Unit - 4

Orthogonal complements

Definition:

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.

Theorem:

Le tW 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

Proof:

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’.

Example: Let V = (R) with the inner product

We compute the orthogonal projection on

Is an orthonormal basis for . For these vectors, we have

And

Hence

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

Key takeaways:

- 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.

2. 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.

For a linear operator T on an inner product space V, we now define a related linear operator on V called the adjoint of T, whose matrix representation with respect to any orthonormal basis for V is .

The analogy between conjugation of complex numbers and adjoints of linear operators will become apparent.

Let V be an inner product space, and let y ∈ V. The function g : V → F defined by g(x) = is clearly linear. More interesting is the fact that if V is finite-dimensional, every linear transformation from V into F is of this form.

Theorem: Let V be a finite-dimensional inner product space over F, and let g: V → F be a linear transformation. Then there exists a unique vector y ∈ V such that g(x) = for all x ∈ V.

Proof:

Let = {v1, v2, . . . , vn} be an orthonormal basis for V, and let

Define h: V → F by h(x) = , which is clearly linear. Furthermore, for

1 ≤ j ≤ n we have

Theorem: 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.

Proof:

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.

Example:

Let T be the linear operator on defined by T(a1, a2) = (2ia1+3a2, a1−a2).

If is the standard ordered basis for , then

So

Hence

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

(a) =

(b) = for any c ∈ F;

(c) =

(d) = T

(e) = I.

Proof:

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

Note-

Let A and B be n × n matrices. Then

(a)

(B) for any c ∈ F;

(c) =

(d) = T

(e) = I.

Key takeaways:

Let V be an inner product space, and let y ∈ V. The function g : V → F defined by g(x) = is clearly linear. More interesting is the fact that if V is finite-dimensional, every linear transformation from V into F is of this form.

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

For example the surveyer 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 surveyer 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,

Thus we reduce this problem to finding the constants c and d that minimize

E.( the line y = ct + d is called the least squares line)

If we suppose,

Then it follows-

We develop a general method for finding an explicit vector that minimizes E; that is, given an m × n matrix A, we find such that for all vectors .

This method not only allows us to find the linear function that best fits the data, but also, for any positive integer n, the best fit using a polynomial of degree at most n.

For , let denote the standard inner product of x and y in . Recall that if x and y are regarded as column vectors, then

First lemma: Suppose then

Second lemma: Suppose , then rank(

Note- If A is an m × n matrix such that rank(A) = n, then is invertible.

Theorem: let then there exists such that and for all .

Furthermore, if rank (A) = n, then

Example: Suppose the data collected by surveyor is (1, 2), (2, 3), (3, 5) and (4, 7), then

Hence

Thus

Therefore,

It follows that the line y = 1.7t is the least squares line. The error E is computed as

There are two types of linear equations-

1. Consistent

2. Inconsistent

Consistent –

If a system of equations has one or more than one solution, it is said be consistent.

There could be unique solution or infinite solution.

For example-

A system of linear equations-

2x + 4y = 9

x + y = 5

Has unique solution,

Whereas,

A system of linear equations-

2x + y = 6

4x + 2y = 12

Has infinite solutions.

Inconsistent-

If a system of equations has no solution, then it is called inconsistent.

Consistency of a system of linear equations-

Suppose that a system of linear equations is given as-

This is the format as AX = B

Its augmented matrix is-

[A:B] = C

(1) Consistent equations-

If Rank of A = Rank of C

Here, Rank of A = Rank of C = n ( no. Of unknown) – unique solution

And Rank of A = Rank of C = r , where r<n - infinite solutions

(2) Inconsistent equations-

If Rank of A ≠ Rank of C

Solution of homogeneous system of linear equations-

A system of linear equations of the form AX = O is said to be homogeneous, where A denotes the coefficients and of matrix and O denotes the null vector.

Suppose the system of homogeneous linear equations is,

It means,

AX = O

Which can be written in the form of matrix as below,

Note- A system of homogeneous linear equations always has a solution if

1. r(A) = n then there will be trivial solution, where n is the number of unknown,

2. r(A) < n, then there will be an infinite number of solution.

Example: Find the solution of the following homogeneous system of linear equations,

Sol. 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,

Example: Find out the value of ‘b’ in the system of homogeneous equations-

2x + y + 2z = 0

x + y + 3z = 0

4x + 3y + bz = 0

Which has

(1) Trivial solution

(2) Non-trivial solution

Sol. (1)

For trivial solution, we already know that the values of x , y and z will be zerp, so that ‘b’ can have any value.

Now for non-trivial solution-

(2)

Convert the system of equations into matrix form-

AX = O

Apply

Respectively, we get the following resultant matrices

For non-trivial solutions, r(A) = 2 < n

b – 8 = 0

b = 8

Solution of non-homogeneous system of linear equations-

Example-1: check whether the following system of linear equations is consistent of not.

2x + 6y = -11

6x + 20y – 6z = -3

6y – 18z = -1

Sol. 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.

Example: Check the consistency and find the values of x , y and z of the following system of linear equations.

2x + 3y + 4z = 11

X + 5y + 7z = 15

3x + 11y + 13z = 25

Sol. Re-write the system of equations in augmented matrix form.

C = [A, B]

That will be,

Apply

Now apply,

We get,

~~

Here rank of A = 3

And rank of C = 3, so that the system of equations is consistent,

So that we can can solve the equations as below,

That gives,

x + 5y + 7z = 15 ……………..(1)

y + 10z/7 = 19/7 ………………(2)

4z/7 = 16/7 ………………….(3)

From eq. (3)

z = 4,

From 2,

From eq.(1), we get

x + 5(-3) + 7(4) = 15

That gives,

x = 2

Therefore the values of x, y , z are 2 , -3 , 4 respectively.

Minimal Solutions to Systems of Linear Equations:

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.

Theorem: Suppose Suppose that Ax = b is consistent. Then the following statements are true.

(a) There exists exactly one minimal solution s of Ax = b, and s ∈ R().

(b) The vector s is the only solution to Ax = b that lies in R(LA∗ ); that is,

If u satisfies ()u = b, then s = u.

Example:

Consider the system

x + 2y + z= 4

x − y + 2z = −11

x + 5y = 19.

In order to find the minimal solution to this system, we must first find some solution u to

So we consider the system

6x + y + 11z= 4

x + 6y − 4z = −11

11x − 4y + 26z= 19

For which one solution is

(Any solution will suffice.) Hence

Is the minimal solution to the given system.

Key takeaways:

1. If a system of equations has no solution, then it is called inconsistent.

2. Consistent equations-

If Rank of A = Rank of C

Here, Rank of A = Rank of C = n ( no. Of unknown) – unique solution

And Rank of A = Rank of C = r , where r<n - infinite solutions

Inconsistent equations-

If Rank of A ≠ Rank of C

3. A system of homogeneous linear equations always has a solution if

1. r(A) = n then there will be trivial solution, where n is the number of unknown,

2. r(A) < n, then there will be an infinite number of solution.

Here we will start the concept with a lemma.

Lemma: Let T be a linear operator on a finite-dimensional inner product space V. If T has an eigenvector, then so does

Proof:

Suppose that v is an eigenvector of T with corresponding eigenvalue . Then for any x ∈ V,

And hence v is orthogonal to the range of . So is not onto

And hence is not one-to-one.

Thus has a nonzero null space, and any nonzero vector in this null space is an eigenvector of with corresponding eigenvalue .

Note-

- A subspace W of V is said to be T-invariant if T(W) is contained in W. If W is Tinvariant, we may define the restriction TW: W → W by TW(x) = T(x) for all x ∈ W. It is clear that TW is a linear operator on W. Recall
- A polynomial is said to split if it factors into linear polynomials.

Theorem: Let T be a linear operator on a finite dimensional inner product space V. Suppose that the characteristic polynomial of T splits. Then there exists an orthonormal basis for V such that the matrix is upper triangular.

Definition:

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 −.

General properties of normal operators:

Let V be an inner product space, and let T be a normal operator on V. Then the following statements are true.

(a) ||T(x)|| = || (x)|| for all x ∈ V.

(b) (b) T − cI is normal for every c ∈ F.

(c) If x is an eigenvector of T, then x is also an eigenvector of T∗. In fact, if T(x) = λx, then (x) = x.

(d) If and 2 are distinct eigenvalues of T with corresponding eigenvectors x1 and x2, then x1 and x2 are orthogonal.

Proof:

(a) For any x ∈ V, we have

(b) Do yourself

(c) Suppose that T(x) = x for some x ∈ V. Let U = T − I. Then U(x) = 0, and U is normal by (b). Thus (a) implies that

(d) Let 1 and 2 be distinct eigenvalues of T with corresponding eigenvectors x1 and x2. Then, using (c), we have

Since = we conclude that = 0.

Definition:

Let T be a linear operator on an inner product space V. We say that T is self-adjoint (Hermitian) if T = . An n × n real or complex matrix A is self-adjoint (Hermitian) if A = .

Note-

- A linear operator on a real inner product space has only real eigenvalues. The lemma that follows shows that the same can be said for self-adjoint operators on a complex inner product space.
- The characteristic polynomial of every linear operator on a complex inner product space splits, and the same is true for self-adjoint operators on a real inner product space.

Lemma: 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.

Proof:

(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.

Theorem: Let T be a linear operator on a finite-dimensional real inner product space V. Then T is self-adjoint if and only if there exists an orthonormal basis β for V consisting of eigenvectors of T.

Proof:

Suppose that T is self-adjoint. By the lemma, we may apply Schur’s theorem to obtain an orthonormal basis β for V such that the matrix A = is upper triangular. But

So A and are both upper triangular, and therefore A is a diagonal matrix.

Thus β must consist of eigenvectors of T.

Key takeaways:

- Let T be a linear operator on a finite-dimensional inner product space V. If T has an eigenvector, then so does
- A polynomial is said to split if it factors into linear polynomials.
- Let T be a linear operator on an inner product space V. We say that T is self-adjoint (Hermitian) if T = . An n × n real or complex matrix A is self-adjoint (Hermitian) if A = .

Let V be an inner product space, and let T: V → V be a projection. We say that T is an orthogonal projection if R= N(T) and N = R(T).

Theorem: 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 = .

Proof:

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 projection, hence we must show that R(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

The spectral theorem

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

(c) = for 1 ≤ i, j ≤ k.

(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.

Note-

The spectral theorem can also be stated as below-

Let T be a normal (symmetric) operator on a complex (real) finite-dimensional inner product space V. Then there exists linear operators ; . . . ; on V and scalars . . . such that

The above linear operators ; . . . ; are projections in the sense that = . Moreover, they are said to be orthogonal projections because they have the additional property that 0 for

Key takeaways:

- Let V be an inner product space, and let T: V → V be a projection. We say that T is an orthogonal projection if R= N(T) and N = R(T).
- Let T be a normal (symmetric) operator on a complex (real) finite-dimensional inner product space V. Then there exists linear operators ; . . . ; on V and scalars . . . such that

References:

1. Rao A R and Bhim Sankaram Linear Algebra Hindustan Publishing house.

2.Gilbert Strang, Linear Algebra and its Applications, Thomson, 2007.

3. Stephen H. Friedberg, Arnold J. Insel, Lawrence E. Spence, Linear Algebra (4th Edition), Pearson, 2018.

4. Linear Algebra, Seymour Lipschutz, Marc Lars Lipson, Schaum’s Outline Series