Unit 4 | unit 4 orthogonal complements

Unit - 4

Orthogonal complements

Q1) What are orthogonal complements?

A1)

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.

Let S be a subset of an inner product space V. The orthogonal complement of S, denoted by (read ‘‘S perp’’) consists of those vectors in V that are orthogonal to every vector u S; that is,

In particular, for a given vector u in V, we have

That is consists of all vectors in V that are orthogonal to the given vector u.

We show that is a subspace of V. Clearly 0 , because 0 is orthogonal to every vector in V. Now suppose v, w . Then, for any scalars a and b and any vector u S, we have

Thus, av + bw , and therefore S? is a subspace of V.

Q2) Let W be a finite-dimensional subspace of an inner product space V, and let y ∈ V. Then there exist unique vectors u ∈ W and z ∈ such that y = u + z. Furthermore, if {} is an orthonormal basis for W, then

A2)

Let {} be an orthonormal basis for W, let u be as defined in the preceding equation, and let z = y − u. Clearly u ∈ W and y = u + z.

To show that z ∈⊥, it suffices to show, that z is orthogonal to each . For any j, we have

To show uniqueness of u and z, suppose that y = u + z = u’ + z’ where

U’ ∈ W and z’ ∈ W⊥. Then u – u’’ z’ − z ∈ W ∩ = {0 }. Therefore,

u = u’ and z = z’.

Q3) What is Bessel’s inequality?

A3)

Bessel’s inequality: Let V be an inner product space, and let S = {v1, v2, . . . , vn} be an orthonormal subset of V. For any x ∈ V we have

Q4) Let V be a finite-dimensional inner product space, and let T be a linear operator on V. Then there exists a unique function T∗ : V → V such that for all x, y ∈ V. Furthermore, T∗ is linear.

A4)

Let y ∈ V. Define g: V → F by g(x) = for all x ∈ V. We

First show that g is linear. Let x1, x2 ∈ V and c ∈ F. Then

Hence g is linear.

To obtain a unique vector y’∈ V such that

g(x) = ; that is, for all x ∈ V. Defining T∗ : V → V

By (y) = y_, we have .

To show that is linear, let y1, y2 ∈ V and c ∈ F. Then for any x ∈ V,

We have

Since x is arbitrary,

Finally, we need to show that is unique. Suppose that U: V → V is linear and that it satisfies for all x, y ∈ V. Then for all x, y ∈ V, So = U.

The linear operator T∗ described in this theorem is called the adjoint of the operator T. The symbol is read “T star.”

Thus T∗ is the unique operator on V satisfying for

All x, y ∈ V. Note that we also have

So for all x, y ∈ V.

For an infinite-dimensional inner product space, the adjoint of a linear operator

T may be defined to be the function T∗ such that for all x, y ∈ V, provided it exists.

Q5) Let V be an inner product space, and let T and U be linear operators on V. Then

(a) =

(b) = for any c ∈ F;

(d) = T

(e) = I.

A5)

Here we prove (a) and (d); the rest are proved similarly. Let x, y ∈ V.

(a) Because

has the property unique to . Hence

(d) Similarly, since

Q6) What is least square method?

A6)

Suppose a survey conducted by taking two measurements such as at times respectively.

For example the surveyor wants to measure the birth rate at various times during a given period.

Suppose the collected data set is plotted as points in the plane.

From this plot, the surveyor finds that there exists a linear relationship between the two variables, y and t, say- y = ct + d, and would like to find the constants c and d so that the line y = ct + d represents the best possible fit to the data collected. One such estimate of fit is to calculate the error E that represents the sum of the squares of the vertical distances from the points to the line; that is,

Q7) Find the solution of the following homogeneous system of linear equations,

A7)

The given system of linear equations can be written in the form of matrix as follows,

Apply the elementary row transformation,

, we get,

, we get

Here r(A) = 4, so that it has trivial solution,

Q8) Check whether the following system of linear equations is consistent of not.

2x + 6y = -11

6x + 20y – 6z = -3

6y – 18z = -1

A8)

Write the above system of linear equations in augmented matrix form,

Apply , we get

Apply

Here the rank of C is 3 and the rank of A is 2

Therefore both ranks are not equal. So that the given system of linear equations is not consistent.

Q9) What do you understand by minimal Solutions to Systems of Linear Equations?

A9)

Even when a system of linear equations Ax = b is consistent, there may be no unique solution. In such cases, it may be desirable to find a solution of minimal norm. A solution s to Ax = b is called a minimal solution if ||s|| for all other solutions u. The next theorem assures that every consistent system of linear equations has a unique minimal solution and provides a method for computing it.

Q10) Define normal.

A10)

Let V be an inner product space, and let T be a linear operator on V. We say that T is normal if T = T. An n × n real or complex matrix A is normal if A= A.

Note- T is normal if and only if is normal, where is an orthonormal basis.

For example: Suppose that A is a real skew-symmetric matrix; that is, = −A. Then A

Is normal because both Aand A are equal to −.

Q11) Let T be a self-adjoint operator on a finite-dimensional inner product space V. Then

(a) Every eigenvalue of T is real.

(b) Suppose that V is a real inner product space. Then the characteristic polynomial of T splits.

A11)

(a) Suppose that T(x) = x for x _= 0 . Because a self-adjoint operator is also normal

x = T(x) = x) = x.

So λ =; that is, is real.

(b) Let n = dim(V), be an orthonormal basis for V, and A = . Then A is self-adjoint. Let be the linear operator on defined by = Ax for all x ∈ . Note that is self-adjoint because = A, where γ is the standard ordered (orthonormal) basis for , So, by (a), the eigenvalues of are real. By the fundamental theorem of algebra,

The characteristic polynomial of splits into factors of the form t − . Since each is real, the characteristic polynomial splits over R. But has the same characteristic polynomial as A, which has the same characteristic polynomial as T. Therefore the characteristic polynomial of T splits.

Q12) Let V be an inner product space, and let T be a linear operator on V. Then T is an orthogonal projection if and only if T has an adjoint and = T = .

A12)

Suppose that T is an orthogonal projection. Since = T because T is a projection, we need to show that T∗ exists and T = . Now V = R(T) ⊕ N(T) and R = N(T). Let x, y ∈ V. Then we can write x = x1 + x2 and y = y1 + y2, where x1, y1 ∈ R(T) and x2, y2 ∈ N(T). Hence

And

for all x, y ∈ V; thus exists and T = .

Now suppose then T is a projections, hence we must show thatR(T) = N and R

Let x ∈ R(T) and y ∈ N(T). Then x = T(x) =, and so

Therefore x ∈ N, from which it follows that R(T) ⊆ N.

Let y ∈ N. We must show that y ∈ R(T), that is, T(y) = y. Now

Since y − T(y) ∈ N(T), the first term must equal zero. But also

Thus y − T(y) = 0; that is, y = T(y) ∈ R(T). Hence R(T) = N

Q13) State and prove spectral theorem.

A13)

Let that T is a linear operator on a finite-dimensional inner product space V over F with the distinct eigenvalues . Assume that T is normal if F = C and that T is self-adjoint if F = R. For each i (1 ≤ i ≤ k), let Wi be the eigenspace of T corresponding to the eigenvalue , and let be the orthogonal projection of V on. Then the following statements are true.

(a) V = W1 ⊕ W2 ⊕ ·· · ⊕ Wk.

(b) If denotes the direct sum of the subspaces for j i, then

(d) I = + + · · · + .

(e) T = + + · · · +

Proof:

(a) As we know that T is diagonalizable,

So that,

V = W1 ⊕ W2 ⊕ ·· · ⊕ Wk

(b)

If x ∈ and y ∈ for some I is not equals to j, then = 0.

It follows easily from this result that

From first result, we have

On the other hand, we have

Hence

Proof of (d)-

Since Ti is the orthogonal projection of V on , it follows from

(b) that N() = R(=

Hence, for x ∈ V, we have x = x1 + x2 + · · · + xk, where (x) = ∈.

Now we will prove the last result,

For x ∈ V, write x = x1 + x2 + · · · + xk, where xi ∈. Then

+ + · · · +

(x) + (x) + · · · + (x)

+ + · · · + )x

The set {, , . . . , } of eigenvalues of T is called the spectrum of T, the sum I = ++· · ·+ in (d) is known as resolution of the identity operator induced by T, and the sum T = + + · · · + in (e) is called the spectral decomposition of T. The spectral decomposition of T is unique up to the order of its eigenvalues.