for 𝑉 to be orthogonal to w,
we must have
< 𝑣 βˆ’ 𝑐𝑀|𝑀 > = 0 or, < 𝑣|𝑀 > βˆ’π‘ < 𝑀|𝑀 β‰₯ 0.
or, < 𝑣|𝑀 > = 𝑐 < 𝑀|𝑀 >
or, 𝑐 = <𝑣|𝑀>
<𝑀|𝑀>
Then, < (𝑣 βˆ’ 𝑐𝑀)|𝑀 > = < 𝑣|𝑀 > βˆ’ 𝑐 < 𝑀|𝑀 >
= < 𝑣 |𝑀 > βˆ’ <𝑣|𝑀>
< 𝑀|𝑀 > = 0
<𝑀|𝑀>
The above scalar c is called the fourier coefficient of v with respect to w or the
component of v along w. cw is the projection of v along w as shown below
Example 2.
Find the coefficient c and the projection cw of
𝑣 = (1, βˆ’1, 2)
𝑀 = (0, 1, 1) in 𝑅3.
Solution: < 𝑣|𝑀 > = 1 and ‖𝑀‖2 = 2
along
Joint initiative of IITs and IISc – Funded by MHRD Page 17 of 28
NPTEL – Physics – Mathematical Physics - 1
𝑐 =
< 𝑣|𝑀 > 1
=
2
‖𝑀‖2
𝑐𝑀 = (0, 1
, 1
) is the projection of v along w.
2 2
Example 3.
Consider the following basis of Euclidean
space
𝑅3: {𝑣1 = (1, 1, 1), 𝑣2 = (0, 1, 1), 𝑣3 = (0, 0, 1)}. Use the Gram-Schmidt procedure to
transform {𝑣𝑖} into an orthonormal basis {𝑒𝑖} in 𝑅3.
Solution: To begin with, set 𝑀1 = 𝑣1 = (1, 1, 1)
Then find 𝑣2 βˆ’
<𝑣2|𝑀1>
‖𝑀 β€–2
1
𝑀1 = (0, 1, 1) βˆ’ 3
(1, 1, 1)
2
= (βˆ’ 2
, 1
, 1
)
3 3 3
Next clear fractions to obtain
𝑀2 = (βˆ’2, 1, 1)
Then find
𝑣3 βˆ’
<𝑣3|𝑀1> <𝑣3|𝑀2>
‖𝑀 β€–2
1 2
βˆ’ ‖𝑀 β€–2
= (0, 0, 1) βˆ’ 1
(1, 1, 1) βˆ’ 1
(βˆ’2, 1,
1)
3 6
= (0, βˆ’ 1
, 1
)
2 2
Again clear fractions to get,
𝑀3 = (0, βˆ’1, 1)
Normally {𝑀1, 𝑀2, 𝑀3} to obtain the following required orthonormal basis of 𝑅3.
{𝑒1 = (1, 1, 1),
1 1
√3
𝑒2 = (βˆ’2, 1, 1), 𝑒3 = (0, βˆ’1, 1)}
Joint initiative of IITs and IISc – Funded by MHRD Page 18 of 28
√6
1
√2

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lec12.ppt

  • 1. NPTEL – Physics – Mathematical Physics - 1 Lecture 12 Gram-Schmidt Orthogonalization (GSO) [continued] Example 1. Two vectors u and v where 𝑒, 𝑣 ∈ 𝑉 are said to be orthogonal if < 𝑒|𝑣 > = 0. Similarly a real matrix P is said to be orthogonal if P is invertible and if 𝑃𝑇 = π‘ƒβˆ’1 or 𝑃𝑃𝑇 = 𝑃𝑇𝑃 = 1. Besides the rows and columns of P form an orthonormal set of vectors. Also det |𝑃| = Β±1. Suppose 𝑀 β‰  0, let v be any vector in V, it needs to be shown that, 𝑐 = <𝑣|𝑀> = <𝑣|𝑀> <𝑀|𝑀> ‖𝑀‖2 is the unique scalar such that 𝑉 = 𝑣 βˆ’ 𝑐𝑀 is orthogonal to w. The proof can proceed as follows. In order for 𝑉 to be orthogonal to w, we must have < 𝑣 βˆ’ 𝑐𝑀|𝑀 > = 0 or, < 𝑣|𝑀 > βˆ’π‘ < 𝑀|𝑀 β‰₯ 0. or, < 𝑣|𝑀 > = 𝑐 < 𝑀|𝑀 > or, 𝑐 = <𝑣|𝑀> <𝑀|𝑀> Then, < (𝑣 βˆ’ 𝑐𝑀)|𝑀 > = < 𝑣|𝑀 > βˆ’ 𝑐 < 𝑀|𝑀 > = < 𝑣 |𝑀 > βˆ’ <𝑣|𝑀> < 𝑀|𝑀 > = 0 <𝑀|𝑀> The above scalar c is called the fourier coefficient of v with respect to w or the component of v along w. cw is the projection of v along w as shown below Example 2. Find the coefficient c and the projection cw of 𝑣 = (1, βˆ’1, 2) 𝑀 = (0, 1, 1) in 𝑅3. Solution: < 𝑣|𝑀 > = 1 and ‖𝑀‖2 = 2 along Joint initiative of IITs and IISc – Funded by MHRD Page 17 of 28
  • 2. NPTEL – Physics – Mathematical Physics - 1 𝑐 = < 𝑣|𝑀 > 1 = 2 ‖𝑀‖2 𝑐𝑀 = (0, 1 , 1 ) is the projection of v along w. 2 2 Example 3. Consider the following basis of Euclidean space 𝑅3: {𝑣1 = (1, 1, 1), 𝑣2 = (0, 1, 1), 𝑣3 = (0, 0, 1)}. Use the Gram-Schmidt procedure to transform {𝑣𝑖} into an orthonormal basis {𝑒𝑖} in 𝑅3. Solution: To begin with, set 𝑀1 = 𝑣1 = (1, 1, 1) Then find 𝑣2 βˆ’ <𝑣2|𝑀1> ‖𝑀 β€–2 1 𝑀1 = (0, 1, 1) βˆ’ 3 (1, 1, 1) 2 = (βˆ’ 2 , 1 , 1 ) 3 3 3 Next clear fractions to obtain 𝑀2 = (βˆ’2, 1, 1) Then find 𝑣3 βˆ’ <𝑣3|𝑀1> <𝑣3|𝑀2> ‖𝑀 β€–2 1 2 βˆ’ ‖𝑀 β€–2 = (0, 0, 1) βˆ’ 1 (1, 1, 1) βˆ’ 1 (βˆ’2, 1, 1) 3 6 = (0, βˆ’ 1 , 1 ) 2 2 Again clear fractions to get, 𝑀3 = (0, βˆ’1, 1) Normally {𝑀1, 𝑀2, 𝑀3} to obtain the following required orthonormal basis of 𝑅3. {𝑒1 = (1, 1, 1), 1 1 √3 𝑒2 = (βˆ’2, 1, 1), 𝑒3 = (0, βˆ’1, 1)} Joint initiative of IITs and IISc – Funded by MHRD Page 18 of 28 √6 1 √2