z = a + bi Addition, subtraction, and multiplication of complex numbers

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1 MAT 332: Frithjof Lutscher 4 Comple numbers Introductory consideration We can easily solve the equation 2 4 The answer is ±2, in particular, is a rational number, even an integer The equation 2 2 is a bit more tricky The solution ± 2 is not a rational number Instead, we have defined the square root of a positive number as the real number that gives the original number back when multiplied by itself But what should we do with the equation 2 +? The answer cannot be a real number (Why?) Can we do the same as above and define a number whose square equals -? Indeed, this is what mathematicians did in the eighteenth century (it was a daring act and caused a lot of controversy), and they call that number i or imaginary (We will see that comple numbers are hardly more imaginary than 2) Definition A comple number z is a number of the form z a + bi with real numbers a, b and the symbol i that satisfies i 2 We call a Re(z) the real part of z and b Im(z) the imaginary part of z The real number a can be considered the comple number a + i A comple number of the form z bi is called purely imaginary Addition, subtraction, and multiplication of comple numbers Comple numbers are easily added, subtracted and multiplied, if we keep the rule i 2 in mind and use the distributive laws Eamples (a + bi) ± (c + di) (a + c) ± (b + d)i (a + bi) (c + di) ac + bci+adi+bdi 2 (ac bd)+(ad + bc)i (3 + 5i) + (2 7i) 5 2i 2 (5+7i) (8 26i) 3+43i 3 ( 3 + 2i) (4 5i) ( 2 ( )) + (5 + 8)i i 4 (2 5i) (3 + 4i) (6 2) + ( 5 + 8)i i 5 (9 + 2i) + 5 (9 + 2i) + (5 + i) 4 + 2i 6 3i + (2 + 3i) ( 3i) + (2 + 3i) 2 + i (3 5i) 6 i 4

2 MAT 332: Frithjof Lutscher 5 8 3i ( + 4i) 2 3i Before we look at inverses and division of comple numbers, we introduce the comple conjugate of a comple number Definition and observation The comple conjugate of z a + bi is z a bi, ie, we simply change the sign of the imaginary part Since the multiplication z z (a + bi)(a bi) a 2 + b 2 always produces a non-negative real number, we can take the square root modulus or absolute value of z a + bi as We define the z z z a 2 + b 2 From the identity z z z 2, we find the inverse of z to be Eample z z z/ z 2 Start with z 3 + 4i and w 2 i The comple conjugates are z 3 4i and w 2+i The absolute values are z 5 and w 5 The inverses are Finally, we can divide z 25 (3 4i), w (2 + i) 5 z w z w w 2 w (2 + i), 5 z w z z 2 (2 i) 25 Eample 2 Start with z 4i and w 5+3i The comple conjugates are z +4i and w 5 3i The absolute values are z 7 and w 37/4 The inverses are Division gives z 7 ( + 4i), w 4 (5 3i) 37 z w z w w ( 5 5i), w z w z z 2 ( 5 + 5i) 5

3 MAT 332: Frithjof Lutscher 6 Geometric interpretation It is very helpful to think of a comple number as a point in the plane with the real part as the -value and the imaginary part as the y-value Hence, we identify the comple number z a + bi with the point (a, b) or with the vector (arrow) from the origin to the point (a, b) (We will talk about vectors in more detail shortly) Then the absolute value of the comple number is simply the distance of the corresponding point from the origin or the length of the vector (arrow), see Figure 3 With this correspondence, the addition of comple numbers become the addition of vectors as it is known from the physics of forces, see Figure 3 To interpret multiplication, we take a slightly different point if view Instead of giving the coordinates of the vector as the endpoint (a, b), we consider its length r and the angle φ it makes with the -ais (counterclockwise) as (r cos φ, r sin φ) This representation is called polar coordinates Then multiplication of two numbers is simply multiplication of the lengths and addition of the angles, see Figure 3 We write z r(cos φ + i sin φ), and w s(cos ψ + i sin ψ) Then we multiply, using the trigonometric identities zw r(cos φ + i sin φ) s(cos ψ + i sin ψ) Observation and definition rs[cos φ cos ψ sin φ sin ψ + i(cos φ sin ψ + cos ψ sin φ)] rs[cos(φ + ψ) + i sin(φ + ψ)] Every comple number of the form z cos φ + i sin φ has absolute value one We introduce the eponential notation (known as Euler s formula) ep(iφ) e iφ cos φ + i sin φ It might look strange at first, but the same rules as for the real eponential function apply This has many advantages First of all, we can write any comple number in polar coordinates as z re iφ And we can easily multiply comple numbers in this form For eample, the calculation above becomes a single step (no need to look up the trig identities) re iφ se iψ rse i(φ+ψ) Eamples The comple number z + i has modulus z 2 and angle φ π/4 Hence z +i 2e iπ/4 2 The comple number w 2 + i has modulus z 5 and angle φ π/6 Hence w +i 5e iπ/6 3 Their product is zw + 3i e i5π/2 6

4 MAT 332: Frithjof Lutscher 7 4 In general, if z a+bi then r z a 2 + b 2 The argument φ is not uniquely defined If we restrict it between π and π the we get φ arctan(b/a) if a> φ arctan(b/a)+ π if a<,b> φ arctan(b/a) π if a<,b< 7

5 MAT 332: Frithjof Lutscher 8 4 y point ( 2,2) comple number 2+2i point (,25) comple number +25i point (3,) comple number 3+i 6 5 z+w3+5i 4 z+3i y 3 2 w2+2i zwrse φ+ψ 5 φ cosφ sinφ y φ+ψ wse ψ zre φ φ ψ Figure 3: Top panel: correspondence between comple numbers and points in the plane Middle panel: Addition of comple numbers Bottom panel: polar coordinates on the unit circle (left) and multiplication of comple numbers using polar coordinates (right) 8

6 MAT 332: Frithjof Lutscher 9 Linear Algebra I - Linear systems of equations Introductory eample Suppose there are two types of food Type I contains g of protein and 5g of carbohydrates per g, type II contains 8g of protein and 2g of carbohydrates per g Easy question Suppose I take 5g of type I and 75g of type II, how much protein and carbohydrates do I get? Answer: First we have to choose units, let s say g Then I take 5 units of type I and 75 units of type II Now let, 2 denote the respective units of food type I and II Then the amount of protein and carbohydrates are given by b , b , grams respectively NOTE: we are given the, 2 and want to find the b,b 2 Harder question Suppose I want to take 6g of protein and 2g of carbohydrates How much of each food type to I have to take? The equations are just as above +8 2 b 6, b 2 2 HOWEVER: this time, with the same notation as above, we are given the b,b 2 and want to get the, 2 SOLUTION: We multiply the second equation by -2: Then we add the two equations: , , which gives 2 3/2 This we put back into one of the original equations to get or 2/5 The eample above is a special case of a linear system of equations More generally, we write a linear system of m equations with n unknowns as a + a a n n b a 2 + a a 2n n b 2 a m + a m a mn n b m, where 9

7 MAT 332: Frithjof Lutscher 2, n are the variables or unknowns, 2 a,a mn are the coefficients, 3 b,b m are the right hand side A solution of such a system is a set of numbers (s,,s n ) that makes all the equations true when we substitute the s i for the i The equations are called linear since each of the variables i appears only linearly (as opposed to higher powers or other nonlinear functions) The subject of linear algebra is to study such linear systems of equations Goal of this chapter In this chapter we learn how to solve linear systems of equations In particular, we answer the three questions: Is there always a solution? 2 Can there be more than one solution? 3 How can we compute all solutions? Fact: A linear system of equations has either eactly one solution, 2 infinitely many solutions, or 3 no solution Cases () and (2) are called consistent whereas case (3) is called inconsistent Eamples Consider the three systems of equations: 2, , The first system has no solution, the second has eactly one solution ( 2, 2 ) and the last system has infinitely many solutions (for all real numbers t, the pair t, 2 t works) We can see graphically, why there are these three cases Each equation is the equation of a line in - 2 -space In the first case, the two lines are parallel and have no point in common In the second case, there is one point of intersection In the third case, the lines are identical, and all points are part of both lines, see Figure 4 2

8 MAT 332: Frithjof Lutscher Figure 4: Graphical interpretation of systems of linear equations Left: no solution, Middle: unique solution, Right: infinitely many solutions The Gaussian Elimination Algorithm Elementary row operations: The solution of a linear system of equations does not change under the following three elementary row operations: multiply a row by a nonzero number, 2 add a multiple of one row to another, 3 echange the order of two rows Eample Find the solution of Answer: ( 5) R R2+R R+R Now, the solution is obvious: 3/2 and 2 2, or (3/2, 2) ( /4) R ( /2) R2 3/2 2 2 Eample 2 Find the solution of

9 MAT 332: Frithjof Lutscher 22 Answer: and finally ( ) R3+R2 So the solution is (, 2, ) Simplify Notation! R R (/2) R2 R R (/2) R3 ( 3) R3+R2 R R ( ) R3+R Looking at the two eamples above, we realize that we do not need to write the variables i all the time, provided we agree and stick to a particular order Similarly, we can do away with the + and the signs The only things that matters are the coefficients in front of the i and the right hand side Hence, we collect these into an array of numbers, which we call matri We can then simply perform the row operations on the rows of the matri For the eample above: We call the matri above the augmented matri, where we have the coefficients, a ij of the linear system together with the right hand side b i, separated by the vertical lines We also call the matri 2 3 the coefficient matri, ie, this matri contains only the coefficients a ij The same steps as in eample 2 above, but in matri notation are: R R2 2 3 ( ) R3+R ( ) R3+R Now, we remember that the first column corresponds to, the second to 2, and the third to 3 Then the solution is, 2 2, 3 or simply (, 2, ) 22

10 MAT 332: Frithjof Lutscher 23 Reduced row-echelon form Now that we have seen several eamples of simple forms of systems where the solutions could easily be read off, we will formalize this process a bit Definition: The leading entry of a row in a matri is the leftmost nonzero coefficient in that row A matri is in row-echelon form if the following three rules are true Rows of zeros are below any nonzero row 2 The leading entry of any row is to the right of any leading entry in any row above it 3 All entries in the column below a leading entry are zero A matri is in reduced row-echelon form if it is in row-echelon form and in addition 4 Each leading entry is 5 All entries in the column above a leading entry are zero Eamples The first two systems are in row-echelon form but not reduced The third and fourth are in reduced row-echelon form, the last one is neither ,,, 3, Eample 3 Find the solution of Answer: Use the matri notation to do the row operations ( 2)R+R R2+2R3 ( ) R2 2 4 Hence, we have the solution 2, 2, 3 4 or (2,, 4)

11 MAT 332: Frithjof Lutscher 24 Eample 4 Find the solution of Answer in matri notation: R+R2 R+R ( /2) R3+R At this point we pause and look at the last row of zeros In the original notation with variables i, this row reads: This is clearly impossible Hence, this system is inconsistent, it has no solution Eample 5 Find the solution of Answer in matri notation: This time, the last row reads 5 R /3 R R2 R Whatever we choose for 3, we can always find an 2 to make the equation true Hence, this equation has infinitely many solutions We set 3 t as a free variable, then we get 2 5 3t We plug this into the first equation and solve for as +2t, or 2t Hence, the infinitely many solutions can be written as the set {( 2t, 5 3t, t): t R} Eample 6 Find the solution of

12 MAT 332: Frithjof Lutscher 25 Answer in matri notation: R R+R2 4 R+R R2+R3 7 4 Again, we have a row of zeros at the bottom However, this time the last row reads: This equation is satisfied for all values of, 2, 3 We do not run into the same problem as in eample 4 We simply continue as in eample 5 The second row 4/7 /7 2 + (4/7) 3 /7 3/7 /7 4/7 /7 has again infinitely many solutions We denote 3 t as the free variable and compute 2 /7 (4/7)t In the first row, we get + (3/7)t /7 or /7 3t/7 Hence, the solution set is {( 7 3t 7, 7 4t 7,t): t R} Eample 7 For which values of h does the system + h have (a) a unique solution, (b) infinitely many solutions, and (c) no solution? Answer: h The last row gives the equation 2 R+R2 (2h 4) 2 h 3 2h 4 If 2h 4, (ie, h 2) then 2 t is a free variable, and 3 h 2 3 2t If, on the other hand, 2h 4 then the only way to satisfy the second row is 2 In this case, the first row gives 3 Hence, if h 2 then there is a unique solution, if h 2 then there are infinitely many solutions For no value of h is there no solution 25

13 MAT 332: Frithjof Lutscher 26 Practice Problems Solve the following systems of equations by bringing them into reduced rowechelon form (e) (a) (c) Problems with parameters (b) (d) (f) (a) For which values of a, b does the system + a b have (i) a unique solution, (ii) infinitely many solutions, or (iii) no solutions? (b) Eplain why the system a + 3 b c is consistent if c a + b but inconsistent in all other cases 3 Application Insects of two species are reared on two types of food Species consumes 5 units of food A and 3 units of food B per day Species 2 consumes 2 units of A and 4 units of B, respectively Every day, 9 units of food A and 96 units of food B are provided How many individuals of each species are reared? 26

14 MAT 332: Frithjof Lutscher 27 Solutions to practice problems (d) (a) (, 2, 3) (b) {(2 2t, 5 27t, t): t R} (c) s + 2 t, s t, s, t : s, t R 4 (f) (,, ) (e) 2, 7 2, t, 4t 26,t 3 : t R 2 (a) Unique solution if a 3/2 infinitely many solutions if a 3/2 and b 2 No solution if a 3/2 and b 2 (b) Add the first and second row 3 Let i be the number of individuals of species i Then the system is and the solution is (, 2 ) (2, 5) , , 27

15 MAT 332: Frithjof Lutscher 28 Linear Algebra II - Vectors and matrices In the last section, we introduced matrices to simplify our life, as a short hand notation for linear systems of equations In this section, we study matrices as objects in their own right We learn when it is possible to add and multiply them and how to do it While there is no direct biological application in this section, the content presented here is the foundation of everything to come It is like learning the grammar of a language so that one can speak it properly later Definition: An m n-matri A is a rectangular array of numbers with m rows and n columns, ie, a a 2 a n a 2 a 22 a 2n A [a ij ] a m a m2 a mn The numbers a ij are called entries If m n then A is a square matri A n-matri is called a row vector: [c,c 2,,c n ] An m -matri is called a column vector: Two matrices A [a ij ],B [b ij ] are said to be equal if they have the same dimension and if for all i, j we have a ij b ij b b 2 b m We note two special (classes of) matrices, the zero matri, and the identity matri I The number of rows and columns of these two matrices is usually clear from the contet If A is a square matri, then the elements a ii are called the diagonal elements and their sum of called the trace of A, ie, 5 2 A has the diagonal elements 5,, -4 and the trace is tr(a)

16 MAT 332: Frithjof Lutscher 29 Matri addition and scalar multiplication If two matrices, A [a ij ] and B [b ij ] are both of the format m n, then we can form the sum of the two by entrywise addition to obtain a matri of the same format C A + B [c ij ][a ij + b ij ] If k is a number and A [a ij ] an m n-matri, then we define the entrywise product which is again an m n-matri Eamples ka [ka ij ], Consider the two 2 3-matrices 2 3 A 3 2 B Then we have A + B 3A +2B The transpose of a matri , 5A , The transpose of a matri A is obtained from A by interchanging rows and columns, or, loosely speaking, by flipping the matri along its diagonal More formally, if A [a ij ], then the transpose is A T [a ji ] If A is of the format m n then A T is of the format n m In particular, the transpose of a column vector is a row vector and vice versa Eamples A v 4 7, A T v T

17 MAT 332: Frithjof Lutscher 3 Matri-vector multiplication If the matri A has n columns and the column vector has n rows, then we can define the product A as the following column vector: a a 2 a n a 2 a 22 a 2n 2 A, a m a m2 a mn a + a a n n a 2 + a a 2n n n A a m + a m a mn n To remember this definition, simply think about linear systems of equations! Note that the resulting vector has the same number of rows as the matri In general, the dimensions work as follows: (m n) (n ) (m ) Eamples

18 MAT 332: Frithjof Lutscher Matri-matri multiplication We use the definition of the matri-vector product to define a product of two matrices Consider a matri A with n columns and a matri B with n rows We may think of each column of B as a column vector of length n We know how to multiply each of these with the matri A and then we put the resulting vectors into one matri More formally, the product of A is given by a a n a 2 a 2n a m a mn b b k, B b n b nk AB Note that the dimensions multiply as follows: Eamples AB AB 2 AB k (m n) (n k) (m k) B B 2 B k Take the following matrices: A , B 4 The product AB is defined and it is AB, C , D The product BA is not defined since the number of rows of B does not equal the number of columns of A However, if we transpose B first, then we can multiply B T A : The product CA is defined, but not vice versa (check the format!): 2 2 CA

19 MAT 332: Frithjof Lutscher 32 Again, if we transpose C, then we can multiply AC T AC T Since A is a square matri, we can form the power A 2 7 AA 5 22 Since A and D are square of the same size, we can multiply them both ways: 7 AD, DA NOTE: The order of the product matters! Matri multiplication is NOT commutative, even if both products are defined Multiplication of the identity matri and the zero matri are just as easy as multiplying the real numbers zero and one: AI IA A, A A As a final, not so obvious eample for matri multiplication, we note (check the formats!) Note: We have seen that many operations with real numbers (addition, multiplication) also work with (some) matrices Two of several differences are The commutative law does not hold 2 The cancellation law does not hold Eample The matri A is an eample where A 2 but A 32

20 MAT 332: Frithjof Lutscher 33 Practice Problems Consider the following matrices: 3 A 2 4, B 2 D E C Compute the following if possible, if not, eplain why (a) D + E (b) D E (c) 5A (d) 7C (e) 2B C (f) 4E 2D (g) 3(D +2E) (h) A A (i) tr(d) (j) tr(d 3E) (k) tr(a) (l) tr(b) (m) 2A T + C (n) D T E T (o) (D E) T (p) B B T 2 Compute the following if possible, if not, eplain why (a) AB (b) BA (c) 3ED (d) (AB)C (e) A(BC) (f) CC T (g) (DA) T (h) (C T B)A T (i) tr(dd T ) (j) B

21 MAT 332: Frithjof Lutscher 34 Solutions to Practice Problems (a) D + E (d) 7C (g) 3(D+2E) (b) D E (c) 5A, (e) 2B C not defined (f) 4E 2D , , (h) A A, the zero matri (i) tr(d) ++4 5, (j) tr(d 3E) , (k) tr(a) not defined (l) tr(b) (m) 2AT +C, (n) D T E T 4 (o) (D E ) T D T E T 2 (a) AB (d) (AB)C (g) (DA) T (p) B B T, (b) BA not defined, (c) 3ED, (e) A(BC) (AB)C, (f) CC T, (h) (C T B)A T (j) B , ,,, (i) tr(dd T ) , 34

22 MAT 332: Frithjof Lutscher 35 Linear Algebra III - Inverses and Determinants We know that for every nonzero real number,, there eists an inverse, / such that the product In the last section, we learned how to multiply matrices (and when this is possible) Here we ask whether inverses eist for matrices as well, and how we can compute them Introductory eample The multiplication of two square matrices can give the identity matri: But of course, not every (square) matri has an inverse The zero matri, for eample, is not invertible Definition: A square matri A is invertible if there eists another square matri B of the same dimension, such that AB I BA We then write A B Note: If a matri is not square, then it cannot be invertible How can we find out whether a given matri has an inverse and what the inverse is? Let s go back to the eample above with 2 5 A 3 We want to find a matri such that AB b A b 2 Hence, we have to solve the two systems b A b 2 b b B 2 b 2 b 22 b2 A b 22, b2, A b 22 We can do this simultaneously, using the three allowed elementary row operations of multiplication, addition and interchange 2 5 R 2R2 2 5 R+5R

23 MAT 332: Frithjof Lutscher 36 5 R, R Once we have reduced the left hand side to the identity matri, we have the inverse of the original matri on the right hand side Algorithm to compute an inverse matri: Given a square matri A, we take the identity matri I and write the system [ A I ] Then we use the three row operations to reduce the left hand side to the identity matri If this is possible, we end up with [ I B ][I A ] If we cannot reduce the left hand side to the identity matri, then the original matri A is not invertible Eample Find the inverse of A We write /2 7/2 /2 3/2 /2 3/2 3/2 7/2 Hence, A is invertible and its inverse is A /2 3/2 Eample 2 Find the inverse of A We write Because we have a row of zeros on the left hand side, we cannot transform the left hand side into the identity matri by elementary row operations Therefore, the matri A is not invertible 36

24 MAT 332: Frithjof Lutscher 37 Eample 3 Find the inverse of A We write /2 2 2 /2 / /2 2 2 /2 /2 Hence, the matri A is invertible and the inverse is given by the right side of the last step above Eample 4 Find the inverse of A We write Since the left half of the above matri is the zero matri, we cannot transform it into the identity matri, using row operations Therefore, A is NOT invertible When is a matri invertible? Sometimes we are not interested in the eact inverse of a matri but only in the question of whether or not the matri is invertible Let s go back to the case of 2 2-matrices When is the matri a a A 2 a 2 a 22 37

25 MAT 332: Frithjof Lutscher 38 b b invertible? We take the potential inverse B 2 b 2 b 22 AB I eplicitly We get the two systems of equations, and write out the equations a b + a 2 b 2 a 2 b + a 22 b 2 a b 2 + a 2 b 22 a 2 b 2 + a 22 b 22 For the system on the left, we multiply the first row by a 2 and the second by a and add the two in order to eliminate b from the second equation a b + a 2 b 2 a 2 a 2 b 2 + a a 22 b 2 a 2 If the epression a a 22 a 2 a 2 is nonzero, we can solve for b 2 as b 2 a 2 a a 22 a 2 a 2 The system on the right for b 2,b 22 is similar: If a a 22 a 2 a 2 then we can solve for b 22 a a a 22 a 2 a 2 Since the epression a a 22 a 2 a 2 is so important, we give it a special name Definition and Result: For a 2 2 matri a a A 2 a 2 a 22 the determinant of A is given by det(a) a a 22 a 2 a 2 If det(a) then A is invertible and A a22 a 2 det(a) a 2 a If det(a) then A is not invertible Eample, revisited The determinant of the matri A is det(a) The inverse according to our new formula is therefore A 2 computed earlier 3 7 3, just as we had 38

26 MAT 332: Frithjof Lutscher 39 Eample 2, revisited The determinant of the matri A is det(a) Hence, the matri is not invertible, as we have already seen before Our theory so far only applies to matrices of size 2 2, therefore we cannot revisit the eamples 3 and 4 We do that later For now, we look at some more eamples of matrices of size 2 2 Eample 5 The determinant of the matri A is det(a) Therefore, the inverse of A eists and is given by A Eample 6 3 The determinant of the matri A is 2 /2 det(a) ( ) 3/2+27/2 Therefore, the inverse of A eists and is given by A 2 / Eample 7 The determinant of the matri A Hence, the matri is not invertible /7 is det(a) ( 7) ( 8/7) Determinants for matrices of size 3 3 Determinants can be defined for square matrices of all sizes, and it is always true that if det(a) then the matri A is invertible We will only consider the case of 3 3-matrices here, since it has a fairly simple form Determinants of bigger matrices can be computed, but it takes time 39

27 MAT 332: Frithjof Lutscher 4 The determinant of the matri A a a 2 a 3 a 2 a 22 a 23 a 3 a 32 a 33 is given by det(a) a a 22 a 33 + a 2 a 23 a 3 + a 3 a 2 a 32 a 3 a 22 a 3 a 32 a 23 a a 33 a 2 a 2 Unfortunately, there is no really simple formula for the inverse of a 3 3 matri (or larger) analogous to the 2 2 case To find the inverse, we still have to use the row-reduction algorithm This determinant formula can be remembered more easily by attaching the first two columns of the matri A as columns 4 and 5 of a larger matri and then taking products along the diagonal down with plus signs and products along the diagonal up with minus signs, ie, Eample 3, revisited The determinant of A a a 2 a 3 a a 2 a 2 a 22 a 23 a 2 a 22 a 3 a 32 a 33 a 3 a 32 is det(a) Hence, the matri is invertible, as we already know Eample 4, revisited The determinant of A is det(a) Hence, the matri is not invertible, confirming our previous result Important Observation: If A is an invertible square matri then the unique solution of the system of linear equations A b is given by A b 4

28 MAT 332: Frithjof Lutscher 4 Eample, revisited The solution of the system is given by

29 MAT 332: Frithjof Lutscher 42 Practice Problems Compute the determinant and find the inverse if it eists (a) 4 A (b) A 2 (c) A (d) A (e) A (f) A Solve the linear system of equations For each of the matrices above solve the system A b where b and b in cases (e) and (f) 2 in cases (a)-(d) 42

30 MAT 332: Frithjof Lutscher 43 Solutions to Practice Problems (a) A 3 2 (b) A /4 (c) A 2/3 /3 (d) and (f) are not invertible (e) A (a) /4, (b) ( + 2t)/3 In case (d) the solution is : t R t (e) 7/3, (c) 2/3 In case (f), there is no solution NOTE: Even if the matri is not invertible, there can still be a solution (see (d)), but it is not unique in that case 5 5 2, 43

31 MAT 332: Frithjof Lutscher 44 Linear Algebra IV: Eigenvalues and Eigenvectors Observation and introductory eample Consider the matri A so that A y A On the other hand, we also have A 3, A, A y For eample 3 In these two cases, the vectors A are simply multiples of the vectors, whereas in y y the eamples above, there was no obvious relationship between A and The goal y y of this section is to find and study these special cases where a vector multiplied by a matri is simply a multiple of the vector Definition: Let A be a square matri A vector v that is not the zero vector and a number λ are called eigenvector and eigenvalue, respectively if Eample The vector 2 Av λv is an eigenvector of the matri A corresponding to the eigenvalue λ 3, since 3 8 Eample 2 The vector is an eigenvector of the matri A λ, since NOTE: λ is allowed, but the eigenvector cannot be the zero vector corresponding to the eigenvalue 44

32 MAT 332: Frithjof Lutscher 45 Eample 3 The vector λ since 2 is an eigenvector of A corresponding to the eigenvalue 2 How to compute eigenvalues and eigenvectors? We are looking for a nonzero solution of the equation Av λv, which is the same as saying that we are looking for a nonzero solution of the homogeneous equation (A λi)v If the matri (A λi) is invertible, then the only solution to the equation is the zero solution, which we do not want Hence, we need to make sure that the matri (A λi) is not invertible In the previous section we saw that a matri is not invertible precisely if its determinant is zero Therefore, we need to find values of λ that make the determinant of (A λi) to zero Result: The number λ is an eigenvalue of the square matri A if and only if it satisfies the equation det(a λi) If λ is an eigenvalue of the square matri A then we find the corresponding eigenvector(s) by solving the linear system of equations (A λi)v Note: Any scalar multiple of an eigenvector is again an eigenvector Eample Find the eigenvalues and eigenvectors of A First we form the matri A λi λ λ Then we calculate its determinant as λ det λ 2 λ λ λ 45

33 MAT 332: Frithjof Lutscher 46 Setting the determinant to zero, we get the two eigenvalues λ ± For λ we solve the system The second row is satisfied, the first row reads + 2 Hence, 2 t is a free variable, and t is the resulting solution, so that v is the eigenvector corresponding to λ Note that any multiple of v is also an eigenvector to the same eigenvalue For λ 2 we solve the system The second row is satisfied, the first row reads + 2 Hence, 2 t is a free variable, and t is the resulting solution, so that v 2 is the eigenvector corresponding to λ 2 Note that any multiple of v is also an eigenvector to the same eigenvalue Eample 2 Find the eigenvalues and eigenvectors of A First we form the matri A λi 2 Then we calculate its determinant as λ λ λ 2 λ 2 λ λ det(a λi) λ(2 λ)( λ) (2 λ) (2 λ)(λ 2 ) Hence the eigenvalues are λ,λ 2,λ 3 2 For λ we solve the system (writing only the coefficient matri and suppressing the right hand column of zeros) 2 We have 3 t as the free variable and solve for 2 2t, and t to get the eigenvector v 2 For λ 2 we solve the system (writing only the coefficient matri and suppressing the right hand column of zeros)

34 MAT 332: Frithjof Lutscher 47 The second row gives the equation 3 2, which means 2 We have 3 t the free variable and solve for t to get the eigenvector v 2 For λ 3 2 we solve the system (writing only the coefficient matri and suppressing the right hand column of zeros) 2 2 The second row gives the equation 2, which means 2 t is free The first equation says 2 + 3, while the third equation says 2 3 The only way to make both of them true is to set 3 Hence we get the eigenvector v 3 Eample 3 Find the eigenvalues and eigenvectors of A First we form the matri 2 A λi 2 3 Then we calculate its determinant as λ λ λ λ 2 2 λ 3 λ det(a λi) λ(2 λ)(3 λ) (2 λ)( 2) (2 λ)(λ 2)(λ ) Hence the eigenvalues are λ 2,λ 2 2,λ 3 For the double eigenvalue λ λ 2 2we solve the system 2 2 The first equation is + 3 The variables 2 s and 3 t are free The solution is 3 t Hence, we get essentially two eigenvectors For one, we set t and s For the other, we set t and s Then v, v 2 For λ 3, we solve the system (suppressing the column of zeros)

35 MAT 332: Frithjof Lutscher 48 In this case, we get 3 t to be the free variable The resulting solution then is 2 3 t 2 and 2 3 2t The eigenvector therefore is v 3 Eample 4 Find the eigenvalues and eigenvectors of A First we form the matri λ λ A λi λ λ Then we calculate its determinant as λ det ( λ) 2 +λ 2 2λ +2 λ Setting the determinant to zero, we get λ 2 2 ± 4 8 ± ± i Hence, we get comple numbers as eigenvalues Just to be sure, we check that λ +i solves the original equation: ( + i) 2 2( + i) i 2 2i + 2 We can compute the corresponding eigenvectors in the same way as above, but the calculation involves comple numbers For λ + i we get i i ( i R+R2) i Then 2 t is a free variable and it, so that the eigenvector is v For λ 2 i we get i i (i R+R2) i Then 2 t is a free variable and it, so that the eigenvector is v i i 48

36 MAT 332: Frithjof Lutscher 49 Practice Problems Find the eigenvalues and eigenvectors of the following matrices (a) A 3, (b) A (e) A (c) A , (f) A (h) A 3 4 2, (d) A , (i) A, , 6 3, (g) A 2 2 3, 49

37 MAT 332: Frithjof Lutscher 5 Solutions For problem (a): det(λi A) det λ 2 λ 3 2 λ λ 2 (λ 3) 4(λ 3) (λ 3)(λ 2 4) So the eigenvalues are λ 3, λ 2 2 and λ 3 2 To find the eigenvectors corresponding to an eigenvalue λ we have to solve the system (λi A) where 2 3 For λ 3 we have to solve The eigenvectors are For λ The eigenvectors are For λ Hence 3 and from the first equation also 5 t t,with t R Then 3 t is free, 2 2t and t t 2t t 2 with t R t

38 MAT 332: Frithjof Lutscher The eigenvectors are For problem (b): We have det(a λi) det t t t 2 2 Then 3 t is free, t and 2 with t R 5 λ 2 3 λ 2 4 λ ( 5 λ)( λ)(4 λ) ()( 2)( 5 λ) ( 5 λ) [( λ)(4 λ) + 2] (5 + λ)(λ 2 5λ + 6) (5 + λ)(λ 2)(λ 3) Thus, the eigenvalues of A are λ 5, λ 2 2 and λ 3 3 The eigenvector v associated with the eigenvalue λ 5 is the solution of the system 2 3 (A λ I)v 6 2 y 9 z The augmented matri is R 2R 3 R and R 2 6R 3 R 2 gives ( /2)R R gives 56 9 R R R 2 and R 3 9R R 3 gives 5

39 MAT 332: Frithjof Lutscher 52 Finally, R 3 R and R 2 R 3 gives The solution is y z and t Thus, the eigenvectors are v s real number ecept Net, for λ 2 2 we have (A λ 2 I)v 2 2 y 2 z Thus R 3 + R 2 R 3 and R +2R 2 R gives ( /7)R R and R 2 R 2 gives 2 The solution is thus z and y 2z With z s, we get v 2 s real number ecept Finally, for λ 3 3 we have (A λ 3 I)v y z 2 where t is any where s is any The augmented matri is

40 MAT 332: Frithjof Lutscher 53 R 2 +2R 3 R 2 and R 2R 3 R gives 8 5 ( /8)R R and R 2 R 3 gives 5/8 The solution is 5z/8 and y z With z r, we get v 3 r real number ecept For problem (c): 5/8 where r is any det(a λi) det 3 λ λ 4 λ (3 λ)( 2 λ)( λ)++ (3 λ)(4) (3 λ)[ 2 λ + λ 2 4] (3 λ)[λ 2 λ 6] (3 λ)(λ + 3)(λ 2) Thus λ 3,λ 2 3,λ 3 2 Eigenvectors are v, v 2 4/3, v 3 /2 /4 For problem (d): 4 λ 2 5 det(a λi) det λ λ (4 λ) [( λ)( 3 λ) 2] (4 λ)(λ 2 +2λ 5) (4 λ)(λ 3)(λ + 5) Thus, the eigenvalues of A are λ 5, λ 2 3 and λ

41 MAT 332: Frithjof Lutscher 54 Eigenvectors are v 7/27 /3, v 2 3, v 3 For problem (e): Eigenvalues are Eigenvectors are For problem (f): Eigenvalues are Eigenvectors are For problem (g): Eigenvalues are Eigenvectors are For problem (h): Eigenvalues are Eigenvectors are v 3 v v v λ 5,λ 2, v 2 λ 5,λ i, v 2 3 λ i,λ 2 i i, v 2 λ 2,λ 2,λ 3 4, v 2 5/3, v 3 For problem (i): Eigenvalues are λ,λ 2 2,λ 3 2 Eigenvectors are v 2, v 2, v 3 54

42 MAT 332: Frithjof Lutscher 55 LinearAlgebra V - Markov chains Eample We consider two ponds, a smaller and a larger one, where a population of a total of 3 ducks lives We observe that a duck on the smaller pond switches to the larger pond the net day with a probability of 4% and stays on the smaller pond with a probability of 6% A duck on the large pond stays with a probability of 8% and moves on to the smaller pond with a probability of 2% Suppose that on the first day of observation, there are 5 ducks on each pond How many ducks are there on each pond the net day, the third day, in the long run? Let s denote n as the number of ducks on the small pond and y (n) the number on the large pond on day n Then at time t wehave () 5,y () 5 To compute the number of ducks on day, we simply add the ducks that stay to the ones that newly arrive, ie, () 6 +2y , y () 4 +8y We can write these equations in matri form as () 6 2 () y () y (2) y () Now it is easy to continue For the numbers on day 2, we simply apply the matri again (2) 6 2 () y () We can continue from here But we cal also ask if there will be a state where the arrivals equal the departures on both ponds so that the number of ducks stays the same day after day If the numbers stay the same, say,y, then we can find them from the equation or y y y, y We see that this is the same as finding an eigenvector of the matri with eigenvalue one The solution is y r 2 for any real number r But we know that the total number of ducks is 3, hence + y ( + 2)r 3, which makes r,,y 2 There are two alternative approaches and interpretations of the same problem 55

43 MAT 332: Frithjof Lutscher 56 If we do not know the total number of ducks to begin with, but only the percentage of ducks on each pond, we can do the same calculation with percentages For eample, initially, both ponds have the same number of ducks, ie, () y () 5% 5 Then () y () () y () The final distribution of ducks in the long run would then be + y 3r, and hence r /3 or /3 and y 2/3 2 If we don t look at the whole population but only at a single duck, we could be interested in the probability that the duck is on the small pond versus on the big pond Then n and y n denote the probability that the duck is on the small or large pond on day n We could assume that the duck is initially on either pond with equal probability as to get y 5% 5 Then the rest is as above, and we compute that in the long run, the duck is twice as likely to be on the big pond than on the small pond A Markov chain is a process where the state of a system at a given time can be predicted by knowing just the state of the system at the previous time A Markov chain with k states is described by the transition matri P [p ij ], i, j k, where p ij is the probability that the system will be in state i after is was in state j at the previous time For each j, we have p j + p 2j + + p kj, since the system has to go to some state A state vector is a vector 2 where i is the probability that the system is in state i In particular k k If (n) is the state vector at time n, then the state vector at the net time, (n+) is given by (n+) P (n) The steady state distribution is given by the equation P 56

44 MAT 332: Frithjof Lutscher 57 Following up on the previous eample: a duck can be in one of two states: on the small pond or on the big pond If it is on the small pond then the transition probability to the big pond is 4 whereas the transition probability to the small pond (ie, the duck remains) is 6 Vice versa, a duck on the big pond has a transition probability of 2 to the small pond and of 8 to the big pond (ie, remaining) The transition matri is therefore Eample 2 P Consider the two soft drinks Choke and Popsi Market research found that in a given year, 7% of Choke drinkers switch to Popsi and 3% stick with Choke, while 5% of Popsi drinkers switch to Choke and 5% keep drinking Popsi Assuming that that a system behaves like a finite Markov chain, find the transition matri P of probabilities for switching from one brand to another, and determine the percentage of Choke and Popsi drinkers in the long run Solution: Denote by the fraction of Choke drinkers and by y the fraction of Popsi drinkers Then the transition matri is given by 3 5 P 7 5 The steady state satisfies y y The solution is 5/7t, y t Since the sum has to equal one, 5/7t + t 2/7t, we get t 7/2 and 5/2,y 7/2 Hence, in the long run, approimately 42% drink Choke and 58% drink Popsi Eample 3: Forest succession (Horn, 98) Suppose that a forest contains three types of trees: Gray Birch (GB), Red Maple (RM) and Beech (BE) By counting saplings under each tree, we can guess the probability that a certain tree is replaced by another tree For eample, under a GB, we find 2 GB, 5 RM and 3 BE saplings for a total of 2 We conclude that the probability of the GB to be replaced by GB is, to be replaced by RM is 75, and to be replaced by BE is 5 Under a RM, we find 2 GB, 3RM, and 5 BE saplings for a total of Hence, the probability of RM to be replaced by GB is, to be replaced by RM is 3 and to be replaced by BE is 5 We choose the state vector to give the probability to be GB, RM, or BE at any given location, ie, Prob of GB y Prob of RM z Prob of BE 57

45 MAT 332: Frithjof Lutscher 58 Then the transition matri is 2 P If all tree species are equally present in the first year, then the distribution in the second year is () 2 () 2 /3 y () y () /3 z () z () / The steady state distribution is given by the solution of y z y z / 3/3 4/3 Equivalently, we solve y z Row reduction gives (we drop the right hand column of zeros) The second equation is y 5z We set z t, the free variable and solve for y 5t/32 The first equation is which we solve to get Hence the solution is 3 + y 5z, or 3 + 5t t 5t 3 32 t 96 y z t /96 5/32 5t, 58

46 MAT 332: Frithjof Lutscher 59 With the condition the + y + z, we get t 5 96 t, so that t 96/5 and y z /5 45/5 96/5 6% 3% 64% 59

47 MAT 332: Frithjof Lutscher 6 Practice Problems There are two weekly Farmers Markets in Ottawa: one on Bank and Heron (B), the other in the Glebe (G) People who shop at B one week return there the net week with a probability of 8% and go to G in 2% of the cases People who go to G return there with a probability of 7% and shop at B the net week with probability 3% Write down the matri that describes this Markov chain (You may assume that the total number of people who get groceries at a Farmers Market is constant) Show that the matri has one eigenvalue equal to one Find the proportion of people who shop at B and G in the long run, respectively 2 In a lab eperiment, there are three chambers: A, B, C, with connecting corridors between A and B, as well as B and C, but not between A and C A rat is placed into chamber A, and its location is recorded every hour The rat stays within a chamber with probability 5 and leaves the chamber with probability 5 If it leaves chamber A or C, it goes to B (it has to, there are no other corridors) If it leaves chamber B, it chooses to go to A or C with equal probability Write down the matri that describes this Markov chain What is the probability to find the rat in chamber A, B, C after the first hour? After the second hour? In the long run, what is the probability to find the rat in A, B, or C, respectively? 6

48 MAT 332: Frithjof Lutscher 6 Solutions to Practice Problems We denote the percent of people at B by and the percent of people at G by y, and consider the vector Then the matri has the form y P Note that the entries in a column add up to one The matri has eigenvalue, if the system 8 3 P y 2 7 y y has a nontrivial solution We solve (P I) y y The solution is y t, free variable 3 2 t 3 2 t Since we are dealing with proportions, we need + y, which gives 5t + t 25t and hence t 2/5 In the long run the percentage of people shopping at B is 3/5 6% and at G is y 2/5 4% 2 We denote by, y, z the probability that the rat is in chamber A, B, C, respectively, and write the vector y Then the matri is z 5 25 P Note again that the columns add up to one If the rat starts at A, then the corresponding vector is After one hour, the probabilities are P Hence, the rat is in A or B with equal probability, but not in C After the second hour the probabilities are / / /8 6

49 MAT 332: Frithjof Lutscher 62 To see what is happening in the long run, we have to solve the system P y z y z or The solution is t y z Together with the condition that + y + z, we get t /4 and the respective probabilities of 25,y 5,z 25 62