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Matrix Analysis Basic Operations A matrix is a rectangular array of elements arranged in horizontal rows and vertical columns, and usually enclosed in brackets. A matrix is real-valued (or, simply, real) if all its elements are real numbers or real-valued functions; it is complex-valued if at least one element is a complex number or a complex-valued function. If all its elements are numbers, then a matrix is called a constant matrix. Matrices are designated by boldface uppercase letters. A general matrix A having r rows and с columns may be written a11 a A 21 a r1 a12 a22 a11 a1c a2c arc where the elements of the matrix are double subscripted to denote location. By convention, the row index precedes the column index, thus, a25 represents the element of A appearing in the second row and fifth column, while a31 represents the element appearing in the third row and first column. A matrix A may also be denoted as [aij], where aij denotes the general element of A appearing in the ith row and jth column. A matrix having r rows and с columns has order (or size) "r by c," usually written rc. Two matrices are equal if they have the same order and their corresponding elements are equal. The transpose of a matrix A, denoted as AT, is obtained by converting the rows of A into the columns of A one at a time in sequence. If A has order mn, then AT has order nm. Vectors and DOT Products A vector is a matrix having either one row or one column. A matrix consisting of a single row is called a row vector; a matrix having a single column is called a column vector. The dot product А-В of two vectors of the same order is obtained by multiplying together corresponding elements of A and В and then summing the results. The dot product is a scalar, by which we mean it is of the same general type as the elements themselves. Matrix Addition and Matrix Subtraction The sum A + В of two matrices A = [aij] and В = [bij] having the same order is the matrix obtained by adding corresponding elements of A and B. That is, A + B = [aij] + [aij] = [aij + bij] Matrix addition is both associative and commutative. Thus, A + (B + C) = (A + В) + С and A + В = В + А The matrix subtraction A – В is denned similarly: A and В must have the same order, and the subtractions must be performed on corresponding elements to yield the matrix [aij – bij]. Scalar Multiplication and Matrix Multiplication For any scalar k, the matrix kA (or, equivalently, Ak) is obtained by multiplying every element of A by the scalar k. That is, kA= k[aij] = [kaij]. Let A=[aij] and B = [bij] have orders rp and pc, respectively, so that the number of columns of A equals the number of rows of B. Then the product AB is defined to be the matrix С = [cij] of order rс whose elements are given by p cij aik bkj i 0, 1, 2, , r; j 0, 1, 2, c k 1 Each element cij of AB is a dot product; it is obtained by forming the transpose of the ith row of A and then taking its dot product with the yth column of B. Matrix multiplication is associative and distributes over addition and subtraction; in general, it is not commutative. Thus, A(BC) = (AB)C A(B + C) = AB + AC (B - C)A = BA - CA but, in general, AB ВА. Also, (AB)T=ATBT Linear Transformation Introduction Suppose S = {u1, u2,..., un} is a basis of a vector space V over a field k and, for v V, suppose v = alu1 + a2u2 + … +апиn Then the coordinate vector of v relative to the basis S, which we write as a column vector unless otherwise specified or implied, is denoted and defined by a1 a vs 2 a1 an a2 an T The mapping v[v]S, determined by the basis S, is an isomorphism between V and the space Kn Matrix Representation of a Linear Operator Let T be a linear operator on a vector space V over a field K and suppose S = {u1, u2, ...,un} is a basis of V. Now T(u1), ..., Т(un) are vectors in V and so each is a linear combination of the vectors in the basis S; say, T(u1) = a11u1 + a12u1 + … + a1nu1 T(u2) = a21u1 + a22u1 + … + a2nu1 …………………………………. T(un) = an1u1 + an2u1 + … + annu1 The following definition applies. Definition: The transpose of the above matrix of coefficients, denoted by mS(T) or [T]S, is called the matrix representation of T relative to the basis S or simply the matrix of T in the basis S; that is, a11 a TS 21 an1 a12 a22 an 2 a1n a2 n ann (The subscript S may be omitted if the basis S is understood.) Remark: Using the coordinate (column) vector notation, the matrix representation of T may also be written in the form mT= [T] = ([T(ul)], [T(u2)],..., [T(un)]) that is, the columns of m(T) are the coordinate vectors ([T(ul)], [T(u2)],..., [T(un)]. Example 1 (a) Let V be the vector space of polynomials in t over R of degree 3, and let D:VV be the differential operator defined by D(p(t)) = d(p(t))/dt. We compute the matrix of D in the basis {1, t, t2, t3}. 2 3 D(1) = 0 = 0 + 0t + 0t + 0t D(t) =1 = 1 + 0t + 0t2 + 0t3 D(t3) = 3t2 = 0 + 0t + 3t2 + 0t3 0 0 D 0 0 1 0 0 0 2 0 0 0 3 0 0 0 [Note that the coordinate vectors D(l), D(t), D(t2), and D(t3) are the columns, not the rows, of [D].] (b) Consider the linear operator F : R2 R2 defined by F(x, y) = (4x2y, 2x+y) and the following bases of R2: (c) S={m,=A. 1),мг = (-1,0)} and E = {e, = A, 0), e2 = @, 1)} (d) We have Eigenvalues and Eigenvectors CHARACTERISTIC EQUATION Consider the problem of solving the system of two first-order linear differential equations, du1/dt = 7u1 − 4u2 and du2/dt = 5u1 − 2u2. In matrix notation, this system is or equivalently, u'=Au Eq(1) u1 u1 7 4 , A and u u u . Because solutions of a single 5 2 2 2 where u equation u' = λu have the form u = αeλt, we are motivated to seek solutions of Eq(1) that also have the form u1 = α1eλt and u2 = α2eλt. Eq(2) Differentiating these two expressions and substituting the results in Eq(1) yields In other words, solutions of Eq(1) having the form Eq(2) can be constructed provided 1 in the matrix equation Ax = λx can be found. Clearly, 2 solutions for λ and x = x=0 trivially satisfies Ax = λx, but x = 0 provides no useful information concerning the solution of Eq(1). What we really need are scalars λ and nonzero vectors x that satisfy Ax = λx. Writing Ax = λx as (A − λI) x = 0 shows that the vectors of interest are the nonzero vectors in N (A − λI) . But N (A − λI) contains nonzero vectors if and only if A − λI is singular. Therefore, the scalars of interest are precisely the values of λ that make A − λI singular or, equivalently, the λ ’s for which det (A − λI) = 0. These observations motivate the definition of eigenvalues and eigenvectors. Eigenvalues and Eigenvectors For an n × n matrix A, scalars λ and vectors xn×1 0 satisfying Ax = λx are called eigenvalues and eigenvectors of A, respectively, and any such pair, (λ, x), is called an eigenpair for A. The set of distinct eigenvalues, denoted by σ (A) , is called the spectrum of A. λ σ (A) A − λI is singular det (A − λI) = 0. Eq(3) {x 0| x N (A − λI)_ is the set of all eigenvectors associated with λ. From now on, N (A − λI) is called an eigenspace for A. Nonzero row vectors y* such that y*(A − λI) = 0 are called lefthand eigenvectors for A Let’s now face the problem of finding the eigenvalues and eigenvectors of the matrix 7 4 appearing in Eq(1). As noted in Eq(3), the eigenvalues are the scalars A 5 2 λ for which det (A − λI) = 0. Expansion of det (A − λI) produces the second-degree polynomial which is called the characteristic polynomial for A. Consequently, the eigenvalues for A are the solutions of the characteristic equation p(λ) = 0 (i.e., the roots of the characteristic polynomial), and they are λ = 2 and λ = 3. The eigenvectors associated with λ = 2 and λ = 3 are simply the nonzero vectors in the eigenspaces N (A − 2I) and N (A − 3I), respectively. But determining these eigenspaces amounts to nothing more than solving the two homogeneous systems, (A − 2I) x = 0 and (A − 3I) x = 0. For λ = 2, For λ = 3, In other words, the eigenvectors of A associated with λ = 2 are all nonzero multiples of x = (4/5 1)T , and the eigenvectors associated with λ = 3 are all nonzero multiples of y = (1 1)T . Although there are an infinite number of eigenvectors associated with each eigenvalue, each eigenspace is one dimensional, so, for this example, there is only one independent eigenvector associated with each eigenvalue. Let’s complete the discussion concerning the system of differential equations u' = Au in Eq(1). Coupling Eq(2) with the eigenpairs (λ1, x) and (λ2, y) of A computed above produces two solutions of u' = Au, namely, It turns out that all other solutions are linear combinations of these two particular solutions. Below is a summary of some general statements concerning features of the characteristic polynomial and the characteristic equation. Characteristic Polynomial and Equation The characteristic polynomial of An×n is p(λ) = det (A − λI). The degree of p(λ) is n, and the leading term in p(λ) is (−1)nλn. The characteristic equation for A is p(λ) = 0. The eigenvalues of A are the solutions of the characteristic equation or, equivalently, the roots of the characteristic polynomial. Altogether, A has n eigenvalues, but some may be complex numbers (even if the entries of A are real numbers), and some eigenvalues may be repeated. If A contains only real numbers, then its complex eigenvalues must occur in conjugate pairs—i.e., if λ σ (A) , then σ (A) . Proof. The fact that det (A − λI) is a polynomial of degree n whose leading term is (−1)nλn follows from the definition of determinant Then is a polynomial in λ. The highest power of λ is produced by the term earlier contained the proof that the eigenvalues are precisely the solutions of the characteristic equation, but, for the sake of completeness, it’s repeated below: The fundamental theorem of algebra is a deep result that insures every polynomial of degree n with real or complex coefficients has n roots, but some roots may be complex numbers (even if all the coefficients are real), and some roots may be repeated. Consequently, A has n eigenvalues, but some may be complex, and some may be repeated. The fact that complex eigenvalues of real matrices must occur in conjugate pairs is a consequence of the fact that the roots of a polynomial with real coefficients occur in conjugate pairs. 1 1 1 1 Example: Determine the eigenvalues and eigenvectors of A Solution: The characteristic polynomial is so the characteristic equation is λ2 − 2λ+2 = 0. Application of the quadratic formula yields so the spectrum of A is σ (A) = {1 + i, 1 − i}. Notice that the eigenvalues are complex conjugates of each other—as they must be because complex eigenvalues of real matrices must occur in conjugate pairs. Now find the eigenspaces. For λ = 1 + i, For λ = 1− i, In other words, the eigenvectors associated with λ1 = 1 + i are all nonzero multiples of x1 = (i 1)T , and the eigenvectors associated with λ2 = 1 − i are all nonzero multiples of x2 = (−i 1)T . In previous sections, you could be successful by thinking only in terms of real numbers and by dancing around those statements and issues involving complex numbers. But this example makes it clear that avoiding complex numbers, even when dealing with real matrices, is no longer possible—very innocent looking matrices, such as the one in this example, can possess complex eigenvalues and eigenvectors.