Skip to main content

Vectors that stay on their line

Most vectors are knocked off their line by a matrix, but a few are only stretched, shrunk or flipped. Meet eigenvectors and eigenvalues as a picture, check a claimed eigenvector with one multiplication, and find them by sweeping round the circle in code.
M03.1.1L2 Intermediate25 min
In review

The maths and code on this page are checked by automated tests on every change, but an independent reader has not signed it off yet. Spot a mistake or a confusing step? Tell us.

Before this lesson: M02.4 Matrices as linear maps M02.5 Matrix multiplication as composition M02.9 Change of basis

What you will be able to do
  • State the definition of an eigenvector and its eigenvalue, and read it as a picture: the map keeps the vector on its own line and stretches it by the eigenvalue.
  • Check a claimed eigenvector with one matrix-vector product, and read off its eigenvalue.
  • Read the eigenvalues of a projection, a reflection and a scaling from their geometry, without algebra.
  • Find the eigenvector directions of a 2 by 2 matrix numerically by sweeping unit vectors round the circle.

The question

Feed a vector through the same matrix again and again, as a deep network does when its layers share one weight matrix, or as PageRank does when it follows links click after click. Most vectors swing round and change direction at every step. Are there directions that the matrix never turns, only stretches? If so, how do we find them, and what do they tell us about the matrix?

Intuition

A matrix moves every vector of the plane (lesson M02.4). Take a vector, draw the line through it and the origin, and look at where the matrix sends the vector. Almost always the image has left that line: the matrix has turned it.

Take the running matrix of this course,

\[ \mat{A} = \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix}. \]

It sends \(\ihat = (1, 0)\) to \((4, 2)\) and \(\jhat = (0, 1)\) to \((1, 3)\). Both are turned. But the vector \((1, 1)\) goes to \({(4 + 1,\; 2 + 3)} = (5, 5)\), which lies on the very same line, five times as far out.

Two square panels on a grid. Left panel, titled Most vectors are turned: blue arrows i-hat and j-hat along the axes, each lying on a pale blue band through the origin. Vermillion arrows show their images A i-hat, pointing to (4, 2), and A j-hat, pointing to (1, 3). Small green curved arrows mark that each image has turned away from the band of its input. A note reads: each image leaves the shaded line of its input. Right panel, titled Eigenvectors stay on their line: two pale orange bands cross at the origin, one along the direction (1, 1) and one along (1, minus 2). Short orange arrows v1 and v2 lie on them. Long vermillion arrows lie on the same bands: A v1 equals 5 v1, reaching (5, 5), and A v2 equals 2 v2, reaching (2, minus 4). Labels lambda 1 equals 5 and lambda 2 equals 2 sit beside the bands, with the note: on their own lines, only stretched.
M03.1.1-F1

The running matrix \(\mat{A}\) turns most vectors off their line (left). Two directions are special: along \(\vect{v}_1 = (1, 1)\) and \(\vect{v}_2 = (1, -2)\) the image stays on the line of the input, stretched 5 times and 2 times (right).

A vector that a matrix keeps on its own line is an eigenvector of the matrix, and the factor it is stretched by is its eigenvalue. “Eigen” is German for “own”: the vector keeps its own line. For \(\mat{A}\) there are two such lines, and on them the whole action of the matrix is a plain stretch. That is why eigenvectors explain a matrix: along them, a complicated map becomes simple multiplication.

Build it up

Step 1. The test is “same line”, not “same arrow”. A vector \(\vect{v}\) and its image \(\mat{A}\vect{v}\) lie on one line through the origin exactly when one is a multiple of the other: \({\mat{A}\vect{v}} = {\lambda\,\vect{v}}\) for some number \(\lambda\), written with the Greek letter lambda by long custom.

Step 2. The number says what happens along the line. If \(\lambda = 2\), the vector doubles. If \(\lambda = \frac12\), it halves. If \(\lambda = -1\), it flips to point the other way along the same line. If \(\lambda = 0\), it is crushed to the origin.

Four rows, each a pale orange band standing for the line of a vector v, with a short tick for the origin. In every row an orange arrow v points right from the origin, and a vermillion arrow A v sits just under it. Row 1, labelled lambda equals 2, stretched: A v is twice as long as v. Row 2, lambda equals one half, shrunk: A v is half as long. Row 3, lambda equals minus 1, flipped: A v has the same length but points left. Row 4, lambda equals 0, sent to the origin: A v is only a dot at the origin.
M03.1.1-F2

What the eigenvalue \(\lambda\) does to \(\vect{v}\), always along the line of \(\vect{v}\): stretch, shrink, flip, or collapse to the origin.

Step 3. Some maps tell you their eigenvalues by their shape. Three maps built on the lines \(y = x\) and \(y = -x\) need no algebra at all.

Three small square panels. First, titled Projection onto y equals x: two pale orange bands along the diagonals. On the band y equals x an orange arrow and its vermillion image coincide, labelled lambda equals 1. On the band y equals minus x an orange arrow is labelled lambda equals 0, and a vermillion dot at the origin is labelled sent to 0. Second, titled Reflection across y equals x: on the band y equals x the arrow and its image coincide, labelled lambda equals 1; on the other band the vermillion image points the opposite way from the orange arrow, labelled flipped and lambda equals minus 1. Third, titled Scaling by 2: four orange arrows in different directions, each with a vermillion image twice as long on the same pale band, and the note every line, lambda equals 2.
M03.1.1-F4

Eigenvalues read off by geometry. A projection keeps its target line (\(\lambda = 1\)) and crushes the line it projects along (\(\lambda = 0\)). A reflection keeps its mirror (\(\lambda = 1\)) and flips the perpendicular line (\(\lambda = -1\)). A scaling by 2 doubles every vector, so every line is an eigenvector line with \(\lambda = 2\).

  • The projection onto the line \(y = x\) leaves a vector on that line where it is, so those vectors have \(\lambda = 1\). It sends a vector on the line \(y = -x\) to the origin, so those have \(\lambda = 0\).
  • The reflection across \(y = x\) leaves the mirror line alone, \(\lambda = 1\), and turns every vector on \(y = -x\) round, \(\lambda = -1\).
  • The scaling by 2 doubles every vector, so every non-zero vector is an eigenvector with \(\lambda = 2\).

The formal statement

NoteDefinition: eigenvector and eigenvalue

Let \(\mat{A}\) be an \({n \times n}\) matrix. A non-zero vector \(\vect{v}\) in \(\R^n\) is an eigenvector of \(\mat{A}\) if there is a number \(\lambda\) with

\[ \mat{A}\vect{v} = \lambda\,\vect{v}. \]

The number \(\lambda\) is the eigenvalue that belongs to \(\vect{v}\). A number \(\lambda\) is an eigenvalue of \(\mat{A}\) when at least one non-zero \(\vect{v}\) satisfies the equation.

The matrix must be square, because \(\mat{A}\vect{v}\) and \(\vect{v}\) have to live in the same space to be compared. The vector must be non-zero, because \({\mat{A}\vect{0}} = \vect{0} = {\lambda\,\vect{0}}\) for every number \(\lambda\): allowing the zero vector would make every number an eigenvalue and the word would mean nothing. The eigenvalue itself may be zero.

Consequence 1: an eigenvector is really a line. If \(\vect{v}\) is an eigenvector with eigenvalue \(\lambda\) and \(c \neq 0\) is any number, then \(c\,\vect{v}\) is one too, with the same eigenvalue:

\[ \begin{aligned} \mat{A}(c\,\vect{v}) &= c\,(\mat{A}\vect{v}) \\ &= c\,(\lambda\,\vect{v}) \\ &= \lambda\,(c\,\vect{v}). \end{aligned} \]

The first step is linearity (M02.4.1), the second uses \({\mat{A}\vect{v}} = {\lambda\vect{v}}\), the third reorders the numbers. So \((1, 1)\), \((2, 2)\) and \((-3, -3)\) are all eigenvectors of \(\mat{A}\) with \(\lambda = 5\). What matters is the line, not the length or the sign.

Consequence 2: in the plane, “same line” is one number. Two vectors \(\vect{v} = (v_1, v_2)\) and \(\vect{w} = (w_1, w_2)\) lie on one line through the origin exactly when the parallelogram they span is flat, that is, when the determinant with them as columns is zero (lesson M02.8.3):

\[ v_1 w_2 - v_2 w_1 = 0 . \]

This number is often called the two-dimensional cross product. With \(\vect{w} = \mat{A}\vect{v}\) it gives a test that a computer can run on any direction.

Consequence 3: the eigenvalue from the vector. If \({\mat{A}\vect{v}} = {\lambda\vect{v}}\), take the dot product of both sides with \(\vect{v}\):

\[ \begin{aligned} \vect{v} \cdot \mat{A}\vect{v} &= \lambda\,(\vect{v} \cdot \vect{v}), \\ \lambda &= \frac{\vect{v} \cdot \mat{A}\vect{v}}{\vect{v} \cdot \vect{v}} . \end{aligned} \]

Division is allowed because \({\vect{v} \cdot \vect{v}} = {\norm{\vect{v}}^2} > 0\) for a non-zero vector. For a unit vector the denominator is 1.

Go deeper: why a straight line through the origin is the natural object

A line through the origin is a one-dimensional subspace: it is closed under adding and scaling. A matrix keeps such a line in place, \({\mat{A}L} \subseteq L\), exactly when the line is spanned by an eigenvector. Axler builds the whole theory on this idea: an operator has a one-dimensional subspace that it maps into itself if and only if it has an eigenvalue (Section 5A). In higher dimensions the same question can be asked of planes and larger subspaces, and the answers lead to the decompositions of the rest of this course.

Worked example

Check that \((1, 1)\) and \((1, -2)\) are eigenvectors of \(\mat{A}\), find their eigenvalues, and show that \((1, 0)\) is not an eigenvector.

Three rows of boxes. Each row starts with a blue box holding an input v, followed by a green arrow labelled multiply by A with the arithmetic underneath, ending in a vermillion box holding the image A v, and then a verdict. Row 1: (1, 1) becomes (4 plus 1, 2 plus 3) equals (5, 5), which equals 5 times (1, 1): eigenvector, lambda equals 5. Row 2: (1, minus 2) becomes (4 minus 2, 2 minus 6) equals (2, minus 4), which equals 2 times (1, minus 2): eigenvector, lambda equals 2. Row 3: (1, 0) becomes (4 plus 0, 2 plus 0) equals (4, 2), which is not a multiple of (1, 0): not an eigenvector.
M03.1.1-F3

One multiplication per claim. An image that is a multiple of the input confirms an eigenvector and names its eigenvalue; any other image refutes the claim.

Each product is “row times column” for each entry (M02.4.2).

\[ \begin{aligned} \mat{A}\begin{bmatrix} 1 \\ 1 \end{bmatrix} &= \begin{bmatrix} 4 \cdot 1 + 1 \cdot 1 \\ 2 \cdot 1 + 3 \cdot 1 \end{bmatrix} \\[4pt] &= \begin{bmatrix} 5 \\ 5 \end{bmatrix} = 5 \begin{bmatrix} 1 \\ 1 \end{bmatrix}. \end{aligned} \]

So \(\vect{v}_1 = (1, 1)\) is an eigenvector with \(\lambda_1 = 5\). Next,

\[ \begin{aligned} \mat{A}\begin{bmatrix} 1 \\ -2 \end{bmatrix} &= \begin{bmatrix} 4 - 2 \\ 2 - 6 \end{bmatrix} = \begin{bmatrix} 2 \\ -4 \end{bmatrix} \\[4pt] &= 2 \begin{bmatrix} 1 \\ -2 \end{bmatrix}, \end{aligned} \]

so \(\vect{v}_2 = (1, -2)\) is an eigenvector with \(\lambda_2 = 2\). Last, \({\mat{A}\,(1, 0)} = (4, 2)\). A multiple of \((1, 0)\) has second entry 0, and this one has 2, so \((1, 0)\) is not an eigenvector. The cross-product test agrees: \({1 \cdot 2 - 0 \cdot 4} = 2 \neq 0\), while for \(\vect{v}_1\) it gives \({1 \cdot 5 - 1 \cdot 5} = 0\).

Consequence 3 recovers the eigenvalue from the vector: \({\vect{v}_2 \cdot \mat{A}\vect{v}_2} = {1 \cdot 2 + (-2)(-4)} = 10\) and \({\vect{v}_2 \cdot \vect{v}_2} = 5\), so \(\lambda_2 = 10 / 5 = 2\).

The geometric examples, checked. The projection onto \(y = x\) and the reflection across it have the matrices

\[ \mat{M} = \frac12\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}, \qquad \mat{N} = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}. \]

Then \({\mat{M}\,(1, 1)} = {\tfrac12\,(2, 2)} = (1, 1)\) and \({\mat{M}\,(1, -1)} = {\tfrac12\,(0, 0)} = (0, 0)\), giving 1 and 0. The reflection swaps the two entries, so \({\mat{N}\,(1, 1)} = (1, 1)\) and \({\mat{N}\,(1, -1)} = (-1, 1)\), giving 1 and \(-1\). The pictures were right.

Code

From scratch: sweep a unit vector round the circle, one degree at a time, and watch the cross-product test of Consequence 2 change sign. Between two neighbouring angles where it changes sign there is an eigenvector line; halving the gap again and again pins it down.

import math

A = [[4, 1], [2, 3]]

def apply(M, v):
    return (M[0][0] * v[0] + M[0][1] * v[1], M[1][0] * v[0] + M[1][1] * v[1])

def cross(degrees):
    """v1 w2 - v2 w1 for the unit vector v at this angle and its image w = A v."""
    a = math.radians(degrees)
    v = (math.cos(a), math.sin(a))
    w = apply(A, v)
    return v[0] * w[1] - v[1] * w[0]

def refine(lo, hi, steps=50):
    """Halve the bracket [lo, hi] until the sign change is pinned to a point."""
    for _ in range(steps):
        mid = (lo + hi) / 2
        if (cross(lo) > 0) == (cross(mid) > 0):
            lo = mid
        else:
            hi = mid
    return (lo + hi) / 2

found = []
for d in range(360):
    if (cross(d) > 0) != (cross(d + 1) > 0):
        angle = refine(d, d + 1)
        a = math.radians(angle)
        v = (math.cos(a), math.sin(a))
        w = apply(A, v)
        lam = v[0] * w[0] + v[1] * w[1]          # Consequence 3, with v of length 1
        found.append((round(angle, 4), round(lam, 6)))
print(found)
[(45.0, 5.0), (116.5651, 2.0), (225.0, 5.0), (296.5651, 2.0)]

Four angles, two lines: \(45^\circ\) and \(225^\circ\) are the two halves of the line of \((1, 1)\), with \(\lambda = 5\), and about \(116.57^\circ\) and \(296.57^\circ\) are the line of \((1, -2)\), with \(\lambda = 2\).

A graph against the direction of v, from 0 to 360 degrees. A reddish purple wave shows the cross product v1 w2 minus v2 w1, ranging from about minus 1 to 2. It crosses zero four times, at 45, about 116.6, 225 and about 296.6 degrees. At each crossing an orange dot sits on the axis inside a pale orange vertical band, labelled lambda equals 5, lambda equals 2, lambda equals 5 and lambda equals 2. A note reads: zero where w equals A v lies on the line of v.
M03.1.1-F6

The sweep of the code. The test is zero exactly where \(\mat{A}\vect{v}\) lies on the line of \(\vect{v}\): twice for each eigenvector line, once for each of its two directions.

NumPy checks a claim in one line, and the same check works for any size of matrix.

import numpy as np

A = np.array([[4.0, 1.0], [2.0, 3.0]])
for v, lam in [((1, 1), 5), ((1, -2), 2), ((1, 0), None)]:
    v = np.array(v, dtype=float)
    image = A @ v
    print(v, "->", image, " eigenvector:", lam is not None and np.allclose(image, lam * v))
[1. 1.] -> [5. 5.]  eigenvector: True
[ 1. -2.] -> [ 2. -4.]  eigenvector: True
[1. 0.] -> [4. 2.]  eigenvector: False

The sweep found the lines without being told them, but it only works in the plane, and it only notices a change of sign. It can miss two lines that are very close together, and it misses a line where the test touches zero without changing sign: for the shear of lesson M03.1.3 the test is \({-v_2^2}\), never positive, so the sweep finds nothing. The next lesson finds eigenvalues exactly, by algebra.

Explore

M03.1-X1

Eigenvector hunter

Turn the unit vector \(\vect{u}\) with the slider, the step buttons or by dragging its tip, and watch its image \(\mat{A}\vect{u}\). When the two line up, the line lights up and the read-out gives \(\lambda\). Choose another matrix with the buttons, or set its entries.

A static picture in place of the interactive: on the left, the basis vectors i-hat and j-hat and their images under A, both turned off their lines; on the right, the eigenvectors (1, 1) and (1, minus 2) of A, whose images 5 times and 2 times as long stay on the same pale orange lines.

Things to try:

  1. With \(\mat{A}\), press Next line-up four times. Which angles do you stop at, and which eigenvalue belongs to each?
  2. Choose S. Its eigenvector lines are perpendicular. Is that true for \(\mat{A}\)?
  3. Choose Shear. How many lines light up as you go once round?
  4. Choose Rotation or T, and press Play. Does anything ever line up?

Pitfalls

Two square panels with the same arrows: an orange arrow v pointing down to the right along the direction (1, minus 1), and a vermillion arrow N v of the same length pointing up to the left, its exact opposite. Left panel, marked with a cross and titled Test: same direction?: a thin dashed line, with the note points the other way, so not an eigenvector. Right panel, marked with a tick and titled Test: same line?: both arrows lie on one pale orange band, with the notes same line, flipped: N v equals minus 1 v, and lambda equals minus 1.
M03.1.1-F5

Wrong beside right. The reflection \(\mat{N}\) sends \((1, -1)\) to \((-1, 1)\). Testing the direction rejects it; testing the line accepts it with \(\lambda = -1\), which is correct.

“An eigenvector must keep its direction.”

It must keep its line. A negative eigenvalue flips the vector to the other side of the origin, and that still counts: the reflection \(\mat{N}\) has the eigenvalue \(-1\). In data science a negative eigenvalue is a real signal, for example the direction in which a saddle-shaped loss curves downwards (M03.4).

“The zero vector is an eigenvector of every matrix.”

It satisfies \({\mat{A}\vect{0}} = {\lambda\vect{0}}\), but for every \(\lambda\) at once, so it carries no information. The definition excludes it. The eigenvalue 0 is allowed, and it means that a whole line is crushed to the origin: the matrix is not invertible.

“Each eigenvalue has one eigenvector.”

Every non-zero multiple of an eigenvector is another one (Consequence 1). Software has to pick one, and it usually picks a vector of length 1, with either sign. When NumPy prints \((-0.71, -0.71)\) where you wrote \((1, 1)\), nothing is wrong: it is the same line.

Check yourself

Recall

1. State the definition of an eigenvector and its eigenvalue.

Show answer

A non-zero vector \(\vect{v}\) with \({\mat{A}\vect{v}} = {\lambda\vect{v}}\) for some number \(\lambda\); that number is the eigenvalue belonging to \(\vect{v}\).

2. Why is the zero vector never called an eigenvector?

Show answer

Because \({\mat{A}\vect{0}} = {\lambda\vect{0}}\) holds for every \(\lambda\), so it would make every number an eigenvalue.

3. What are the eigenvalues of the projection onto a line, and of the reflection across a line, in the plane?

Show answer

The projection: 1 (vectors on the line) and 0 (vectors it projects along). The reflection: 1 (the mirror line) and \(-1\) (the perpendicular line).

4. How do you check, with one multiplication, that \(\vect{v}\) is an eigenvector?

Show answer

Compute \(\mat{A}\vect{v}\) and see whether it is a multiple of \(\vect{v}\). If it is, the multiple is the eigenvalue.

5. Which test tells, in the plane, whether \(\vect{v}\) and \(\mat{A}\vect{v}\) lie on one line?

Show answer

With \(\vect{w} = \mat{A}\vect{v}\), the cross product \({v_1 w_2 - v_2 w_1}\) is zero: the determinant with columns \(\vect{v}\) and \(\vect{w}\) vanishes.

Exercises

1. ★ For \(\mat{S} = \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}\), decide which of \((1, 1)\), \((1, -1)\) and \((2, 1)\) are eigenvectors, and give the eigenvalues.

Show solution

\({\mat{S}\,(1, 1)} = {(2 + 1,\; 1 + 2)} = (3, 3) = {3\,(1, 1)}\): an eigenvector with \(\lambda = 3\). \({\mat{S}\,(1, -1)} = {(2 - 1,\; 1 - 2)} = (1, -1)\): an eigenvector with \(\lambda = 1\). \({\mat{S}\,(2, 1)} = {(4 + 1,\; 2 + 2)} = (5, 4)\). The cross product is \({2 \cdot 4 - 1 \cdot 5} = 3 \neq 0\), so \((2, 1)\) is not an eigenvector.

2. ★★ A matrix \(\mat{A}\) has the eigenvector \(\vect{v}\) with eigenvalue 3. Show that \(\vect{v}\) is an eigenvector of \(\mat{A}^2 = \mat{A}\mat{A}\) and find its eigenvalue. Check on the running \(\mat{A}\) with \(\vect{v} = (1, -2)\), using \(\mat{A}^2 = \begin{bmatrix} 18 & 7 \\ 14 & 11 \end{bmatrix}\).

Show solution

Apply \(\mat{A}\) twice and use the eigenvector equation each time:

\[ \begin{aligned} \mat{A}^2\vect{v} &= \mat{A}(\mat{A}\vect{v}) = \mat{A}(3\,\vect{v}) \\ &= 3\,\mat{A}\vect{v} = 3 \cdot 3\,\vect{v} = 9\,\vect{v}. \end{aligned} \]

So \(\vect{v}\) is an eigenvector of \(\mat{A}^2\) with eigenvalue 9. On the running \(\mat{A}\) the eigenvalue of \((1, -2)\) is 2, so the same steps predict \(2^2 = 4\), and indeed \({\mat{A}^2\,(1, -2)} = {(18 - 14,\; 14 - 22)} = (4, -8) = {4\,(1, -2)}\).

3. ★★★ Show that a rotation of the plane by \(90^\circ\), \(\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}\), has no real eigenvector, first by geometry and then with the cross-product test.

Show solution

Geometry. The rotation turns every non-zero vector by a right angle, so its image is perpendicular to it and never on its line.

Algebra. The rotation sends \((v_1, v_2)\) to \(\vect{w} = (-v_2, v_1)\). The cross product is

\[ \begin{aligned} &v_1 w_2 - v_2 w_1 \\ &\quad= v_1 v_1 - v_2 (-v_2) \\ &\quad= v_1^2 + v_2^2 , \end{aligned} \]

which is the squared length of \(\vect{v}\) and so is positive for every non-zero \(\vect{v}\). It is never zero, so no non-zero vector stays on its line. Lesson M03.1.4 finds the eigenvalues this rotation does have: complex ones.

Summary card

An eigenvector is a direction a matrix only stretches; the eigenvalue is the stretch.

The anchor figure again: on the left the basis vectors are turned off their lines by A; on the right the eigenvectors (1, 1) and (1, minus 2) stay on their pale orange lines, stretched 5 times and 2 times.
M03.1.1-F1

Kept on their line, only stretched.

  • Definition. \({\mat{A}\vect{v}} = {\lambda\vect{v}}\) with \(\vect{v} \neq \vect{0}\).
  • A line, not an arrow. Every \(c\,\vect{v}\) with \(c \neq 0\) has the same \(\lambda\).
  • Check. One product \(\mat{A}\vect{v}\); in the plane, \({v_1 w_2 - v_2 w_1} = 0\).
  • Eigenvalue from vector. \(\lambda = \frac{\vect{v} \cdot \mat{A}\vect{v}}{\vect{v} \cdot \vect{v}}\).
  • Running \(\mat{A}\). \((1, 1)\) with 5 and \((1, -2)\) with 2.

Sources and next steps

Sources. Each was opened and checked against this lesson on 15 November 2026.

  • Gilbert Strang, Eigenvalues and eigenvectors, MIT OpenCourseWare 18.06SC, lecture 21, and its lecture summary. Eigenvectors are the vectors for which \(Ax\) is parallel to \(x\); a projection has eigenvalues 1 and 0, and the matrix with rows \((0, 1)\) and \((1, 0)\) has eigenvectors \((1, 1)\) and \((1, -1)\) with eigenvalues 1 and \(-1\).
  • Dan Margalit and Joseph Rabinoff, Interactive Linear Algebra, Section 5.1 (open textbook, Georgia Tech). The definition, with eigenvectors non-zero by definition, and the reflection whose mirror line has eigenvalue 1 and whose perpendicular line has eigenvalue \(-1\).
  • David Austin, Understanding Linear Algebra, An introduction to eigenvalues and eigenvectors (open textbook). An eigenvector and its image are scalar multiples, so they lie on the same line.
  • Grant Sanderson, Eigenvectors and eigenvalues, 3Blue1Brown, Linear Algebra, chapter 14. Eigenvectors as vectors that stay on their own span; a scaling makes every vector an eigenvector, and a rotation of the plane has none.
  • Sheldon Axler, Linear Algebra Done Right, 4th edition (open access), Section 5A. Definition 5.8 requires an eigenvector to be non-zero, and the discussion before it links eigenvalues to one-dimensional subspaces that the operator maps into themselves.

Next. M03.1.2 The characteristic polynomial replaces the sweep with algebra that finds every eigenvalue exactly.

This lesson unlocks

M03.1.2 The characteristic polynomial

Your progress is saved in this browser only. No account, no tracking.
Back to top