r r Prof. Inder K. Rana Room 112 B Department of Mathematics - - PowerPoint PPT Presentation
r r Prof. Inder K. Rana Room 112 B Department of Mathematics IIT-Bombay, Mumbai-400076 (India) Email: ikr@math.iitb.ac.in Lecture 10 Prof. Inder K. Rana Department of Mathematics, IIT - Bombay Recall
Room 112 B Department of Mathematics IIT-Bombay, Mumbai-400076 (India) Email: ikr@math.iitb.ac.in Lecture 10
Department of Mathematics, IIT - Bombay
Definition (Orthonormal bases) Let B = {v1, v2, ..., vk} be a basis of a vector space V ⊆ I
is an orthonormal basis of V, if B is an orthonormal set. Theorem Let S = {v1, v2, ..., vk} be a set of non-zero vectors in I
vectors are mutually orthogonal, then S is a linearly independent set. Theorem Let B = {u1, u2, ..., uk} be an orthonormal basis of a vector space V and x ∈ V. Then x = x, u1u1 + x, u2u2 + · · · + x, ukuk.
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How to construct orthonormal basis Suppose B = {v1, v2, ..., vk} is a a spanning set/basis of V. Step 1:Drop zero vectors, if any. Step 2: Define w1 := v1. w2 := v2 − v2, w1 w12 w1. Step 3:· · · having constructed {w1, ..., wj−1}, (dropping zero vectors if any) define wj = vj −
j−1
vj, wℓ wℓ2 wℓ Step 4:Continue till all the vectors v1, v2, ..., vk have been used. Step 5:Normalize the vectors w1, w2, ..., wj obtained to get an
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Theorem Let {u1, u2, ..., uk} is an orthonormal set in V and v ∈ V Then
k
v, uℓ2 ≤ v, v . Equality holds if and only if {u1, u2, ..., uk} be an orthonormal basis (called Parseval’s Identity).
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Definition Let V be an inner product space. A linear transformation T : V → V is called an isometry on V if Tv, Tw = v, w for all v, w ∈ V. That is, T is a linear map which preserves the angles between the vectors. Theorem Let V be an inner product space and T : V → V be a linear map. Then the following statements are equivalent: (i) T is an isometry. (ii) T preserves length, i.e., Tv = v for all v ∈ V.
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Proof: Assume that T is an isometry. Then, for all u ∈ V, Tu2 = Tu, Tu = u, u = u. Conversely, suppose that Tv = v ∀ v ∈ V. Then for all u, w ∈ V, T(u + w), T(u + w) = Tu + Tw, Tu + Tw = Tu, Tu + Tw, Tu + Tu, Tw + Tw, Tw = u, u + 2Tu, Tw + w, w . . . (1). Also, since T is an isometry, T(u + w), T(u + w) = u + w, u + w = u, u + 2u, w + w, w. . . . (2 From (1) and (2) it follows that Tu, Tw = u, w ∀ u, w ∈ V.
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Example 1 Consider the map T : I R → I R, T(x) = αx, for x ∈ I R, where α ∈ I R is fixed. Then T is a linear map and T(x) = x, i.e., |α||x| = |x| if and only if |α| = 1. Thus, T is an isometry if and only if |α| = 1. Example 2 Let T : I R2 → I R2 be defined by T x y
−x sin θ + y cos θ
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Then, Tx2 = (x cos θ + y sin θ)2 + (−x sin θ + y cos θ)2 = x2(cos2 θ + sin2 θ) + y2(sin2 θ + cos2 θ) = x2 + y2 = x2. Hence, T is an isometry on I R2.
Department of Mathematics, IIT - Bombay
Theorem Let T : V → V be a linear map and B be an ordered orthonormal basis
(i) T is an isometry if and only if the column vectors of [T]B form an
(ii) T is an isometry if and only if row vectors of [T]B form an
Note: If C1, A = [C2, ..., Cn] in the column form, then AtA = [C1, C2, ..., Cn]t [C1, C2, ..., Cn] = [Ci, Cj]. Definition A real matrix A = [aij]n×n is said to be orthogonal if the column vectors
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Theorem Let A be a n × n real matrix. Then the following are equivalent: (i) A is orthogonal. (ii) AtA = In. (iii) AAt = In. Note: Thus, orthonormal matrices arise in matrix representation of isometries with respect to orthonormal basis.
Department of Mathematics, IIT - Bombay
Consider the matrix: A = −5 −7 2 4
Then A
−1
−5 −7 2 4 1 −1
−2
−1
Thus the matrix A scales the vector
−1
Question: Given a matrix A how to find all the vectors that are scaled by it? How to find all the scaling factors for a given matrix? Why it is important to answer these questions?
Department of Mathematics, IIT - Bombay
For the matrix A as above we saw A
−1
−1
That is (A − 2I)
−1
Let V = {X ∈ I R2 | (A − 2I)X = 0}. Then V is a vector subspace of I R2
R2. Let us find its dimension: For that we have to find its nullity: A − 2I = −5 − 2 −7 2 4 − 2
Clearly det(A − 2I) = 0, and hence A − 2I is invertible.
Department of Mathematics, IIT - Bombay
Hence V = {X ∈ I R2 | (A − 2I)X = 0} has dimension 1 and is a basis
Question: Does there exist some other scalar λ and another nonzero vector X ∈ I R2 such that (A − λI)X = 0? Note that this implies that the matrix (A − λI) is not invertible! Hence det(A − λI) = 0. That is A − λI = −5 − λ −7 2 4 − λ
{
−1
Hence λ2 + λ − 6 = 0, ⇒ λ = 2, −3.
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Let us find a vector X 0 ∈ I R2 such that (A + 3I)X = 0. Since A + 3I = −5 + 3 −7 2 4 + 3
−2 −7 2 7 −2 −7
W := {X ∈ I R2 | (A + 3I)X = 0} = [ −7 2
The relations: A
−1
−1
A −7 2
−7 2
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Let C1 =
−1
−7 2
A[C1 C2] = −5 −7 2 4 1 −7 −1 2
−7 −1 2 2 −3
2 −3
Department of Mathematics, IIT - Bombay
Noting that the matrix P :=
−7 −1 2
P−1 A P = 2 −3
Question: Given a matrix A when does there exist an invertible matrix P such that P−1AP will be a diagonal matrix, and how to find P?
Department of Mathematics, IIT - Bombay
This motivates our next definition Definition
1
The scalar λ is called an eigenvalue of matrix A if there exists a nonzero vector ]box such that AX = λX.
2
The vector X in the above is called an eigenvector of A corresponding to the eigenvalue λ. Eigenvalue Problem (EVP): Given a matrix A, find its eigenvalues and eigenvectors corresponding to them
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Rewrite the EVP as (A − λI)v = 0, v = 0. By the theory of homogeneous linear equations, a solution exists ⇐ ⇒ the rank ρ(A − λI) < n which happens ⇐ ⇒ the determinant |(A − λI)| = 0. Definition (Characteristic polynomial) The determinant DA(λ) = D(λ) =
a12 · · · a1n a21 a22 − λ · · · a2n . . . . . . . . . an1 an2 · · · ann − λ
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Definition (Characteristic roots/eigenvalues) The possibly complex roots of the characteristic polynomial DA(λ) are called the characteristic roots or the eigenvalues of A. There are n such roots when counted with multiplicity. Example 1: Find the positive eigenvalues and corresponding eigenvectors for A = 9 5 0.4 0.4 . Solution: pause −D(λ) = λ3 − 18 5 λ − 4 5 = ⇒ λ = 2 is a root. The other roots are from − D(λ) (λ − 2) = λ2 + 2λ + 2 5 = 0 which are negative and hence discarded.
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For λ = 2 we solve the homogeneous system (A − 2I) x y z = −2 9 5 0.4 −2 0.4 x y z = . By row operations we get
9 5 0.4 −2 x y z = = ⇒ x y z = 25z 5z z . Hence λ = 2 is the only positive eigenvalue and a corresponding eigenvector is 25 5 1 .
Department of Mathematics, IIT - Bombay
Theorem (Characteristic roots/eigenvalues) The characteristic polynomial and therefore the eigenvalues of the transposed matrix AT are identical to that of A. Proof: DAT (λ) =
T| =
The corresponding eigenvectors may NOT be same.
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Example 2: Consider the matrix B = 0.8 0.1 0.1 0.1 0.7 0.2 0.1 0.9 .Find a unit row-vector u ∈ I R3 s.t. uB = u. Solution: The problem is equivalent to the eigenvalue problem BTuT = uT. We note that BT is a Stochastic matrix and hence has 1 as an eigenvalue. B 1 1 1 = 1 1 1 = ⇒ 1 is an eigenvalue of B and hence also of BT. So solve (BT − I)v = 0 directly for v = uT. The corresponding homogeneous system is: −0.2 0.1 0.1 −0.3 0.1 0.1 0.2 −0.1 x y z = .
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Applying GEM or ERO′s −.2 .1 .1 −.3 .1 .1 .2 −.1 E32(−1) − → −.2 .1 .1 −.3 .1 .5 −.2 E12(2) − → −.5 .2 .1 −.3 .1 .5 −.2 P12 − → .1 −.3 .1 −.5 .2 .5 −.2 E32(1) − → .1 −.3 .1
.2 . Hence .2z .4z z is an eigenvector and for unit vector we choose z = ± 5 √ 30 . This gives u = ± 1 √ 30 [1, 2, 5]. Called equilibrium point since repeated applications leaves u unchanged: u BB...B
N times
= u.
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Find the characteristic polynomial and its roots. Find eigenvectors corresponding to each root by solving the resulting singular homogeneous linear system of equations. Remarks:
1
If λ is a real characteristic root, then Eλ = N(A − λI) is a vector subspace of I Rn.
2
If λ is complex then the eigenvectors will be in I Cn (assuming A ∈ Mn(I R)).
3
For odd n, at least one eigenvalue will be real.
Department of Mathematics, IIT - Bombay