Start from an eigenvalue
The characteristic polynomial of A = (4 −1; 2 1) is , so its eigenvalues are 3 and 2. Each eigenvalue has its own eigenvector, and each is found the same way: solve .
Start with . Take 3 off each entry of the main diagonal of A: 4 − 3 = 1 and 1 − 3 = −2, while −1 and 2 stay. So A − 3I = (1 −1; 2 −2).
Multiply it out
An eigenvector for 3 is a column v = (x; y) with A v = 3v, which is the same as (A − 3I) v = 0: it is whatever A − 3I sends to zero. Multiply out: (1 −1; 2 −2) times (x; y) is (x − y; 2x − 2y), and this must be (0; 0). That gives two equations, x − y = 0 and 2x − 2y = 0.
The two rows say one thing
The second equation is twice the first, so both reduce to y = x. This is no accident. The eigenvalue 3 was chosen to make det(A − 3I) = 1 × (−2) − (−1) × 2 = 0, and a 2 × 2 matrix has determinant 0 exactly when one row is a multiple of the other.
It has to work out this way. If the two rows gave two different conditions, such as x − y = 0 and x + y = 0, the only column satisfying both would be x = 0, y = 0, and the zero column is never an eigenvector. So if your two rows disagree, check the eigenvalue and the subtraction before going on.
Choose the simplest one
Every point on the line y = x satisfies the condition, (1, 1), (2, 2), (−5, −5) and so on, as long as it is not the origin. An eigenvector names a direction, so any one of them will do; take the simplest, v = (1, 1). Check it: A sends (1; 1) to (4 − 1; 2 + 1) = (3; 3), which is 3 times (1; 1).
The other eigenvalue
Repeat at . Take 2 off the diagonal: 4 − 2 = 2 and 1 − 2 = −1, so A − 2I = (2 −1; 2 −1). Both rows read 2x − y = 0, so y = 2x.
The simplest point on y = 2x is (1, 2), so v = (1, 2). Check it: A sends (1; 2) to (4 − 2; 2 + 2) = (2; 4), which is 2 times (1; 2), so its eigenvalue is 2.
The eigenvectors for fill the line y = x, and those for fill the steeper line y = 2x. The dots are the two chosen, (1, 1) and (1, 2).
A second matrix
Take B = (2 3; 2 1). Its diagonal total is 3 and its determinant is 2 × 1 − 3 × 2 = −4, so its characteristic polynomial is , and its eigenvalues are 4 and −1.
At , B − 4I = (−2 3; 2 −3). The first row says −2x + 3y = 0, so 2x = 3y, and the second row says the same. A quick way to satisfy a row (p q) is the column (q, −p): from (−2, 3) take (3, 2), and −2 × 3 + 3 × 2 = 0. Check: B sends (3; 2) to (6 + 6; 6 + 2) = (12; 8), which is 4 times (3; 2).
At , take −1 off the diagonal, which adds 1: B + I = (3 3; 2 2). Both rows say x + y = 0, so v = (1, −1). Check: B sends (1; −1) to (2 − 3; 2 − 1) = (−1; 1), which is −1 times (1; −1). The eigenvalue is negative, so this direction is kept and reversed.
The usual mistakes
Taking off all four entries. A − 3I is (1 −1; 2 −2); taking 3 off every entry gives (1 −4; −1 −2), whose determinant is not 0, and then only the zero column solves it.
Writing the coordinates the wrong way round. y = 2x is satisfied by (1, 2), not by (2, 1): A sends (2, 1) to (7, 5), which is not a multiple of (2, 1).
Pairing an eigenvector with the wrong eigenvalue. (1, 1) belongs to 3 and (1, 2) belongs to 2; check by multiplying.
Giving (0, 0) as the answer. It solves for every , and an eigenvector must be non-zero.
A wildflower, a metal disc and two gyms
In the applications below, the eigenvector of the larger eigenvalue gives the mix of seedlings and flowering plants a wildflower settles to. The two eigenvectors of a press give the long and short axes of the oval it makes from a disc. And the eigenvector for the eigenvalue 1 gives the numbers of members two gyms settle to.
Worked example: A Wildflower Spreading Across a Meadow, and the Mix of Seedlings and Flowering Plants It Settles To
Question 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.
1.det(L − λ I) = −λ12121 − λ = −λ(1 − λ) − 12 × 12 = 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. 2.(a) λ2 − λ − 6 = (λ − 3)(λ + 2) = 0, so the eigenvalues are λ = 3 and λ = −2.
(a) (λ − 3)(λ + 2) = 0: the eigenvalues are 3 and −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.
For λ = 3, s = 4f: the eigenvector is 41, the dashed level of 4. 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.
(b) 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.
The counts from 10 flowering plants give 12, 1.7, 6.5, 2.8, 5.0, … seedlings for each, settling toward 4.
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
Common mistakes
- 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.
More eigenvalues and eigenvectors problems, worked step by step →
Worked example: A Metal Disc Pressed into an Oval, and the Two Directions It Is Stretched Along Without Turning
Question 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.
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.
det(M − λ I) = λ2 − 7λ + 6 = (λ − 6)(λ − 1): 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.
M21 = 621: the direction 21 stays on its own dashed line. 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.
M1−2 = 1−2: the direction 1−2 stays on its line and keeps its length. 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.
(a) The oval's long axis runs along (2, 1) with radius 6 dm; its short axis runs along (1, −2) with radius 1 dm. 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.
(b) The two axes are at right angles, and the area is multiplied by 6 × 1 = 6 = det M: from π to 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
Common mistakes
- 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.
More eigenvalues and eigenvectors problems, worked step by step →
Worked example: Two Gyms in a Town Trading Members Each Year, and the Numbers They Settle To
Question 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.
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.
det(T − λ I) = λ2 − 1.4λ + 0.4. The points are this year's members at A and B. 2.This factorizes as (λ − 1)(λ − 0.4), so 1 is an eigenvalue, and the other eigenvalue is 0.4.
(λ − 1)(λ − 0.4) = 0: 1 is an eigenvalue, and the other 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.
For λ = 1, a = 2b: the 1200 members shared in the ratio 2 : 1 are 800 and 400, the dashed levels. 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.
(a) T800400 = 800400: the gyms settle to 800 members at A and 400 at B. 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.
The start is 800400 − 2501−1, and T multiplies the 250 by 0.4 each year. 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.
(b) A has 700 members after one year, 760 after two and 784 after three, closing on 800.
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
Common mistakes
- 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.
More eigenvalues and eigenvectors problems, worked step by step →