Chapter 6 Eigenvalues Basil Hamed

Chapter 6 Eigenvalues Basil Hamed
Engineering Analysis Chapter 6 Eigenvalues Basil Hamed

we will be concerned with the equation Ax = λx
we will be concerned with the equation Ax = λx. This equation occurs in many applications of linear algebra. If the equation has a nonzero solution x, then λ is said to be an eigenvalue of A and x is said to be an eigenvector belonging to λ. Wherever there are vibrations, there are eigenvalues, the natural frequencies of the vibrations. If you have ever tuned a guitar, you have solved an eigenvalue problem. When engineers design structures, they are concerned with the frequencies of vibration of the structure. The eigenvalues of a boundary value problem can be used to determine the energy states of an atom or critical loads that cause buckling in a beam. Basil Hamed

6.1 Eigenvalues and Eigenvectors
Many application problems involve applying a linear transformation repeatedly to a given vector. The key to solving these problems is to choose a coordinate system or basis that is in some sense natural for the operator and for which it will be simpler to do calculations involving the operator. With respect to these new basis vectors (eigenvectors) we associate scaling factors (eigenvalues) that represent the natural frequencies of the operator. Basil Hamed

Definition EXAMPLE 2 Let Solution Let A be an n×n matrix. A scalar λ is said to be an eigenvalue or a characteristic Value of A if there exists a nonzero vector x such that A x = λ x. The vector x is said to be an eigenvector or a characteristic vector belonging to λ. it follows that λ = 3 is an eigenvalue of A and x = (2, 1 ) 𝑇 is an eigenvector belonging to λ. Actually, any nonzero multiple of x will be an eigenvector, because Basil Hamed

For example, (4, 2 ) 𝑇 is also an eigenvector belonging to λ = 3 The equation Ax = λx can be written in the form Thus, λ is an eigenvalue of A if and only if (1) has a nontrivial solution. Basil Hamed

Equation (1) will have a nontrivial solution if and only if A − λI is singular, or, equivalently det(A − λI) =0 (2) If the determinant in (2) is expanded, we obtain an nth-degree polynomial in the variable λ: This polynomial is called the characteristic polynomial, and equation (2) is called the characteristic equation, for the matrix A. The roots of the characteristic polynomial are the eigenvalues of A. If roots are counted according to multiplicity, then the characteristic polynomial will have exactly n roots Thus, A will have n eigenvalues, some of which may be repeated and some of which may be complex numbers. Basil Hamed

Let A be an n ×n matrix and λ be a scalar. The following statements are equivalent: (a) λ is an eigenvalue of A. (b) (A − λI) x = 0 has a nontrivial solution. (c) N (A − λI) ≠ {0} (d) A − λI is singular. (e) det(A − λI) = 0 EXAMPLE 3 Find the eigenvalues and the corresponding eigenvectors of the matrix Basil Hamed

Solution The characteristic equation is Thus, the eigenvalues of A are λ1 = 4 and λ2 = −3. To find the eigenvectors belonging to λ1 = 4, we must determine the null space of A − 4I. Solving (A − 4I)x = 0, we get Hence, any nonzero multiple of (2, 1 ) 𝑇 is an eigenvector belonging to λ1, and {(2, 1 ) 𝑇 } is a basis for the eigenspace corresponding to λ1. Similarly, to find the eigenvectors for λ2, we must solve Basil Hamed

(A + 3I)x = 0 In this case, {(−1, 3 ) 𝑇 } is a basis for N(A + 3I) and any nonzero multiple of (−1, 3 ) 𝑇 is an eigenvector belonging to λ2. Example The linear system can be written in matrix form as Find the eigenvalues and corresponding eigenvectors of the matrix A Basil Hamed

Solution The characteristic equation of A is The factored form of this equation is , so the eigenvalues of A are λ=−2 𝑎𝑛𝑑 λ=5. eigenvector of A (λI – A) x =0 Basil Hamed

If λ = 2 then Solving this system yields 𝑥 1 =-t, 𝑥 2 = t the eigenvectors of A corresponding to λ = -5 solve this system Basil Hamed

Example Find the eigenvalues of Solution The characteristic polynomial of A is The eigenvalues of A must therefore satisfy the cubic equation Thus the eigenvalues of A are Basil Hamed

EXAMPLE 4 Let Find the eigenvalues and the corresponding eigenspaces Solution Thus, the characteristic polynomial has roots λ1 = 0, λ2 = λ3 = 1. The eigenspace corresponding to λ1 = 0 is N(A), which we determine in the usual manner: Basil Hamed

Setting x3 = 𝛼, we find that x1 = x2 = x3 = 𝛼. Consequently, the eigenspace corresponding to λ1 = 0 consists of all vectors of the form 𝛼(1, 1, 1)T . To find the eigenspace corresponding to λ = 1, we solve the system (A − I)x = 0: Setting x2=𝛼 and x3=β, we get x1=3𝛼−β. Thus, the eigenspace corresponding to λ = 1 consists of all vectors of the form Basil Hamed

Example Find bases for the eigenspaces of Solution The characteristic equation of matrix A is , or, in factored form thus the eigenvalues of A are λ 1 =1 and λ 2,3 =2 , so there are two eigenspaces of A. is an eigenvector of A corresponding to λ if and only if x is a nontrivial solution of —that is, of (3) Basil Hamed

If , λ = 2 then 3 becomes Solving this system using Gaussian elimination yields Since are linearly independent, these vectors form a basis for the eigenspace corresponding to λ = 2 . Basil Hamed

If λ = 1 , then 3 becomes Solving this system yields Thus the eigenvectors corresponding to λ = 1 are the nonzero vectors of the form is a basis for the eigenspace corresponding to λ = 1 . . Basil Hamed

EXAMPLE Let Find the eigenvalues and the corresponding eigenvector Solution Basil Hamed

Basil Hamed

EXAMPLE 5 Let Compute the eigenvalues of A and find bases for the corresponding eigenspaces. Solution The roots of the characteristic polynomial are λ1 = 1 + 2i, λ2 = 1 − 2i. It follows that {(1, i ) 𝑇 } is a basis for the eigenspace corresponding to λ1 = 1 + 2i. Similarly, and {(1,−i ) 𝑇 } is a basis for N(A − λ2I). Basil Hamed

A square matrix A is invertible if and only if λ= 0 is not an eigenvalue of A. Basil Hamed

Complex Eigenvalues If A is an n × n matrix with real entries, then the characteristic polynomial of A will have real coefficients, and hence all its complex roots must occur in conjugate pairs. Thus, if λ = a + bi (b ≠ 0) is an eigenvalue of A, then λ = a − bi must also be an eigenvalue of A. The sum of the diagonal elements of A is called the trace of A and is denoted by tr(A). Basil Hamed

EXAMPLE 6 If Then det(A) = − = 13 and tr(A) = 5 − 1 = 4 The characteristic polynomial of A is given by and hence the eigenvalues of A are λ1 = 2 + 3i and λ2 = 2 − 3i. Note that Basil Hamed

If A is an n x n triangular matrix (upper triangular, lower triangular, or diagonal), then the eigenvalues of A are the entries on the main diagonal of A Example By inspection, the eigenvalues of the lower triangular matrix Basil Hamed

Similar Matrices We close this section with an important result about the eigenvalues of similar matrices. Recall that a matrix B is said to be similar to a matrix A if there exists a nonsingular matrix S such that B = 𝑆 −1 AS. Theorem Let A and B be n×n matrices. If B is similar to A, then the two matrices have the same characteristic polynomial and, consequently, the same eigenvalues. EXAMPLE 7 Given Basil Hamed

It is easily seen that the eigenvalues of T are λ1 = 2 and λ2 = 3. If we set A = 𝑆 −1 TS, then the eigenvalues of A should be the same as those of T. We leave it to the reader to verify that the eigenvalues of this matrix are λ1 = 2 and λ2 = 3 Basil Hamed

6.2 Systems of Linear Differential Equations
A differential equation is an equation that involves an unknown function and its derivatives. An important, simple example of a differential equation is where r is a constant. The idea here is to find a function x(t) that will satisfy the given differential equation. This differential equation is discussed further subsequently. linear algebra is helpful in the formulation and solution of differential equations. Homogeneous Linear Systems We consider the first-order homogeneous linear system of differential equations, Basil Hamed

We can write above eq. in matrix form by letting Basil Hamed

Eigenvalues play an important role in the solution of systems of linear differential equations. In this section, we see how they are used in the solution of systems of linear differential equations with constant coefficients. We begin by considering systems of first-order equations of the form Basil Hamed

then the system can be written in the form Let us consider the simplest case first. When n = 1, the system is simply y Clearly, any function of the form Basil Hamed

satisfies equation (1). A natural generalization of this solution for the case n > 1 is to take where x = (x1, x2, , xn ) 𝑇 . To verify that a vector function of this form does work, we compute the derivative Now, if we choose λ to be an eigenvalue of A and x to be an eigenvector belonging to λ, then Basil Hamed

Hence, Y is a solution of the system. Thus, if λ is an eigenvalue of A and x is an eigenvector belonging to λ, then 𝑒 λ𝑡 x is a solution of the system 𝑌 ′ = AY. This will be true whether λ is real or complex. EXAMPLE 1 Solve the system Solution is a solution of the system. Basil Hamed

Example For the system the matrix has eigenvalues λ1 = 2 and λ2 = 3 with respective associated eigenvectors These eigenvectors are automatically linearly independent, since they are associated with distinct eigenvalues Hence the general solution to the given system is Basil Hamed

In terms of components, this can be written as Example Consider the following homogeneous linear system of differential equations: Solution The characteristic polynomial of A is Basil Hamed

so the eigenvalues of A are λ1 =1, λ2=2, and λ3=4. Associated eigenvectors are respectively. The general solution is then given by where b1, b2, and b3 are arbitrary constants. Basil Hamed

EXAMPLE For the linear system of previous Example, solve the initial value problem determined by the initial conditions x1(0)=4, x2(0)= 6, and x3(0) = 8. Solution We write our general solution in the form x = Pu as or Therefore, the solution to the initial value problem is Basil Hamed

Complex Eigenvalues Let A be a real n × n matrix with a complex eigenvalue λ = a + bi, and let x be an eigenvector belonging to λ. The vector x can be split up into its real and imaginary parts Since the entries of A are all real, it follows that λ = a − bi is also an eigenvalue of A with eigenvector Basil Hamed

and hence 𝑒 λ𝑡 x and 𝑒 λ𝑡 𝑥 are both solutions of the first-order system 𝑌 ′ = AY. Any linear combination of these two solutions will also be a solution. Thus, if we set then the vector functions Y1 and Y2 are real-valued solutions of 𝑌 ′ = AY. Taking the real and imaginary parts of we see that Basil Hamed

EXAMPLE 2 Solve the system Solution Let Basil Hamed

Basil Hamed

Higher Order Systems Given a second-order system of the form we may translate it into a first-order system by setting Basil Hamed

and then and The equations can be combined to give the 2n × 2n first-order system Basil Hamed

EXAMPLE 3 Solve the initial value problem Solution Basil Hamed

The coefficient matrix for this system, has eigenvalues Corresponding to these eigenvalues are the eigenvectors Basil Hamed

Thus, the solution will be of the form We can use the initial conditions to find c1, c2, c3, and c4. For t = 0, we have or, equivalently, Basil Hamed

The solution of this system is c = (2, 1, 1, 0 ) 𝑇 , and hence the solution to the initial value problem is Therefore, Basil Hamed

In general, if we have an mth-order system of the form where each Ai is an n×n matrix, we can transform it into a first-order system by setting We will end up with a system of the form Basil Hamed

6.3 Diagonalization In this section, we consider the problem of factoring an n×n matrix A into a product of the form XD 𝑋 −1 , where D is diagonal. Theorem 6.3.1 Definition An n × n matrix A is said to be diagonalizable if there exists a nonsingular matrix X and a diagonal matrix D such that 𝑋 −1 AX = D We say that X diagonalizes A. Theorem An n × n matrix A is diagonalizable if and only if A has n linearly independent eigenvectors. Basil Hamed

6.3 Diagonalization Remarks
If A is diagonalizable, then the column vectors of the diagonalizing matrix X are eigenvectors of A and the diagonal elements of D are the corresponding eigenvalues of A. The diagonalizing matrix X is not unique. Reordering the columns of a given diagonalizing matrix X or multiplying them by nonzero scalars will produce a new diagonalizing matrix. If A is n × n and A has n distinct eigenvalues, then A is diagonalizable. If the eigenvalues are not distinct, then A may or may not be diagonalizable depending on whether A has n linearly independent eigenvectors. If A is diagonalizable, then A can be factored into a product XD 𝑥 −1 . Basil Hamed

6.3 Diagonalization It follows from remark 4 that and, in general
EXAMPLE 1 Let Basil Hamed

6.3 Diagonalization The eigenvalues of A are λ1 = 1 and λ2 = −4. Corresponding to λ1 and λ2, we have the eigenvectors x1 = (3, 1 ) 𝑇 and x2 = (1, 2 ) 𝑇 . Let It follows that and Basil Hamed

6.3 Diagonalization EXAMPLE 2 Let
It is easily seen that the eigenvalues of A are λ1 = 0, λ2 = 1, and λ3 = 1. Corresponding to λ1 = 0, we have the eigenvector (1, 1, 1 ) 𝑇 , and corresponding to λ = 1, we have the eigenvectors (1, 2, 0 ) 𝑇 and (1, 0, 1 ) 𝑇 . Let It follows that = A Basil Hamed

6.3 Diagonalization Even though λ = 1 is a multiple eigenvalue, the matrix can still be diagonalized since there are three linearly independent eigenvectors. Note also that If an n × n matrix A has fewer than n linearly independent eigenvectors, we say that A is defective. It follows from Theorem that a defective matrix is not diagonalizable EXAMPLE 3 Let The eigenvalues of A are both equal to 1. Any eigenvector corresponding to λ = 1 must be a multiple of x1 = (1, 0 ) 𝑇 . Thus, A is defective and cannot be diagonalized. Basil Hamed

6.3 Diagonalization Since A is upper triangular the eigenvalues are λ 1,2 =1 The eigenvectors are Basil Hamed

6.3 Diagonalization The preceding theorem guarantees that an n x n matrix A with n linearly independent eigenvectors is diagonalizable, and the proof provides the following method for diagonalizing A. Step 1. Find n linearly independent eigenvectors of A, say p1 ,p2 , …,pn . Step 2. Form the matrix P having p1 ,p2 , …,pn as its column vectors. Step 3. The matrix 𝑃 −1 𝐴𝑃will then be diagonal with λ1,λ2 , …λn, as its successive diagonal entries, where is the eigenvalue corresponding to pi for i =1, 2,……….,n Basil Hamed

6.3 Diagonalization EXAMPLE Finding a Matrix P That Diagonalizes a Matrix A Solution and we found the following bases (in section 6.1) for the eigenspaces: There are three basis vectors in total, so the matrix A is diagonalizable and Basil Hamed

6.3 Diagonalization diagonalizes A. As a check, the reader should verify that Basil Hamed

6.3 Diagonalization EXAMPLE A Matrix That Is Not Diagonalizable Find a matrix P that diagonalizes Solution The characteristic polynomial of A is so the characteristic equation is Basil Hamed

6.3 Diagonalization Thus the eigenvalues of A are λ1 =1, λ2,3=2 and . We leave it for the reader to show that bases for the eigenspaces are Since A 3 x3 is a matrix and there are only two basis vectors in total, A is not diagonalizable Basil Hamed

6.3 Diagonalization EXAMPLE Let
Then the characteristic polynomial of A is p(A) = λ(λ - 1 ) 2 , so the eigenvalues of A are λ1 = 0, λ2 = 1, and λ3 = 1; thus λ 2 = 1 is an eigenvalue of multiplicity 2. We now consider the eigenvectors associated with the eigenvalues λ2 = λ3 = 1. They are computed by solving the homogeneous system (λI - A)x = 0 Basil Hamed

6.3 Diagonalization The solutions are the vectors of the form
where r is any number, so the dimension of the solution space of (I λ 3 - A)x = 0 is 1 (Why?), and we cannot find two linearly independent eigenvectors. Thus A cannot be diagonalized Basil Hamed

6.3 Diagonalization EXAMPLE 4 Let A and B both have the same eigenvalues The eigenspace of A corresponding to λ1=4 is spanned by e2, and the eigenspace corresponding to λ=2 is spanned by e3. Since A has only two linearly independent eigenvectors, it is defective. On the other hand, the matrix B has eigenvector x1 = (0, 1, 3 ) 𝑇 corresponding to λ1 = 4 and eigenvectors x2 = (2, 1, 0 ) 𝑇 and e3 corresponding to λ=2. Thus, B has three linearly independent eigenvectors and consequently is not defective. Even though λ = 2 is an eigenvalue of multiplicity 2, the matrix B is nondefective, since the corresponding eigenspace has dimension 2. Basil Hamed

6.3 Diagonalization The Exponential of a Matrix Given a scalar a, the exponential 𝑒 𝑎 can be expressed in terms of a power series Similarly, for any n × n matrix A, we can define the matrix exponential 𝑒 𝐴 in terms of the convergent power series The matrix exponential (6) occurs in a wide variety of applications. In the case of a diagonal matrix Basil Hamed

6.3 Diagonalization the matrix exponential is easy to compute:
It is more difficult to compute the matrix exponential for a general n × n matrix A. If, however, A is diagonalizable, then Basil Hamed

6.3 Diagonalization EXAMPLE 6 Compute 𝑒 𝐴 for Solution The eigenvalues of A are λ1 = 1 and λ2 = 0 with eigenvectors x1 = (−2, 1 ) 𝑇 and x2 = (−3, 1 ) 𝑇 . Thus Basil Hamed

6.3 Diagonalization The matrix exponential can be applied to the initial value problem studied in Section 6.2. In the case of one equation in one unknown, the solution is We can generalize this and express the solution of (7) in terms of the matrix exponential 𝑒 𝑡𝐴 . In general, a power series can be differentiated term by term within its radius of convergence. Since the expansion of 𝑒 𝑡𝐴 has infinite radius of convergence, we have Basil Hamed

6.3 Diagonalization If, as in (8), we set then and
Thus, the solution of is simply Basil Hamed

6.3 Diagonalization The form of this solution looks
where xi was an eigenvector belonging to λi for i = 1, , n. The ci’s that satisfied the initial conditions were determined by solving a system with coefficient matrix X = (x1, , xn). If A is diagonalizable, we can write (9) in the form Basil Hamed

6.3 Diagonalization Thus To summarize, the solution to the initial value problem (7) is given by If A is diagonalizable, this solution can be written in the form Basil Hamed

6.3 Diagonalization EXAMPLE 7 Use the matrix exponential to solve the initial value problem where (This problem was solved in Example 1 of Section 6.2.) Solution The eigenvalues of A are λ1 = 6 and λ2 = −1, with eigenvectors x1 = (4, 3 ) 𝑇 and x2 = (1,−1 ) 𝑇 . Thus, Basil Hamed

6.3 Diagonalization and the solution is given by Basil Hamed

6.3 Diagonalization EXAMPLE 8 Use the matrix exponential to solve the initial value problem where Solution Since the matrix A is defective, we will use the definition of the matrix exponential to compute etA. Note that 𝐴 3 = O, so Basil Hamed

6.3 Diagonalization The solution to the initial value problem is given by Basil Hamed

6.7 Positive Definite Matrices
In this section we will study the symmetric matrix (i.e., a matrix A for which A = 𝐴 𝑇 ). We restrict our attention to symmetric matrices, because they are easier to handle than general matrices and because they arise in many applied problems. we’re going to talk about a special type of symmetric matrix , called a positive definite matrix. A positive definite matrix is a symmetric matrix with all positive eigenvalues. Note that as it’s a symmetric matrix all the eigenvalues are real, so it makes sense to talk about them being positive or negative. A symmetric n × n matrix A is positive definite if 𝑥 𝑇 Ax > 0 for all nonzero vectors x in 𝑅 𝑛 . Basil Hamed

Quadratic Forms Up to now, we have been interested primarily in linear equations—that is, in equations of the form The expression on the left side of this equation, is a function of n variables, called a linear form. In a linear form, all variables occur to the first power, and there are no products of variables in the expression. Here, we will be concerned with quadratic forms, which are functions of the form Basil Hamed

(1) For example, the most general quadratic form in the variables 𝑥 1 and 𝑥 2 is and the most general quadratic form in the variables 𝑥 1 , 𝑥 2 , and 𝑥 3 is (2) (3) The terms in a quadratic form that involve products of different variables are called the cross-product terms. Thus, in 2 the last term is a cross-product term, and in 3 the last three terms are cross-product terms Basil Hamed

If we follow the convention of omitting brackets on the resulting 1x1 matrices, then 2 can be written in matrix form as (4) and 3 can be written as (5) Note that the products in 4 and 5 are both of the form 𝑋 𝑇 𝐴𝑋, where x is the column vector of variables, and A is a symmetric matrix whose diagonal entries are the coefficients of the squared terms and whose entries off the main diagonal are half the coefficients of the cross-product terms. Basil Hamed

More precisely, the diagonal entry in row i and column i is the coefficient of 𝑥 𝑖 2 , and the off-diagonal entry in row i and column j is half the coefficient of the product 𝑥 𝑖 𝑥 𝑗 . Here are some examples. EXAMPLE 1 Matrix Representation of Quadratic Forms Basil Hamed

Symmetric matrices are useful, but not essential, for representing quadratic forms. For example, the quadratic form , which we represented in Example 1 as 𝑋 𝑇 𝐴𝑋 with a symmetric matrix A, can also be written as where the coefficient 6 of the cross-product term has been split as 5+1 rather than 3+3 , as in the symmetric representation. However, symmetric matrices are usually more convenient to work with. DEFINITION Basil Hamed

Now, it’s not always easy to tell if a matrix is positive definite. Quick,is this matrix? Hard to tell just by looking at it. One way to tell if a matrix is positive definite is to calculate all the eigenvalues and just check to see if they’re all positive calculating eigenvalues can be a real pain. Especially for large matrices. So we’re going to learn some easier ways tot ell if a matrix is positive definite. A matrix is positive definite if it’s symmetric and all its pivots are positive Basil Hamed

Pivots are, in general, way easier to calculate than eigenvalues. Just perform elimination and examine the diagonal terms If we perform elimination (subtract 2× row 1 from row 2) we get The pivots are 1 and −3. In particular, one of the pivots is −3, and so the matrix is not positive definite. Basil Hamed

THEOREM A symmetric matrix A is positive definite if and only if all the eigenvalues of A are positive. EXAMPLE Showing That a Matrix Is Positive Definite Solution The characteristic equation of A is Basil Hamed

Thus the eigenvalues of A are λ 1,2 = 2 and λ 3 =8 . Since these are positive, the matrix A is positive definite, and for all , x≠0 Our next objective is to give a criterion that can be used to determine whether a symmetric matrix is positive definite without finding its eigenvalues. To do this, it will be helpful to introduce some terminology. If Basil Hamed

is a square matrix, then the principal submatrices of A are the submatrices formed from the first r rows and r columns of A for These submatrices are THEOREM A symmetric matrix A is positive definite if and only if the determinant of every principal submatrix is positive. If det(A)=0, then A may be Indefinite or what is known Positive Semi definite or Negative Semi definite. Basil Hamed

EXAMPLE Working with Principal Submatrices The matrix is positive definite since all of which are positive. Thus we are guaranteed that all eigenvalues of A are positive and 𝑋 𝑇 𝐴𝑋 >0 for all x≠0 . Basil Hamed

Example- Is the following matrix positive definite? Basil Hamed

Remark A symmetric matrix A and the quadratic form are called positive semidefinite if 𝑋 𝑇 𝐴𝑋 ≥0 for all x negative definite if for 𝑋 𝑇 𝐴𝑋 <0 x≠0 negative semidefinite if 𝑋 𝑇 𝐴𝑋 ≤0 for all x Indefinite if 𝑋 𝑇 𝐴𝑋 has both positive and negative values Theorem Let A be a symmetric n × n matrix. The following are equivalent. A is positive definite. det(A) > 0. The leading principal submatrices A1, . . , An all have positive determinants. A is nonsingular All the eigenvalues of A are positive Basil Hamed

If A is an n × n diagonal matrix Then A is: Positive definite if and only if di>0 for i= 1,2,...,n, Negative definite if and only if di<0 for i= 1,2,...,n, Positive semi definite if and only if di≥0 for i= 1,2,...,n, Negative semi definite if and only if di≤0 for i= 1,2,...,n, Indefinite if and only if di>0 for some indices i, 1≤i≤n, and negative for other indices Basil Hamed

Example Consider the matrix Then and we have Therefore, even though all of the entries of A are positive, A is not positive definite . Basil Hamed

Example Consider the matrix Then which can be seen to be always nonnegative. Furthermore, QA(x; y) = 0 if and only if x = y and y = 0, so for all nonzero vectors (x; y), QA(x; y) > 0 and A is positive definite, even though A does not have all positive entries. Basil Hamed

Example Consider the diagonal matrix Then which can plainly be seen to be positive except when (x; y; z) = (0; 0; 0). Therefore, A is positive definite. Basil Hamed

Example All of the principal minors are nonnegative (1,0,0) A is not positive Semi definite; it is actually indefinite because we have eigenvalues positive and negative Basil Hamed

Example Let We have which can be confirmed to be positive definite by examination of the principal minors: d1 = 2, d2 = 3, d3 = 4. λ 1 = 4, λ 2 = λ 3 =1 Basil Hamed

Example det (A) =0 The leading principal submatrices all have nonnegative determinants: det(A1) = 1, det(A2) = 0, det(A3) = 0 λ_1=0 λ_2=-1 λ_3=8 However, A is not positive semidefinite, since it has a negative eigenvalue λ =−1. Indeed, x = (1, 1, 1)T is an eigenvector belonging to λ = −1 and 𝑥 𝑇 Ax = −3 Basil Hamed

Example find the quadratic form Example determine if the following matrices are positive definite a) b) Basil Hamed

The eigenvalues of this matrix are the roots of the equation λ2 – 10λ + 24 = 0. They are λ = 6 and λ = 4 which are both positive. Therefore the matrix is positive definite The characteristic equation of this matrix is λ 3 – 3λ + 2 = (λ + 1 ) 2 (λ – 2). Since two of the eigenvalues are negative, the matrix is not positive definite. Example In each part, classify the quadratic form as positive definite, positive semidefinite, negative definite, negative semidefinite, or indefinite 𝑥 𝑥 2 2 ( 𝑥 𝑥 2 ) 2 𝑥 𝑥 2 2 Basil Hamed

Since 𝑥 𝑥 2 2 > 0 unless x1 = x2 = 0, the form is positive definite. Since ( 𝑥 𝑥 2 ) 2 ≥ 0, the form is positive semidefinite. It is not positive definite because it can equal zero whenever x1 = x2 even when x1 = x2 ≠ 0. If |x1| > |x2|, then the form has a positive value, but if |x1| < |x2|, then the form has a negative value. Thus it is indefinite. Basil Hamed

Chapter 6 Eigenvalues Basil Hamed

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