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Topic 13 Notes Jeremy Orloff 13 Vector Spaces, matrices and linearity 13.1 Matlab We will use Matlab for computation and visualization. It will allow us to work with large matrices. There is a student edition available for free from MIT. See http: //ist.mit.edu/matlab/all. We will only use a tiny subset of Matlab’s enormous set of functions. I’ll post some simple (and short) tutorials on its use. A free substitute for Matlab is Octave. It has the advantage that it loads much faster and doesn’t spread digital rights management files all around your computer. The disadvantage is that it can be a little harder to install, especially on the Mac. Look at http://www.gnu.org/software/octave/download.html. I can help you get it installed if you want to try. 13.2 Linearity and vector spaces We’ve seen before the importance of linearity when solving differential equations P (D)x = f (t). To remind you: linearity meant that P (D)(c1 f + c2 g) = c1 P (D)f + c2 P (D)g for functions f , g and constants c1 , c2 . Matrix multiplication is also linear. If A is a matrix and v1 , v2 are vectors the A · (c1 v1 + c2 v2 ) = c1 A · v1 + c2 A · v2 Example 13.1. In this example we write out the calculation to emphasize the linearity. 6(3 + 4) + 5(7 + 8) 6·3+5·7+6·4+5·8 6 5 3 6 5 4 6 5 3+4 = = = + 1 2 7+8 (3 + 4) + 2(7 + 8) 1·3+2·7+1·4+2·8 1 2 7 1 2 8 Superposition Exactly like solving linear differential equations, solving linear systems of algebraic equations involves finding a particular solution and supersitioning with the homogeneous solution. 2 1 3 x Example 13.2. Solve 4 12 1 = 8 x2 6 3 9 1 13 Vector Spaces, matrices and linearity answer: For this example will use ad hoc methods to find particular and homogeneous solutions. Later we will learn systematic methods. The main point here is that the solutions can be superimposed. 2 By inspection we can see one solution is xp = . Just as valid would be to take 0 −1 5 xp = or xp = . 1 −1 Next we have to solve the associated homogeneous equation: 1 3 0 4 12 x1 = 0 x2 3 9 0 This expands to three equations in two unknowns. You can easily check that the 3 general solution is xh = c . −1 By superposition, the solution to the original equation is −1 3 x = xp + xh = +c . 1 −1 If this is unclear you should check the solution by substitution. Vector spaces The key properties of vectors are that they can be added and scaled. A vector space is any set with the following properties. 1. Closure under addition: We can add any two elements in the set and get another member. 2. Closure under scalar multiplication: We can scale any element in the set and get another member. The formal definition requires some more technical properties but this definition will suffice for 18.03. If the scalars are required to be real numbers we say we have a real vector space. If we allow them to be complex numbers then we have a complex vector space. In 18.02 you learned about vectors in the plane and vectors in space: • We call the plane R2 . It is the set of all pairs (x, y). • We call space R3 . It is the set of all triples (x, y, z). • The powers indicate the dimension of each space. Likewise we can work with high dimensional vector spaces like R1000 which consists of all lists of 1000 numbers. • In 18.03 we have used the fact that functions can be added and scaled. That is, the set of functions of t forms a vector space. (It happens to be infinite dimensional, but we can talk about that later.) 2 13 Vector Spaces, matrices and linearity 13.3 Connection to DEs We will give two examples showing directly how matrices arise in differential equations. The companion matrix Here we are going to convert a higher order differential equation into a system of first order equations. Later we will see how this technique allows us to understand DEs in a new way and also how it allows us to use numerical techniques on higher order equation. Consider the second order linear differential equation x00 + 8x0 + 7x = 0. We’ve worked this example many times. The general solution is x = c1 e−t + c2 e−7t . To convert it to a matrix system we introduce a new variable: y = x0 . The original equation is equivalent to y 0 + 8y + 7x = 0. Altogether we have the system of two first order linear DE’s: x0 = y y 0 = −7x − 8y This can be rewritten in matrix form: 0 0 1 x x = y0 −7 −8 y Notice two things: x 0 1 1. If we write this abstractly with x = and A = it looks like y −7 −8 x0 = Ax. Ignoring the fact that x is a vector and A is a matrix this looks like our most important DE: x0 = ax. 2. If we solve the original equation we’ll have found x and hence y = x0 . Conversely if we solve the matrix system we’ll have found both x and y. Since we already know the solution to the DE we have the solution to the system is x x c1 e−t + c2 e−7t 1 1 −t −7t = = = c1 e + c2 e 0 −t −7t y x −c1 e + −7c2 e −1 −7 Notice that the modal solutions are of the form ert v where v is a constant vector. Later we will use the method of optimism to guess solutions of this form. The matrix A of coefficients that arises from this technique will be called the companion matrix to the original DE. Example 13.3. Heat Flow. In this example we will set up a model for heat flow. We won’t solve it for a few days. Suppose we have a metal rod where different parts are at different temperatures. We divide it into 3 regions and imagine that each region exchanges heat with the adjacent 3 13 Vector Spaces, matrices and linearity regions. The regions on either end also exchange heat with the environment. We assume that the top and bottom of the rod are insulated so that heat can only flow out of the bar at the ends. We assume that the heat transfer follows Newton’s law and the rate constant is k at each interface. The figure below shows the the metal bar divided into 3 regions and insulated above and below. The temperature of each region and the temperature of the environment on the left and right ends are indicated in the figure. EL T1 T2 T3 ER Using Newton’s law we can write a DE for the temperature of each region. kEL T10 = −k(T1 − EL ) − k(T1 − T2 ) = −2kT1 + kT2 + kT1 − 2kT2 + kT3 T20 = −k(T2 − T1 ) − k(T2 − T3 ) = T30 = −k(T3 − T2 ) − k(T3 − ER ) = kT2 − 2kT3 + kER We can write this in matrix form 0 −2k k 0 x1 kEL T1 T20 = k −2k k x2 + 0 0 k −2k x3 kER T30 Remark: This particular coefficient matrix occurs quite often in applications. You should make sure you know how to modify the equation if we use n divisions of the rod instead of 3. 13.4 Matrix Multiplication Combination of columns We can view the result of multiplying a matrix times a vector as a linear combination of the columns of the matrix. We will use this again and again, so you should internalize it now! We illustrate with an example: Example 13.4. Consider the following product 6 5 3 6·3+5·4 6 1 · = =3 +4 1 2 4 1·3+2·4 5 2 Notice that the result is a linear combination of the columns of the matrix. To express this abstratcly we write a matrix as v v v v v A= 1 2 3 4 5 4 13 Vector Spaces, matrices and linearity Here each vj is a vector representing a column of A. We then have c1 c2 v1 v2 v3 v4 v5 · c3 = c1 v1 + c2 v2 + c3 v3 + c4 v4 + c5 v5 c4 c5 That is, the product is a linear combination of the columns of A. Block matrices and multiplication 0 0 Consider the following matrix A = 6 1 0 0 A= 6 1 0 0 5 2 1 0 0 0 0 0 5 2 1 0 0 0 0 1 . We can divide this into blocks 0 0 0 0 1 I = 0 B 0 0 As long as the sizes of the blocks are compatible block matrices multiply just like matrices: A B E AE + BF · = C D F CE + DF To convince yourself of this look at the following product and see that the blocks in the first column on the left only touch the top block on the right etc. a b 0 0 1 0 0 0 0 1 c d 6 5 0 0 · e f 1 2 0 0 g h 5