Homogeneous Liner Systems with Constant Coefficients
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1 Homogeneos Liner Systems with Constant Coefficients Jly, 06 The object of stdy in this section is where A is a d d constant matrix whose entries are real nmbers. As before, we will look to the exponential fnction for soltions. x (t = Ax(t ( x(t = e rt where 0 is a d-dimensional colmn vector and r is a nmber. Sbstitting into (, we find that Therefore, re rt = Ae rt = e rt A r = A and (A ri = 0 For 0, we mst have non-trivial soltions to this algebraic eqation and conseqently the determinant det(a ri = 0. Let s give names to the concepts that this calclations generate. For a d d constant matrix A, The eigenvales or characteristic vales of A are those (possibly complex vales r for which has at least one non-trivial soltion. (A ri = 0 The corresponding soltions are called the eigenvectors or characteristic vectors of A associated with r. Note that any non-zero scalar mltiple of a eigenvector is an eigenvector. The determinant p(r = det(a ri is a polynomial of degree d. p(r is called the characteristic polynomial of A and p(r = 0 is called the characteristic eqation of A. We will see that ehe characteristic eqation plays a role for systems similar to the role played by the axiliary eqation for scalar eqations. Example. Let A(t be the constant matrix the characteristic polynomial p(r = det(a ri = det (, ( r r = ( r 4
2 The characteristic eqation has soltions ( r 4 = 0 r = and r = 3 To find eigenvectors, we find non-trivial soltions to ( r r = 0 and Ths, soltions are Ths, is a general soltion. ( = 0 ( = Exercise. Find a general soltion to ( for and ( r r = 0, ( = 0, ( and = x(t = c e t + c e 3t ( Give the fndamental matrix and determine the soltion that satisfies ( x(0 = x 0 = We now give an examplw with d = 3. Example 3. To determine a general soltion to ( for 3 0 we first compte the characteristic polynomial. r p(r = det(a ri = det r 3 r = ( ( r 3 3 r det det r r ( r + det = r (( r( r ( 3 (( r ( 3( + ( ( r( = r( 4 + r + 3 ( r 3 + ( r = r(r + (r + (r + = r(r = r(r (r +
3 The characteristic eqation p(r = 0 has roots, r =, r = 0, and r 3 = For the eigenvectors, (A r I = r r 3 r = 3 3 Check that we can take = (. For r = 0, (A r I = A. Check that we can take = ( For the eigenvectors, (A r 3 I = r 3 r 3 3 = 3 3 r 3 Check that we can take 3 = (0. The general soliion is Complex Eigenvales x(t = e t + + c 3 e t 3. To start with some notation. With z = a + ib is a complex nmber, the we write the complex conjgate z = a ib. Exercise 4. A constant c is real if and only if c = c z z = z z. A constant a is real is to have ā = a. We extend this notation to vectors and matrices. Ths, a matrix A has real-valed entries if and only if Ā = A. Sppose that two soltions to the characteristic eqation p(r = 0 are r ± = α ± iβ. Ths, r = r +. If = a + ib is an eigenvector for A with eigenvale r +. Here a and b are real-valed vectors. Then, 0 = (A r + I(a + ib. Now take the complex conjgate of this eqation, noting that Ā = A and Ī = I. 0 = (A r + I(a + ib = (A r I(a ib, showing that a ib is an eigenvector for A with eigenvale r. Ths, we have two linearly independent 3
4 soltions x (t = e (α+iβt (a + ib = e αt (cos βt + i sin βt(a + ib = e αt ((cos βt a sin βt b + i(sin βt a + cos βt b = x (t + ix (t x (t = e (α iβt (a ib = e αt (cos βt i sin βt(a ib = e αt ((cos βt a sin βt b i(sin βt a + cos βt b = x (t ix (t where x (t = e αt ((cos βt a + sin βt b and x (t = e αt ((cos βt a sin βt b. Now x (t and x (t are linear combinations of x (t and x (t, and ths are soltions to ( Note that x (t is the real part of x (t and that x (t is the imaginary part. Exercise 5. Check that x (t and x (t are linearly independent Example 6. Find a general soltion to ( for ( 3 4 the characteristic polynomial ( 3 r p(r = det(a ri = det = (3 r( r+8 = r r 3+8 = r r+5 = (r r Ths, the root of the characteristic eqation r ± = ± i. As noted above, we need only find the eigenvector to r +, ( ( 3 r+ i 0 = z = 4 r + 4 i z We have an eigenvector z = ( + i = ( ( + i 0 = a + ib. x (t = e t ((cos t a + sin t b and x (t = e t ((cos t a sin t b. ( ( e x (t = t (cos t + sin t e e t and x cos t (t = t (cos t sin t e t. cos t 4
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4. Linear Systems 4A. Review of Matrices 0 2 0 0 0 4A-. Verify that 0 2 =. 2 3 2 6 0 2 2 0 4A-2. If A = and B =, show that AB BA. 3 2 4A-3. Calculate A 2 2 if A =, and check your answer by showing that
Ph.D. Katarína Bellová Page 1 Mathematics 2 (10-PHY-BIPMA2) EXAM - Solutions, 20 July 2017, 10:00 12:00 All answers to be justified.
PhD Katarína Bellová Page 1 Mathematics 2 (10-PHY-BIPMA2 EXAM - Solutions, 20 July 2017, 10:00 12:00 All answers to be justified Problem 1 [ points]: For which parameters λ R does the following system