![]() ![]() Note that since P is an orthogonal matrix, by Property 6 of Orthogonal Vectors and Matrices, P -1 = P T, and so equivalently A = PDP T. Proof: Let c be the eigenvalue associated with X and d be the eigenvalue associated with Y, with c ≠ d.īut since c ≠ d, it follows that X ∙ Y = 0.ĭefinition 1: A square matrix A is orthogonally diagonalizable if there exist an orthogonal matrix P and a diagonal matrix D such that A = PDP -1. ![]() Property 2: If A is a symmetric matrix and X and Y are eigenvectors associated with distinct eigenvalues of A, then X and Y are orthogonal. Thus, λ( X′ ∙ X) = λ′( X′ ∙ X), but since X′ ∙ X is a positive real number we conclude that λ = λ′, which can only happen if λ is a real number. Using the observations made above and the fact that since A is symmetric, we have Thus there are an infinite number of non-trivial solutions to the equation ( A – λI) X = 0, which can be found using Gaussian Elimination (extended to complex numbers if necessary). By definition, this means that det( A – λI) = 0. Proof (Property 1): Let λ be an eigenvalue of the n × n matrix A. AX = λX), then it is easy to show that X ′ is an eigenvector which corresponds to λ ′ (i.e. If X is an eigenvector which corresponds to λ (i.e. Thus if λ is an eigenvalue of an n × n matrix A, then the conjugate λ′ is also an eigenvalue. In general, if z is a root of the polynomial p( x), then the conjugate of z is also a root of p( x). If X ≠ 0, then clearly the dot product is a positive real number. the dot product of a vector and its conjugate is a non-negative real number. If we consider vectors that contain complex numbers as elements, then the conjugate X′ of vector X is a vector with the same shape as X consisting of the conjugates of the elements in X.įirst note that if X =, then X′ ∙ X = | x i| 2 ≥ 0, i.e. Addition, subtraction and multiplication are performed as for real numbers (noting that i times i = -1). b = 0) its conjugate is simply z, but when z is an imaginary number, then its conjugate is distinct from z. If z = a + bi is a complex number, then z′ = a – bi is called the conjugate of z and |z| = is called the absolute value of z. Observation: Before we proceed with the proof of this property, we quickly state a few properties of complex numbers. ![]() We only consider matrices all of whose elements are real numbers. Property 1: All the eigenvalues of a symmetric (real) matrix are real We look at this case next, and then we won’t need to consider imaginary numbers any further. those where A = A T), it turns out that all the eigenvalues are real. Fortunately, in the cases we are most interested in, namely symmetric matrices (i.e. This means that every n × n matrix has n eigenvalues, but unfortunately, it is not necessarily the case that all of the eigenvalues are real numbers. Since det( A – λI) = 0 can be expressed as an nth degree polynomial, as described in Eigenvalues and Eigenvectors, it has n roots, i.e. By definition, λ is an eigenvalue of the n × n matrix A if det( A – λI) = 0. This discussion is relevant to eigenvalues. those where b = 0), but also include non-real numbers, called imaginary numbers (namely those where b ≠ 0). These are numbers of the form a + bi where a and b are real numbers and i is an abbreviation for. In order to capture the roots of all polynomials, even those with only real coefficients, such as x 2 + 1, we have to deal with what are called complex numbers. Unfortunately, as we know the square root of a negative number is not one of the numbers we ordinarily deal with, namely the decimal numbers, also called the real numbers. The 2 nd degree polynomial x 2 + 2 x + 1 also has two roots, namely -1 and -1, and so can be expressed as ( x + 1)( x + 1) or ( x + 1) 2.īut what about the polynomial x 2 + 1? It too has two roots, namely and –. In fact, if r 1, …, r n are the n roots, then the polynomial can be expressed as a n įor example, the 2 nd degree polynomial x 2 – 1 has two roots, namely 1 and -1, and in fact this polynomial is equivalent to ( x – 1)( x + 1). Observation: By the fundamental theorem of algebra any nth degree polynomial p( x) has exactly n roots, i.e. ![]()
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