Describing Eigenvalues and EigenvectorsDate: 04/05/98 at 16:10:40 From: Charlie Higgins Subject: Eigenvalues After working 30 years in industry, I went through an alternative Teacher Certification program and got a teaching certificate. I now teach at a high school. I do not have a degree in mathematics and my degree is in Chemistry, but I do have a minor in math with enough hours to qualify for the math certificate. The new Algebra II books have a chapter on matrices (basic). One of my students asked me the other day "What is an Eigenvalue?" I did not take linear algebra in college and could not answer him. Could someone explain to me in terms where I can explain it to my student what an Eigenvalue is? Warmest Regards, Charlie Higgins Date: 04/06/98 at 08:13:58 From: Doctor Jerry Subject: Re: Eigenvalues Hi Charlie, If A is an n by n matrix, then a real or complex number w is an eigenvalue of A if there is a nonzero n by 1 matrix Y for which A*Y = w*Y. The vector Y is an eigenvector. If you think about 2 by 1 matrices X as like vectors in an (x,y)-plane and A is a given matrix, then multiplication by A is like a deformation of the plane. The point with position vector X is transformed to AX. If A is a rotation matrix, for example, which rotates each point X about the origin, then original position is X and, after rotation, AX. Or A can be thought about as an elastic deformation of the plane, where the plane is an elastic sheet. X is the position vector of a point before deformation and AX is the position vector after deformation. Let: A = [ 2 2 ] [ -1 5 ] and let: X = [ x ] [ y ] Then: AX = [ 2x+2y ] [ -x+5y ] One can ask: under this elastic deformation, is there a direction along which there is no rotation, just pure stretching? So, one wants a direction X for which: AX = wX where w is a scalar. wX would be pure stretching, since the direction wouldn't change. So, we want X and w so that (A-wI)X = 0, where I is 2 by 2 identity. This system of equations has a nontrivial solution only if det(A-wI) = 0. A-wI = [ 2-w 2 ] [ -1 5-w ] det(A-wI) = (2-w)(5-w)+2 = (w-3)(w-4) So, eigenvalues are 3 and 4. This part of linear algebra is very important to engineers, physicists, mathematicians, statisticians, and others. -Doctor Jerry, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 04/06/98 at 08:33:43 From: Doctor Anthony Subject: Re: Eigenvalues An easy way to visualize what eigenvalues and eigenvectors are is to think in 2 dimensions and consider a matrix transformation in the plane. A matrix M, say: M = [a b] [c d] operating on a vector u will in general change u by enlarging, rotating, reflecting, or shearing u to some new vector v. Now suppose we want to look at the SPECIAL vector u which apart from enlargement or contraction remains in same direction as u after the transformation by the matrix M. What we require is: M*u = k*u where k is some scalar factor (the enlargement factor) So: [a b][x] = k[x] [c d][y] [y] and ax + by = kx cx + dy = ky (a-k)x + by = 0 cx + (d-k)y = 0 Now we have two homogeneous equations (no constant terms only terms in x and y) with two unknowns, and we can only have solutions other than x = 0, y = 0 if: a-k b ---- = ---- c d-k then: (a-k)(d-k) - bc = 0 (Equation 1) This condition is usually given in the form: determinant |a-k b| = 0 | c d-k| If we multiply out equation (1) we get: ad - k(a+d) + k^2 - bc = 0 k^2 - k(a+d) + ad - bc = 0 This is a quadratic in k of the form: k^2 - k(Trace M) + Det M = 0 (Trace = sum of leading diagonal of M, Det M = determinant of M) The two values of k which are solutions to this equation are the EIGENVALUES of M. Suppose we call them k1 and k2; then we can also find the unique vectors u1 and u2 that remain unchanged (apart from enlargement) when matrix M is used to transform them. We have: (a-k1)x + by = 0 (a-k1)x = - by x -b --- = ---- y a-k1 and we get: u1 = |x| = | b | |y| |k1-a| and so associated with the eigenvalue k1 we get the eigenvector u1 given by: u1 = | b | |k1-a| and there will be another eigenvector u2 associated with the eigenvalue k2. If we express any other vector in terms of u1 and u2 as unit base vectors, then the transformation represented by M can be given much more simply as the matrix [k1 0] = D [ 0 k2] This has many applications, and one that you might meet fairly soon is finding powers of matrices. You 'diagonalize' a matrix M by expressing it in terms of u1, u2, and the above matrix, D, with leading diagonal k1 and k2. If P = matrix made up of u1 and u2 as columns of P then: M*P = P*D which follows from M*u1 = k1*u1 and M*u2 = k2*u2. So: M = P*D*P^(-1) where P^(-1) is inverse matrix of P. Then: M^2 = P*D*P^(-1)*P*D*P^(-1) = P*D*D*P^(-1) = P*D^2*P^(-1) Similarly: M^3 = P*D^3*P^(-1) and: M^n = P*D^n*P^(-1) and of course D^n is simply: [k1^n 0] [ 0 k2^n] I have just touched on the subject, but you will find that understanding this basic work on eigenvalues and eigenvectors will make the work of transformation geometry a lot clearer than it might otherwise be. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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