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21. Eigenvalues and Eigenvectors

The video “21. Eigenvalues and Eigenvectors” by MIT OpenCourseWare as 6 lessons, with the idea explained, the code from the screen, steps to try in the practice terminal, practice problems with solutions and a quiz that leads back to the moment.

6lessons
4pieces of code
11formulas
19exercises
51min of video
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21. Eigenvalues and Eigenvectors, video by MIT OpenCourseWare▶ 51:22 · MIT OpenCourseWare
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0:006 lessons across the whole video51:22

You will be able to

  • Define eigenvalues and eigenvectors by the equation Ax = λx and check whether a given vector is an eigenvector.
  • Find eigenvalues by solving det(A − λI) = 0 for small matrices.
  • Compute the eigenvectors for each eigenvalue by solving (A − λI)x = 0.
  • Relate eigenvalues to the trace and determinant of a matrix.
  • Find eigenvalues of triangular matrices by reading the diagonal.
  • Recognize eigenvectors geometrically in examples such as projections and rotations.

Good to know first

  • Matrix-vector multiplication
  • Determinants of 2×2 and 3×3 matrices
  • Solving linear systems and finding null spaces

Syllabus

6 lessons and a final project

  1. 1

    Eigenvectors and the equation Ax = λx Free

    0:00–7:007 min of video

    An eigenvector of a square matrix A is a vector that A only stretches or shrinks, never turns, so Ax is parallel to x. The number by which it is stretched is the eigenvalue λ, and the equation Ax = λx captures this idea.

    3 build steps3 exercises3 questions
  2. 2

    Eigenvectors of projections and of a swap matrix Beyond Pass

    7:00–15:008 min of video

    An eigenvector is a vector that keeps its direction when a matrix acts on it, and its eigenvalue is the factor it gets stretched by. We find the eigenvectors and eigenvalues of a projection matrix (eigenvalues 1 and 0) and of the 2 by 2 swap matrix (eigenvalues 1 and -1) by asking which vectors stay on their own line.

    2 formulas3 build steps3 exercises3 questions
  3. 3

    Finding eigenvalues with det(A − λI) = 0 Beyond Pass

    15:00–25:0010 min of video

    An eigenvalue λ makes A − λI singular, so its determinant must be zero. Solving det(A − λI) = 0 gives the eigenvalues, and we can then find the eigenvectors from each one.

    1 code4 build steps3 exercises3 questions
  4. 4

    Computing eigenvectors for a 2×2 matrix Beyond Pass

    25:00–35:0010 min of video

    You learn the two-step recipe for finding eigenvalues and eigenvectors of a 2×2 matrix: first solve the determinant equation for λ, then solve (A − λI)x = 0 for each λ. The worked example uses A = [[3,1],[1,3]] and shows how the eigenvectors come out.

    2 code3 formulas6 build steps3 exercises3 questions
  5. 5

    Trace, determinant and the eigenvalues of a 2×2 matrix Beyond Pass

    35:00–43:208 min of video

    The trace (sum of diagonal entries) and the determinant of a matrix give two quick checks on its eigenvalues: they add up to the trace and multiply to the determinant. The chapter applies this to the 90° rotation matrix, whose eigenvalues turn out to be complex numbers, i and −i.

    3 formulas3 exercises3 questions
  6. 6

    Worked examples: eigenvalues of a rotation and a shear-like matrix, and the closing summary Beyond Pass

    43:20–51:228 min of video

    This lesson works through two eigenvalue computations from start to finish: the 90° rotation matrix, which has no real eigenvalues, and the matrix [[3,1],[0,3]], which has a repeated eigenvalue with only one independent eigenvector. It ends with the professor's closing checks using the trace and determinant.

    1 code3 formulas3 build steps4 exercises3 questions

Final project

Build a 2×2 Eigen-Solver and Verify It by Hand

Write a small program that finds the eigenvalues and eigenvectors of any 2×2 matrix with the recipe from the course: solve det(A − λI) = 0, then solve (A − λI)x = 0. Test it on the course examples: swap, projection, symmetric, rotation and the repeated-eigenvalue matrix. Then explain each result geometrically.

Its 6 tasks open with Beyond Pass.

Free preview · Lesson 1 in full

Eigenvectors and the equation Ax = λx

An eigenvector of a square matrix A is a vector that A only stretches or shrinks, never turns, so Ax is parallel to x. The number by which it is stretched is the eigenvalue λ, and the equation Ax = λx captures this idea.

0:00 to 7:00 in the video

Scratchpad

Draw and write next to the video.

The idea

Most vectors change direction when a matrix multiplies them. Multiplying by A takes a vector x to a new vector Ax, and usually Ax points in a different direction from x. The lecture asks: are there special vectors x that keep their direction? For those vectors, Ax lands on the same line through the origin as x. Such vectors are called eigenvectors, and the word 'eigen' is German for 'own' or 'characteristic' of the matrix.

If x is an eigenvector, then Ax is parallel to x, so Ax is just x multiplied by some number. We call that number λ (lambda), the eigenvalue. Writing this as an equation gives the central equation of the chapter: Ax=λx. The number λ tells you how much x is stretched (λ bigger than 1), shrunk (between 0 and 1), flipped (negative), or squashed to zero (λ = 0).

Here is a concrete picture. Imagine a matrix that stretches everything in the horizontal direction by 2 and leaves vertical things alone. A horizontal vector (1, 0) becomes (2, 0): same direction, so it is an eigenvector with λ = 2. A vertical vector (0, 1) stays (0, 1): an eigenvector with λ = 1. A diagonal vector such as (1, 1) becomes (2, 1), which points somewhere new, so it is not an eigenvector.

Why does this matter? Eigenvectors show the directions in which a matrix acts in the simplest way: pure stretching. Many problems, from vibrations to population growth, become easy once you find these directions. The lecture also points to a useful check: if A is singular (its columns do not give full information, so Ax = 0 has a nonzero solution), then λ = 0 is an eigenvalue, because Ax = 0x = 0 for that x.

The lecture finishes by posing a question to work on: for a projection matrix, which vectors x and which numbers λ are the eigenvectors and eigenvalues? Think about a projection onto a line: vectors on the line stay put, and vectors perpendicular to the line are sent to zero. Keep this picture in mind, because it gives the answer in the next part of the course.

Key points

  • An eigenvector x of A satisfies Ax = λx: A only stretches or shrinks it, so Ax is parallel to x. 2:41
  • The eigenvalue λ is the factor by which x is stretched: bigger than 1 stretches, between 0 and 1 shrinks, negative flips, and 0 squashes x to zero. 3:30
  • Eigenvectors are the special directions in which a matrix acts in the simplest way. 3:00
  • If A is singular, then λ = 0 is an eigenvalue, since Ax = 0 = 0x for a nonzero x. 5:21

Build along in the practice terminal

  1. On paper, draw the vector x = (1, 0) as an arrow from the origin, then draw A x for the matrix A = [[2, 0], [0, 1]]. Check whether the two arrows point in the same direction.
  2. Compute A times (1, 0): you get (2, 0). Write Ax = λx and find λ: 2 times (1, 0) gives λ = 2. So (1, 0) is an eigenvector with eigenvalue 2.
  3. Repeat with x = (1, 1). Compute A times (1, 1) and see whether the result is a multiple of (1, 1). It is not, so (1, 1) is not an eigenvector.

Watch out

  • The zero vector is never called an eigenvector, because A0 = λ0 for every λ. Eigenvectors must be nonzero.
  • λ = 0 is allowed as an eigenvalue (the eigenvector is then sent to zero), but the eigenvector itself is still not the zero vector.
  • A vector that changes direction, like (1, 1) for A = [[2, 0], [0, 1]], is not an eigenvector. Ax must be a multiple of x.

Practice

Calculate Solution checked

A matrix A stretches the vector x = (1, 2) to A x = (3, 6). What is the eigenvalue λ for this eigenvector?

Hint

Find the number k such that (3, 6) = k(1, 2).

Solution
  1. Check that Ax is a multiple of x: (3, 6) = 3 times (1, 2).
  2. So Ax = 3x, which means λ = 3.

3

Explain Solution checked

In your own words, explain what it means for x to be an eigenvector of A, and what the eigenvalue tells you.

Solution
  1. An eigenvector is a nonzero vector whose direction does not change when A multiplies it. A x is just a stretched or shrunk copy of x, and the eigenvalue λ is the stretch factor (negative means it flips direction).

An eigenvector x satisfies Ax = λx: A keeps its direction and only scales it by λ.

Do it Solution checked

Using the matrix A = [[2, 0], [0, 1]] from the steps, find one vector that is an eigenvector with eigenvalue 1 and check it by computing Ax.

Solution
  1. Try x = (0, 1).
  2. A x = (2·0 + 0·1, 0·0 + 1·1) = (0, 1) = 1 times x.
  3. So (0, 1) is an eigenvector with λ = 1.

x = (0, 1) with eigenvalue 1

On your scratchpad

Draw a horizontal arrow and a vertical arrow from the origin. Then draw a diagonal arrow and apply the matrix that doubles horizontal lengths: show where each arrow goes and label which ones keep their direction.

Quiz

  1. If Ax = λx with x not zero, what does this say about Ax?

    1. Ax is perpendicular to x
    2. Ax is parallel to x
    3. Ax is the zero vector
    4. Ax has length zero
    Show the answer

    B. Ax is parallel to x · The equation says Ax equals λ times x, so Ax lies on the same line through the origin as x.

    2:41 watch the moment

  2. For the matrix that doubles the horizontal direction and leaves the vertical alone, which vector is an eigenvector with eigenvalue 1?

    1. (1, 0)
    2. (0, 1)
    3. (1, 1)
    4. (1, -1)
    Show the answer

    B. (0, 1) · (0, 1) is sent to itself, so λ = 1. (1, 0) gets eigenvalue 2 instead, and (1, 1) is not an eigenvector at all.

    3:30 watch the moment

  3. Why is λ = 0 an eigenvalue of a singular matrix A?

    1. Because Ax = 0 for some nonzero x, which is 0 times x
    2. Because the trace is zero
    3. Because A is the identity
    4. Because the determinant of A is one
    Show the answer

    A. Because Ax = 0 for some nonzero x, which is 0 times x · If A is singular, some nonzero x has Ax = 0 = 0·x, which is exactly the eigenvector equation with λ = 0.

    5:21 watch the moment

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