Eigenvalues and Eigenvectors · applications

Applications: Eigenvalues and Eigenvectors

10 question types · Secondary 4 · each worked step by step with a figure that follows the steps

01

Three Roads on a Town Map Redrawn for a Poster, and the Two That Keep Their Direction

methodMultiply Each Road's Direction by the Matrix, Call It an Eigenvector When the Image Is a Multiple of It, and Read the Multiple as the Stretch Along That Road

A street map is redrawn for a poster by the transformation M = 4123, with the town hall at the origin. Three straight roads run out from the town hall along the vectors a = 11, b = 12 and c = 1−2. (a) Show that the road along a stays on its own line on the poster, and find the factor by which it is stretched. (b) Decide which of the roads along b and c also stays on its own line, and find its stretch factor.

xy26−448MaaMa =55= 5 ×11
M11 = 55: the image of a lies on the dashed line through a itself.
Multiply a by the matrix: M11 = 4 × 1 + 1 × 12 × 1 + 3 × 1 = 55.
step 1 of 5

A vector v that a matrix sends to a multiple of itself, Mv = λv, is an eigenvector of the matrix, and the number λ is its eigenvalue. A road along an eigenvector stays on its own line through the origin, and λ is the factor by which every length along it is stretched. A road in any other direction is turned onto a new line.

  1. Multiply a by the matrix: M11 = 4 × 1 + 1 × 12 × 1 + 3 × 1 = 55.
  2. (a) The image is 55 = 511, a multiple of a. So a is an eigenvector of M with eigenvalue 5: the road stays on its own line, and every length along it is stretched by a factor of 5.
  3. M12 = 4 + 22 + 6 = 68. A multiple of b has a second entry twice its first, and 8 is not twice 6. So the road along b is turned onto a new line, and b is not an eigenvector.
  4. M1−2 = 4 − 22 − 6 = 2−4 = 21−2, a multiple of c.
  5. (b) The road along c stays on its own line and is stretched by a factor of 2. Check: the two eigenvalues multiply to 5 × 2 = 10 = det M = 4 × 3 − 1 × 2, and they add to 5 + 2 = 7, the sum 4 + 3 of the diagonal entries.

answer(a) Ma = 55 = 5a, so the road along a stays on its line and is stretched by a factor of 5; (b) the road along c, since Mc = 2−4 = 2c: it is stretched by a factor of 2, while b goes to 68, off its line

techniqueEigenvalues and Eigenvectors

Common pitfalls

  • Calling b an eigenvector because its image 68 is longer than b. All three roads are lengthened; what matters is whether the image lies on the same line through the origin, and (6, 8) is not on the line through (1, 2).
  • Comparing only the first entries: 6 = 6 × 1 suggests an eigenvalue of 6 for b, but then the second entry would have to be 6 × 2 = 12, not 8. Both entries must be multiplied by the same number.
02

Two Strains of Bacteria in a Laboratory Culture, and the Growth Rates Found from the Characteristic Polynomial

methodExpand the Determinant of M Minus Lambda I, Set It to Zero and Factorize for the Eigenvalues, Then Keep the Eigenvector Whose Entries Can Be Counts

In a laboratory culture, two strains of a bacterium turn into each other as they divide. Each hour, every cell of strain A is replaced by 3 cells of strain A and 2 of strain B, and every cell of strain B is replaced by 1 cell of strain A and 2 of strain B. So if there are a thousand cells of A and b thousand of B, an hour later there are Mab thousand, where M = 3122. (a) Find the characteristic polynomial of M, and hence its two eigenvalues. (b) Find an eigenvector for each eigenvalue. Which of them describes a mix of cells that can really occur, and what happens to a culture with that mix each hour?

λλ2− 5λ + 4det(M − λI) = (3 − λ)(2 − λ) − 1 × 2= λ2− 5λ + 4
det(M − λ I) = (3 − λ)(2 − λ) − 2 = λ2 − 5λ + 4, drawn as a graph of λ.
Subtract λ from each diagonal entry and take the determinant: det(M − λ I) = 3 − λ122 − λ = (3 − λ)(2 − λ) − 1 × 2 = λ2 − 5λ + 4.
step 1 of 5

The eigenvalues of a matrix are the roots of its characteristic polynomial, det(M − λ I): the equation Mv = λv has a solution other than v = 0 only when M − λ I has determinant 0. For each eigenvalue, the equations (M − λ I)v = 0 then give an eigenvector.

  1. Subtract λ from each diagonal entry and take the determinant: det(M − λ I) = 3 − λ122 − λ = (3 − λ)(2 − λ) − 1 × 2 = λ2 − 5λ + 4.
  2. (a) The characteristic polynomial is λ2 − 5λ + 4. Setting it to zero, (λ − 1)(λ − 4) = 0, so the eigenvalues are λ = 4 and λ = 1. Check: 4 + 1 = 5 = 3 + 2, the sum of the diagonal entries, and 4 × 1 = 4 = det M.
  3. For λ = 4: (M − 4I)v = −112−2ab = 0 gives −a + b = 0, so b = a and v = 11. Check: M11 = 44.
  4. For λ = 1: (M − I)v = 2121ab = 0 gives 2a + b = 0, so v = 1−2. Check: M1−2 = 3 − 22 − 4 = 1−2.
  5. (b) A culture cannot hold a negative number of cells, so 1−2 describes no real culture. The eigenvector 11 is a culture with equal numbers of the two strains, and each hour it is multiplied by 4: 2 thousand cells of each strain become 8 thousand of each.

answer(a) λ2 − 5λ + 4, so λ = 4 or λ = 1; (b) 11 for λ = 4 and 1−2 for λ = 1; only 11 can be a culture, and a culture with equal numbers of A and B is multiplied by 4 each hour

techniqueThe Characteristic Polynomial · Finding an Eigenvector

Common pitfalls

  • Writing det(M − λ I) = (3 − λ)(2 − λ) + 1 × 2, which gives λ2 − 5λ + 8 and no real roots. A 2 × 2 determinant subtracts the product of the other two entries: − 1 × 2.
  • Subtracting λ from every entry of M instead of only the diagonal. λ I has λ on the diagonal and 0 elsewhere, so the entries 1 and 2 off the diagonal stay as they are.
03

A Wildflower Spreading Across a Meadow, and the Mix of Seedlings and Flowering Plants It Settles To

methodFind the Eigenvalues from the Characteristic Polynomial, Take the Eigenvector of the Larger One as the Long-Run Mix and Its Eigenvalue as the Yearly Growth, and Check Against the Counts Year by Year

A wildflower on a meadow is counted each summer in two stages: seedlings s and flowering plants f. Each flowering plant produces 12 seedlings that are counted the next summer, half of the seedlings survive to flower the next year, and the flowering plants live on from year to year. So next summer's counts are Lsf, where L = 012121. The meadow starts with 10 flowering plants and no seedlings. (a) Find the eigenvalues of L. (b) Find an eigenvector for the larger eigenvalue, and use it to state the long-run growth of the wildflower each year and the long-run number of seedlings for each flowering plant.

years/f24684812det(L − λI) = −λ(1 − λ) − 12 × 1/2= λ2− λ − 6
det(L − λ I) = −λ(1 − λ) − 6 = λ2 − λ − 6. The points are the seedlings for each flowering plant, year by year.
det(L − λ I) = −λ12121 − λ = −λ(1 − λ) − 12 × 12 = λ2 − λ − 6.
step 1 of 5

Any count can be written as a mix of the two eigenvectors. Each year L multiplies the part along each eigenvector by that eigenvector's eigenvalue, so the part whose eigenvalue is larger in size soon outweighs the other. In the long run the counts grow by that eigenvalue each year and stay in the ratio of its eigenvector.

  1. det(L − λ I) = −λ12121 − λ = −λ(1 − λ) − 12 × 12 = λ2 − λ − 6.
  2. (a) λ2 − λ − 6 = (λ − 3)(λ + 2) = 0, so the eigenvalues are λ = 3 and λ = −2.
  3. For λ = 3, the first row of (L − 3I)v = 0 reads −3s + 12f = 0, so s = 4f and v = 41. Check: L41 = 122 + 1 = 123 = 341.
  4. (b) The eigenvalue 3 is larger in size than −2, so the part of the counts along 41 outweighs the rest by a factor of 32 more each year. In the long run the wildflower grows by a factor of 3 each year, with 4 seedlings for every flowering plant.
  5. Check against the counts, starting from 010: they run 12010, 12070, 840130, 1560550, 66001330, so the seedlings for each flowering plant are 12, 1.7, 6.5, 2.8, 5.0. They approach 4 from either side, because the other eigenvalue, −2, is negative.

answer(a) λ = 3 and λ = −2; (b) 41: in the long run the wildflower grows by a factor of 3 each year, with 4 seedlings for every flowering plant

techniqueFinding an Eigenvector · The Characteristic Polynomial · Eigenvalues and Eigenvectors

Common pitfalls

  • Reading the long-run mix from the eigenvector of −2, which is −61. A negative number of seedlings cannot occur, and it is the eigenvalue of larger size, 3, whose part grows fastest and sets the mix.
  • Reading the long-run mix from a single year's count, such as 6600 seedlings to 1330 flowering plants, about 5 to 1. The ratio swings either side of 4 for many years; the eigenvector gives the value it settles to.
04

A Metal Disc Pressed into an Oval, and the Two Directions It Is Stretched Along Without Turning

methodFind the Eigenvalues and Their Eigenvectors: the Eigenvectors Are the Directions Left on Their Own Lines, the Eigenvalues the Stretch Along Each, and Their Product the Area Factor

A press flattens a round metal disc of radius 1 dm, centered at the origin, into an oval. The change of shape is modeled by M = 5222. (a) Find the eigenvalues and eigenvectors of M, and describe the two directions through the center that stay on their own lines. (b) Show that these two directions are at right angles, and that the product of the two stretch factors is the factor by which the press multiplies the area of the disc.

xy−55det(M − λI) = (5 − λ)(2 − λ) − 2 × 2= (λ − 6)(λ − 1): λ = 6 or 1
det(M − λ I) = λ2 − 7λ + 6 = (λ − 6)(λ − 1): the eigenvalues are 6 and 1.
det(M − λ I) = (5 − λ)(2 − λ) − 2 × 2 = λ2 − 7λ + 6 = (λ − 6)(λ − 1), so the eigenvalues are 6 and 1.
step 1 of 5

The directions a transformation stretches without turning are its eigenvectors, and each eigenvalue is the stretch along its own direction. When the matrix is symmetric, with the same entry above and below the diagonal, the two directions are at right angles: they are the long and the short axis of the oval. The area factor, det M, is the product of the eigenvalues.

  1. det(M − λ I) = (5 − λ)(2 − λ) − 2 × 2 = λ2 − 7λ + 6 = (λ − 6)(λ − 1), so the eigenvalues are 6 and 1.
  2. For λ = 6: (M − 6I)v = −122−4xy = 0 gives x = 2y, so v = 21. Check: M21 = 126 = 621.
  3. For λ = 1: (M − I)v = 4221xy = 0 gives 2x + y = 0, so v = 1−2. Check: M1−2 = 5 − 42 − 4 = 1−2.
  4. (a) Along 21 the disc is stretched by a factor of 6, so its radius of 1 dm becomes 6 dm. Along 1−2 the eigenvalue is 1, so the radius stays 1 dm. These are the long and the short axis of the oval.
  5. (b) The scalar product of the two directions is 2 × 1 + 1 × (−2) = 0, so they are at right angles. The product of the stretches is 6 × 1 = 6 = det M = 5 × 2 − 2 × 2: the disc's area of π square decimeters becomes 6π ≈ 18.8 square decimeters.

answer(a) λ = 6 with eigenvector 21, the long axis, where the radius becomes 6 dm, and λ = 1 with eigenvector 1−2, the short axis, where it stays 1 dm; (b) the scalar product is 2 − 2 = 0, and 6 × 1 = 6 = det M, the area factor, so the area becomes 6π ≈ 18.8 square decimeters

techniqueFinding an Eigenvector · The Characteristic Polynomial · Eigenvalues and Eigenvectors

Common pitfalls

  • Taking the stretch directions to be the x- and y-axes, with factors 5 and 2 read off the diagonal. M sends 10 to 52, which is off the x-axis; only a diagonal matrix stretches along the axes.
  • Thinking that an eigenvalue of 1 does not count as a stretch direction. Mv = 1v says the direction stays on its line and keeps its length: the short axis of the oval is still 1 dm.
05

A Faulty Copier That Distorts a Poster, Written as Two Stretches Along Two Fixed Directions

methodFind the Eigenvalues and an Eigenvector for Each, Put the Eigenvectors in the Columns of P and the Eigenvalues Down D in the Same Order, and Check That P Times D Times the Inverse of P Is the Matrix

A faulty copier distorts every poster it copies. With the center of the poster at the origin, the distortion is modeled by M = 3102. A technician wants to describe it as two simple stretches along two fixed directions. (a) Find the eigenvalues of M and an eigenvector for each. (b) Write M = PDP−1, where D is a diagonal matrix, giving P, D and P−1, and check the product.

det(M − λI) = (3 − λ)(2 − λ)λ = 3 or λ = 2
The copier turns the square poster into the green parallelogram. det(M − λ I) = (3 − λ)(2 − λ), so λ = 3 or 2.
det(M − λ I) = (3 − λ)(2 − λ) − 1 × 0 = (3 − λ)(2 − λ), so the eigenvalues are λ = 3 and λ = 2.
step 1 of 5

If M has eigenvectors v1 and v2 with eigenvalues λ1 and λ2, put the eigenvectors in the columns of P and the eigenvalues down the diagonal of D, in the same order. Then MP = PD, so M = PDP−1: P−1 describes a point by its amounts along the two eigenvectors, D stretches each amount, and P puts the point back together.

  1. det(M − λ I) = (3 − λ)(2 − λ) − 1 × 0 = (3 − λ)(2 − λ), so the eigenvalues are λ = 3 and λ = 2.
  2. For λ = 3: (M − 3I)v = 010−1xy = 0 gives y = 0, so v = 10: horizontal lengths are stretched by a factor of 3.
  3. (a) For λ = 2: (M − 2I)v = 1100xy = 0 gives x + y = 0, so v = 1−1. Check: M1−1 = 3 − 1−2 = 2−2 = 21−1.
  4. Put the eigenvectors in the columns: P = 110−1 and D = 3002. det P = 1 × (−1) − 1 × 0 = −1, so P−1 = 1−1−1−101 = 110−1.
  5. (b) PD = 320−2, and PDP−1 = 320−2110−1 = 33 − 200 + 2 = 3102 = M. The copier stretches the poster by a factor of 3 along the horizontal and by a factor of 2 along the slanting direction 1−1.

answer(a) λ = 3 with eigenvector 10, and λ = 2 with eigenvector 1−1; (b) P = 110−1, D = 3002 and P−1 = 110−1, and PDP−1 = 3102 = M

techniqueDiagonalizing a 2 × 2 Matrix · Finding an Eigenvector

Common pitfalls

  • Putting the eigenvalues down D in a different order from the eigenvectors in P, such as D = 2003 with the same P. Then PDP−1 = 2−103, which is not M: each column of P must stand above its own eigenvalue.
  • Writing the eigenvectors in the rows of P instead of the columns. MP = PD multiplies each column of P by M, so it is each column that must be an eigenvector.
06

Rabbits Breeding Month by Month, Counted Ten Months Ahead by Diagonalization

methodDiagonalize the Breeding Matrix, Raise Only the Diagonal Matrix to the Power, and Multiply Back to Find the Counts Far Ahead

A breeder models young pairs y and adult pairs a of rabbits month by month. Each month every adult pair produces 2 young pairs, each young pair grows up into an adult pair, and the adult pairs live on. So next month's counts are Mya, where M = 0211. The breeder starts with one adult pair. (a) Write M = PDP−1, with D a diagonal matrix. (b) Use M10 = PD10P−1 to find the numbers of young and adult pairs after 10 months.

det(M − λI) = −λ(1 − λ) − 2 × 1= (λ − 2)(λ + 1): λ = 2 or −1
det(M − λ I) = λ2 − λ − 2 = (λ − 2)(λ + 1): the eigenvalues are 2 and −1.
det(M − λ I) = −λ(1 − λ) − 2 × 1 = λ2 − λ − 2 = (λ − 2)(λ + 1), so the eigenvalues are 2 and −1.
step 1 of 6

Multiplying M by itself ten times takes nine matrix products. Diagonalizing it first makes the power cheap: Mn = (PDP−1)(PDP−1) ⋯ (PDP−1) = PDnP−1, because each P−1P in the middle is I, and a diagonal matrix is raised to a power by raising each diagonal entry to that power.

  1. det(M − λ I) = −λ(1 − λ) − 2 × 1 = λ2 − λ − 2 = (λ − 2)(λ + 1), so the eigenvalues are 2 and −1.
  2. For λ = 2, the first row of (M − 2I)v = 0 reads −2y + 2a = 0, so y = a and v1 = 11. For λ = −1 it reads y + 2a = 0, so v2 = 2−1. Check: M2−1 = −21 = −1 × 2−1.
  3. (a) P = 121−1 and D = 200−1. det P = 1 × (−1) − 2 × 1 = −3, so P−1 = −13−1−2−11 = 13121−1, and M = PDP−1.
  4. D10 = 21000(−1)10 = 1024001. Work from the right: P−101 = 132−1, and then D10 gives 132048−1.
  5. Finally P gives 132048 − 22048 + 1 = 1320462049 = 682683.
  6. (b) After 10 months there are 682 young pairs and 683 adult pairs, 1365 pairs in all. Check by stepping month by month: 21, 23, 65, 1011, …, 342341, 682683.

answer(a) P = 121−1, D = 200−1 and P−1 = 13121−1; (b) 682 young pairs and 683 adult pairs after 10 months

techniqueMatrix Powers by Diagonalization · Diagonalizing a 2 × 2 Matrix

Common pitfalls

  • Writing M10 = P10D10(P−1)10. Only the diagonal matrix is raised to the power: each P−1 meets the P of the next factor and they cancel, which leaves one P at the front and one P−1 at the back.
  • Taking (−1)10 as −1. An even power of −1 is 1; with −1 the counts come out as 1320502047, which is not a whole number of pairs.
07

Voles and Barn Owls on a Farm, Predicted Four Years Ahead and in the Long Run

methodWrite the Starting Counts as a Mix of the Two Eigenvectors, Multiply Each Part by Its Own Eigenvalue Once for Each Year, and Let the Larger Eigenvalue Decide the Long-Run Ratio

An ecologist models the field voles and barn owls on a farm. If one spring there are x hundred voles and y tens of owls, the next spring there are Mxy, where M = 4−211: the voles breed fast but the owls eat them, and the owls do better when there are more voles. This spring there are 600 voles and 20 owls. (a) Find the eigenvalues and eigenvectors of M, and write the starting counts as a combination of the eigenvectors. (b) Find the numbers of voles and owls in four years' time, and the number of voles for each owl in the long run.

yearvoles per owl1234102030det(M − λI) = (4 − λ)(1 − λ) + 2= (λ − 3)(λ − 2): λ = 3 or 2
det(M − λ I) = λ2 − 5λ + 6 = (λ − 3)(λ − 2): the eigenvalues are 3 and 2.
det(M − λ I) = (4 − λ)(1 − λ) − (−2) × 1 = λ2 − 5λ + 6 = (λ − 3)(λ − 2), so the eigenvalues are 3 and 2.
step 1 of 6

Once the starting counts are written as pv1 + qv2, a year multiplies each part by its own eigenvalue, so after n years the counts are pλ1nv1 + qλ2nv2. This is Mn = PDnP−1 worked on one vector, and the part with the larger eigenvalue decides the ratio in the long run.

  1. det(M − λ I) = (4 − λ)(1 − λ) − (−2) × 1 = λ2 − 5λ + 6 = (λ − 3)(λ − 2), so the eigenvalues are 3 and 2.
  2. For λ = 3 the first row gives x − 2y = 0, so v1 = 21. For λ = 2 it gives 2x − 2y = 0, so v2 = 11. Check: M21 = 8 − 22 + 1 = 63 = 321.
  3. (a) Solve 62 = p21 + q11: 2p + q = 6 and p + q = 2, so p = 4 and q = −2. The starting counts are 421 − 211.
  4. Each year multiplies the first part by 3 and the second by 2, so after n years the counts are 4 × 3n21 − 2 × 2n11. For n = 4 this is 32421 − 3211 = 616292.
  5. (b) In four years there are 616 hundred voles, which is 61,600, and 292 tens of owls, which is 2920. As n grows, 3n outweighs 2n, so the counts approach the direction 21: 2 hundred voles for each ten owls, which is 20 voles for each owl.
  6. Check year by year: M62 = 208, then 6428, 20092 and 616292, and the voles for each owl run 30, 25, 22.9, 21.7, 21.1, approaching 20.

answer(a) λ = 3 with eigenvector 21 and λ = 2 with eigenvector 11, and 62 = 421 − 211; (b) 616 hundred voles, which is 61,600, and 292 tens of owls, which is 2920, after four years, and 20 voles for each owl in the long run

techniqueMatrix Powers by Diagonalization · Finding an Eigenvector

Common pitfalls

  • Multiplying the starting counts by 34 = 81, as if every part grew by the larger eigenvalue, which gives 486 hundred voles. The part along 11 grows by only 2 each year and is subtracted, so the answer is 616 hundred.
  • Reading the long-run ratio as 2 voles for each owl straight from the eigenvector 21. The voles are counted in hundreds and the owls in tens, so 2 hundred voles to 1 ten owls is 20 voles for each owl.
08

An Arrow in a Logo Animation Kept on Its Own Line by Choosing One Entry of the Matrix

methodWrite M Times the Arrow as a Multiple of the Arrow, Read the Eigenvalue from the Entries You Know and the Missing Entry from the Rest, Then Use the Characteristic Polynomial for the Second Eigenvalue

An animation program moves every point of a logo by the matrix M = 1k43, with the center of the logo at the origin. The designer wants the arrow in the logo, which points along 12, to stay on its own line as the logo moves. (a) Find k, and the factor by which the arrow is stretched. (b) Find the other eigenvalue of M and an eigenvector for it, and describe what happens to a line of the logo along that eigenvector.

xy5510vM ×12=1 + 2k10= λ ×12
The arrow v = 12 keeps its dashed line when Mv = λv, and Mv = 1 + 2k10.
The arrow stays on its line when M12 = λ12 for some number λ. Multiply: M12 = 1 + 2k4 + 6 = 1 + 2k10.
step 1 of 5

The arrow stays on its own line exactly when 12 is an eigenvector: M12 = λ12 for some number λ. Comparing entries gives two equations, one for λ and one for k. Once M is known, the characteristic polynomial gives its second eigenvalue.

  1. The arrow stays on its line when M12 = λ12 for some number λ. Multiply: M12 = 1 + 2k4 + 6 = 1 + 2k10.
  2. The second entries give 10 = 2λ, so λ = 5. The first entries then give 1 + 2k = 5, so k = 2.
  3. (a) k = 2, and the arrow is stretched by a factor of 5. Check: 124312 = 510 = 512.
  4. det(M − λ I) = (1 − λ)(3 − λ) − 2 × 4 = λ2 − 4λ − 5 = (λ − 5)(λ + 1), so the other eigenvalue is −1.
  5. (b) For λ = −1: (M + I)v = 2244xy = 0 gives x + y = 0, so v = 1−1, and M1−1 = −11. A line of the logo along 1−1 stays on its own line and keeps its length, but every point on it is sent to the other side of the center: the line is reversed.

answer(a) k = 2, and the arrow is stretched by a factor of 5; (b) λ = −1 with eigenvector 1−1: a line along it stays on its own line with its length unchanged, but is reversed through the center

techniqueEigenvalues and Eigenvectors · The Characteristic Polynomial

Common pitfalls

  • Setting M12 equal to 12 itself, which asks for the arrow not to move at all. The second entries then give 10 = 2, which is false, and it looks as if no k works. The arrow may be stretched; it only has to stay on its line, so the image is λ12.
  • Reading an eigenvalue of −1 as a direction that is not kept. Mv = −v puts the image on the same line through the origin, pointing the opposite way: a negative eigenvalue reverses a direction without turning it off its line.
09

Two Gyms in a Town Trading Members Each Year, and the Numbers They Settle To

methodShow That 1 Is an Eigenvalue, Share the Total Membership in the Ratio of Its Eigenvector for the Long Run, and Read the Other Eigenvalue as the Factor by Which the Gap Shrinks Each Year

A town has two gyms, A and B, with 1200 members between them. Each year 80% of A's members stay and 20% move to B, while 60% of B's members stay and 40% move to A. So if the gyms have a and b members, a year later they have Tab, where T = 0.80.40.20.6. This year A has 550 members and B has 650. (a) Show that 1 is an eigenvalue of T, and find the numbers of members the two gyms settle to. (b) Find the other eigenvalue, and use it to find the number of members at A after one, two and three years.

yearmembers12345400800ABdet(T − λI) = (0.8 − λ)(0.6 − λ) − 0.4 × 0.2= λ2− 1.4λ + 0.4
det(T − λ I) = λ2 − 1.4λ + 0.4. The points are this year's members at A and B.
det(T − λ I) = (0.8 − λ)(0.6 − λ) − 0.4 × 0.2 = λ2 − 1.4λ + 0.48 − 0.08 = λ2 − 1.4λ + 0.4.
step 1 of 6

An eigenvector with eigenvalue 1 is a pair of numbers that T leaves exactly as they are: the membership has settled. The other eigenvector moves members from one gym to the other without changing the total, and its eigenvalue is the factor by which that difference from the settled numbers shrinks each year.

  1. det(T − λ I) = (0.8 − λ)(0.6 − λ) − 0.4 × 0.2 = λ2 − 1.4λ + 0.48 − 0.08 = λ2 − 1.4λ + 0.4.
  2. This factorizes as (λ − 1)(λ − 0.4), so 1 is an eigenvalue, and the other eigenvalue is 0.4.
  3. For λ = 1: (T − I)v = 0 gives −0.2a + 0.4b = 0, so a = 2b and v = 21. Shared out in the ratio 2 : 1, the 1200 members give 800400.
  4. (a) The gyms settle to 800 members at A and 400 at B. Check: T800400 = 640 + 160160 + 240 = 800400: each year 160 members move each way.
  5. For λ = 0.4: 0.4a + 0.4b = 0, so v = 1−1, a move of members from one gym to the other. The starting numbers are 550650 = 800400 − 2501−1, and each year T multiplies the 250 by 0.4.
  6. (b) A is 250, then 100, then 40, then 16 members short of 800: it has 700 members after one year, 760 after two and 784 after three. Check the first year directly: 0.8 × 550 + 0.4 × 650 = 440 + 260 = 700.

answer(a) T21 = 21, so 1 is an eigenvalue, and the gyms settle to 800 members at A and 400 at B; (b) λ = 0.4: A has 700 members after one year, 760 after two and 784 after three

techniqueFinding an Eigenvector · Eigenvalues and Eigenvectors · The Characteristic Polynomial

Common pitfalls

  • Multiplying the eigenvector 21 by 1200 to get 2400 and 1200 members. The settled numbers must add up to the 1200 members there are, so 1200 is shared in the ratio 2 : 1: 800 and 400.
  • Writing A's fractions, 0.8 and 0.2, along the first row instead of down the first column. The new number at A is 0.8a + 0.4b, so the first row must hold 0.8 and 0.4; with the rows and columns swapped the matrix sends 550650 to 570610, and 20 members vanish.
10

Two Trolleys Joined by Springs Between Two Walls, and the Two Ways They Swing

methodFind the Eigenvalues and Eigenvectors of the Spring Matrix: Each Eigenvector Is a Way of Swinging That Keeps Its Shape, and the Square Roots of the Eigenvalues Compare How Fast They Swing

Two identical trolleys on a straight track are joined to each other and to two walls by three identical springs. When the trolleys are pushed x cm and y cm to the right of their resting places, the springs push back in proportion to Kxy, where K = 2−1−12. A way of swinging in which the trolleys keep the same shape of movement is called a mode: its shape is an eigenvector of K, and its frequency is proportional to the square root of its eigenvalue. (a) Find the eigenvalues of K and an eigenvector for each. (b) Describe the two modes, and find how many times as fast the faster mode swings, to 2 decimal places.

ABdet(K − λI) = (2 − λ)2− 1= (λ − 1)(λ − 3): λ = 1 or 3
det(K − λ I) = (2 − λ)2 − 1 = (λ − 1)(λ − 3): the eigenvalues are 1 and 3.
det(K − λ I) = (2 − λ)2 − (−1) × (−1) = λ2 − 4λ + 3 = (λ − 1)(λ − 3), so the eigenvalues are 1 and 3.
step 1 of 5

In a mode, the push back Kv is a multiple of the displacement v itself, so both trolleys are pulled back in step and the shape of the movement never changes: v is an eigenvector of K. A larger eigenvalue means a stronger push back for the same displacement, and so a faster swing.

  1. det(K − λ I) = (2 − λ)2 − (−1) × (−1) = λ2 − 4λ + 3 = (λ − 1)(λ − 3), so the eigenvalues are 1 and 3.
  2. (a) For λ = 1: (K − I)v = 0 gives x − y = 0, so v = 11. For λ = 3: −x − y = 0, so v = 1−1. Check: K1−1 = 3−3 = 31−1.
  3. In the mode 11 the trolleys move the same distance the same way. The middle spring keeps its length, so only the two outer springs push back, and the eigenvalue is the smaller one, 1.
  4. In the mode 1−1 the trolleys move the same distance in opposite ways. The middle spring is stretched or squashed by twice that distance, so the push back is stronger, and the eigenvalue is 3.
  5. (b) The frequencies are in the ratio √3 : √1, so the mode in which the trolleys move in opposite ways swings √3 ≈ 1.73 times as fast as the mode in which they move together.

answer(a) λ = 1 with eigenvector 11, and λ = 3 with eigenvector 1−1; (b) in one mode the trolleys swing together and in the other they swing in opposite ways, and the second swings √3 ≈ 1.73 times as fast

techniqueThe Characteristic Polynomial · Finding an Eigenvector

Common pitfalls

  • Comparing the eigenvalues directly and saying that the faster mode swings 3 times as fast. The frequency is proportional to the square root of the eigenvalue, so the ratio is √3 ≈ 1.73.
  • Taking det(K − λ I) = (2 − λ)2 + 1 because of the two minus signs. The product of the entries off the diagonal is (−1) × (−1) = 1, and it is subtracted, which gives (2 − λ)2 − 1 and the eigenvalues 1 and 3.
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