The Directions a Transformation Doesn't Turn

01 / Visual model

Three arrows through the same grid

02 / Full working

Walk through one example

For A = [[3, 0], [0, 2]], compare v = (1, 0), w = (0, 1), and u = (1, 1). Which inputs are eigenvectors?

1. Start with the first arrowA(1, 0) = (3×1 + 0×0, 0×1 + 2×0)

   The first row makes the first output coordinate; the second row makes the second.

2. Compare it with a scaled copyA(1, 0) = (3, 0) = 3(1, 0)

   One number, 3, multiplies both coordinates. v stays on the horizontal line, so v is an eigenvector with eigenvalue 3. Its length changes; its line does not.

3. Check the vertical arrowA(0, 1) = (3×0 + 0×1, 0×0 + 2×1) = (0, 2) = 2(0, 1)

   w stays on the vertical line. It is an eigenvector with eigenvalue 2.

4. Test the diagonal arrowA(1, 1) = (3, 2)

   To equal λ(1, 1) = (λ, λ), both output coordinates would need to be the same number. They are 3 and 2. No one scalar works, so u is not an eigenvector.

ResultThe horizontal and vertical inputs keep their respective lines, with factors 3 and 2. The diagonal input moves off its line.

03 / Name the rule

Now name the rule we used

Av = λv, with v ≠ 0

Read aloudApply the square matrix A to the nonzero vector v. The answer is the same vector multiplied by one number, λ (lambda). If that happens, v is an eigenvector of A and λ is its eigenvalue.

Objects in the expression

A

A square matrix: a rule that takes a vector to another vector with the same number of coordinates. In this note, A has two rows and two columns.

v

A nonzero input vector. We write its two coordinates as (x, y), shorthand for a two-entry column vector. The zero vector cannot be an eigenvector.

Av

The output after the matrix acts on v. Matrix–vector multiplication gives each output coordinate from a row of A.

λv

One scalar λ multiplies every coordinate of v. λ is a number, not a second matrix.

04 / A tempting wrong turn

Where a reading can go astray

A is diagonal, so every input must keep its direction.

The two coordinate directions are scaled by different factors. For (1, 1), A gives (3, 2), which is not a scaled copy of (1, 1). An eigenvector needs one common factor for all its coordinates.

05 / Independent practice

Try these before opening the answers

The questions move from recognizing a symbol to applying and transferring the idea. Each answer shows the reasoning.

1. recognize

   A square matrix gives Av = v for a nonzero v. Is v an eigenvector? What is λ?

   Show a hint

   Compare Av with λv. Which number leaves every coordinate as it was?

   Reveal worked answer

   Yes. λ = 1.

   Because v is nonzero and A is square, the definition can apply.

   Av = v = 1v, so one common scalar works for every coordinate. The input is unchanged, not merely kept on the same line.

2. recognize

   For A = [[3, 0], [0, 2]], someone says (1, 1) is an eigenvector because A(1, 1) = (3, 2) still points generally up and right. Check the claim.

   Show a hint

   Could (3, 2) equal λ(1, 1) for one number λ?

   Reveal worked answer

   No. (1, 1) is not an eigenvector of this A.

   If (3, 2) = λ(1, 1), the first coordinate demands λ = 3 while the second demands λ = 2.

   One scalar cannot be both. A nearby-looking direction does not satisfy Av = λv.

3. apply

   Keep A = [[3, 0], [0, 2]]. Test v = (0, −3). Give Av and the eigenvalue if v qualifies.

   Show a hint

   Use the two rows of A. Then compare the output with a multiple of (0, −3).

   Reveal worked answer

   Av = (0, −6) = 2v, so v is an eigenvector with λ = 2.

   The first output coordinate is 3×0 + 0×(−3) = 0.

   The second is 0×0 + 2×(−3) = −6. Because (0, −6) = 2(0, −3), one common scalar works.

4. apply

   Let D = [[−2, 0], [0, 1]] and v = (3, 0). Find Dv, decide whether v is an eigenvector, and describe what the sign of λ does.

   Show a hint

   Find Dv first. A negative scalar puts the output on which side of the origin?

   Reveal worked answer

   Dv = (−6, 0) = −2v. Yes: λ = −2, so the arrow reverses along the same line and doubles in length.

   The first coordinate is (−2)×3 + 0×0 = −6; the second is 0×3 + 1×0 = 0.

   One scalar, −2, scales both coordinates. Its negative sign reverses orientation; its absolute value 2 doubles length.

5. transfer

   Let P = [[0, 0], [0, 1]] and v = (2, 0). P sends v to the origin. Does v count as an eigenvector? Would the zero vector itself count?

   Show a hint

   The definition restricts the input v, not the output Pv or the value of λ.

   Reveal worked answer

   Yes: Pv = (0, 0) = 0v, so v is an eigenvector with λ = 0. The zero vector is not an eigenvector.

   The input (2, 0) is nonzero. Multiplication gives (0×2 + 0×0, 0×2 + 1×0) = (0, 0).

   The scalar 0 works: 0(2, 0) = (0, 0). The output has no direction, but the eigenvector definition still holds. The zero input is excluded because it would satisfy A0 = λ0 for every λ.

6. transfer

   R = [[0, −1], [1, 0]] rotates every plane arrow 90° counterclockwise. Test (1, 0), then decide whether any nonzero real vector can be an eigenvector of R.

   Show a hint

   A real scalar multiple stays on the input's line, reverses along it, or is zero. What does a quarter-turn do?

   Reveal worked answer

   R(1, 0) = (0, 1), not a multiple of (1, 0). No nonzero real vector is an eigenvector of this 90° rotation.

   R(1, 0) = (0×1 + (−1)×0, 1×1 + 0×0) = (0, 1). The input lies on the horizontal line and the output on the vertical line.

   The same quarter-turn moves every nonzero real arrow off its line, while preserving nonzero length. Real scaling cannot make that turn, and λ = 0 cannot describe a nonzero output. Thus no real λ and nonzero real v satisfy Rv = λv.

06 / Take it forward

Where this appears again

In the next note, applying A repeatedly to an eigenvector gives Aⁿv = λⁿv. The vector keeps its line while the scalar factor changes with each application. That statement concerns this one eigenvector; it does not by itself describe every possible input.