01 / Visual model
One line, an arrow that flips and shrinks
02 / Full working
Walk through one example
Let A = [[-1/2, 0], [0, 2]] and v = (1, 0). What are the first three outputs, and why does the shortcut work?
- First applicationAv = [[-1/2, 0], [0, 2]](1, 0) = (-1/2, 0) = (-1/2)v
The result stays on the horizontal line through v. That verifies v is an eigenvector and its eigenvalue is -1/2.
- Feed the output back inA^2v = A(Av) = A((-1/2)v) = (-1/2)Av = (1/4)v = (1/4, 0)
Linearity lets the scalar come outside A. Two negative factors make a positive one.
- One more stepA^3v = A((1/4)v) = (1/4)Av = (-1/8)v = (-1/8, 0)
The arrow reverses again and is now one eighth of its original length.
- Read the patternv, (-1/2)v, (1/4)v, (-1/8)v, ... ; A^k v = (-1/2)^k v
The magnitudes 1, 1/2, 1/4, 1/8 shrink; the signs +, -, +, - alternate. No finite term is zero because (-1/2)^k is never zero for finite k.
ResultFor this eigenvector, three applications give A^3v = (-1/8, 0). Repeated application approaches the zero vector along this line, alternating sides.
03 / Name the rule
Now name the rule we used
At step zero, A^0v = v. If Av = λv, then A^k v = λ^k v for whole-number steps k ≥ 1.
Read aloudAt step zero, no application has happened: A^0 is the identity, so the starting vector is still v. If A sends the nonzero vector v to lambda times v, then applying A k times, for k at least one, sends v to lambda raised to k times v. The exponent on A counts applications of the matrix; the exponent on lambda counts how many times its scale factor is used. Separating step zero also makes the rule clear when lambda is zero.
Objects in the expression
- A and A^k
- A is a square matrix representing one linear step. A^k is A composed with itself k times: A^2 = AA, A^3 = AAA. These are matrix products, not entry-by-entry powers.
- v
- A nonzero eigenvector of A. The formula follows this particular direction; an arbitrary starting vector may have components that change at different rates.
- λ
- The scalar multiplying v in one step. Its absolute value controls the length of this eigenvector's iterates; its sign controls whether their direction reverses.
- k
- A whole-number step counter. It changes from 0 to 1 to 2 and onward; it is not a vector component index.
How do other one-step multipliers behave? Scroll the table sideways to see every step.
| Multiplier | start | once | twice | three times |
|---|---|---|---|---|
| 2 | v | 2v | 4v | 8v |
| ½ | v | v/2 | v/4 | v/8 |
| −2 | v | −2v | 4v | −8v |
| −½ | v | −v/2 | v/4 | −v/8 |
| 1 | v | v | v | v |
| −1 | v | −v | v | −v |
| 0 | v | 0 | 0 | 0 |
04 / A tempting wrong turn
Where a reading can go astray
To find A^2, square every number inside A, or assume A^k w = λ^k w for any vector w.
A^2 means AA. For example, M = [[1, 1], [0, 1]] gives M^2 = [[1, 2], [0, 1]], while squaring each entry would leave M unchanged. The λ^k shortcut applies to a verified eigenvector v. A vector made from several directions can carry several scale factors at once.
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.
- recognize
Suppose Av = 3v for a nonzero vector v. Read A^4v aloud, then state its value as a multiple of v.
Show a hint
Count four applications of A, starting from v.
Reveal worked answer
A^4v = 3^4v = 81v.
The equation Av = 3v verifies that v is an eigenvector with eigenvalue 3.
Each application multiplies this same vector direction by 3; four applications use four factors of 3.
Read A^4v as 'apply A four times to v,' so the result is 81v.
- recognize
A nonzero eigenvector v has λ = -2. List its positions at steps k = 0, 1, 2, 3 as multiples of v. Does its length grow or shrink with each application, and does it point to the same side of the origin?
Show a hint
Track absolute value and sign separately: (-2)^0, (-2)^1, (-2)^2, (-2)^3.
Reveal worked answer
Its length doubles each step and its direction alternates: v, -2v, 4v, -8v.
The multipliers at steps 0 through 3 are 1, -2, 4, and -8.
Their absolute values are 1, 2, 4, and 8, so the arrow grows.
The signs alternate, so the arrow switches sides of the origin every step.
- apply
A nonzero eigenvector v has λ = -1/2. Write its positions at steps k = 0, 1, 2, 3, and 4 as multiples of v. Does any of these steps land on zero?
Show a hint
Multiply the previous coefficient by -1/2 each time.
Reveal worked answer
v, -v/2, v/4, -v/8, v/16. None is zero.
Begin with coefficient 1 at k = 0.
Successive multiplication by -1/2 gives -1/2, 1/4, -1/8, and 1/16.
Every listed coefficient is nonzero. Indeed, (-1/2)^k remains nonzero for every finite whole-number k.
- apply
Let A = [[3, 0], [0, -1]] and v = (0, 2). Verify Av = λv, identify λ, and find A^2v.
Show a hint
Multiply A by v once; compare the entire output with v.
Reveal worked answer
Av = (0, -2) = -v, so λ = -1 and A^2v = v = (0, 2).
Matrix-vector multiplication gives Av = (3·0 + 0·2, 0·0 + (-1)·2) = (0, -2).
Since v = (0, 2), the output is -1 times v.
The second application uses (-1)^2 = 1, so A^2v = (0, 2).
- transfer
Let A = [[2, 0], [0, 1/2]] and w = (1, 1). Compute Aw and A^2w. Why can you not choose one λ and write A^2w = λ^2w here?
Show a hint
Look at what happens to the first and second coordinates separately after one step.
Reveal worked answer
Aw = (2, 1/2) and A^2w = (4, 1/4). The first coordinate doubles while the second halves, so w is not an eigenvector and has no single λ for this rule.
One application gives Aw = (2·1, (1/2)·1) = (2, 1/2).
A second application gives A(Aw) = (2·2, (1/2)·(1/2)) = (4, 1/4).
If Aw were λw, both coordinates would equal the same λ. They are 2 and 1/2, so the eigenvector shortcut does not apply to w.
- transfer
For M = [[1, 1], [0, 1]], compare the true matrix square M^2 with the array obtained by squaring each entry. Are they equal?
Show a hint
The top-right entry of MM is row 1 · column 2.
Reveal worked answer
M^2 = MM = [[1, 2], [0, 1]], while entry-by-entry squaring gives [[1, 1], [0, 1]]. They are different.
For MM, the top-right entry is 1·1 + 1·1 = 2; the other entries are 1, 0, and 1.
Squaring the four entries separately gives 1, 1, 0, and 1.
A matrix power represents composition through matrix multiplication, which combines rows and columns.
FIELD NOTES / RECALL
Cards from note 005
Read the front, decide on an answer, then reveal the reasoning.
Card 1 of 6
Suppose Av = 3v for a nonzero vector v. Read A4v aloud, then state its value as a multiple of v.
Back / answer
A4v = 3^4v = 81v.
The equation Av = 3v verifies that v is an eigenvector with eigenvalue 3. Each application multiplies this same vector direction by 3; four applications use four factors of 3. Read A4v as 'apply A four times to v,' so the result is 81v.
Card 1 of 6: Suppose Av = 3v for a nonzero vector v. Read A^4v aloud, then state its value as a multiple of v.. Answer hidden.
Card 1 / recognize
Front / question
Suppose Av = 3v for a nonzero vector v. Read A^4v aloud, then state its value as a multiple of v.
Back / answer
A^4v = 3^4v = 81v.
The equation Av = 3v verifies that v is an eigenvector with eigenvalue 3. Each application multiplies this same vector direction by 3; four applications use four factors of 3. Read A^4v as 'apply A four times to v,' so the result is 81v.
Card 2 / recognize
Front / question
A nonzero eigenvector v has λ = -2. List its positions at steps k = 0, 1, 2, 3 as multiples of v. Does its length grow or shrink with each application, and does it point to the same side of the origin?
Back / answer
Its length doubles each step and its direction alternates: v, -2v, 4v, -8v.
The multipliers at steps 0 through 3 are 1, -2, 4, and -8. Their absolute values are 1, 2, 4, and 8, so the arrow grows. The signs alternate, so the arrow switches sides of the origin every step.
Card 3 / apply
Front / question
A nonzero eigenvector v has λ = -1/2. Write its positions at steps k = 0, 1, 2, 3, and 4 as multiples of v. Does any of these steps land on zero?
Back / answer
v, -v/2, v/4, -v/8, v/16. None is zero.
Begin with coefficient 1 at k = 0. Successive multiplication by -1/2 gives -1/2, 1/4, -1/8, and 1/16. Every listed coefficient is nonzero. Indeed, (-1/2)^k remains nonzero for every finite whole-number k.
Card 4 / apply
Front / question
Let A = [[3, 0], [0, -1]] and v = (0, 2). Verify Av = λv, identify λ, and find A^2v.
Back / answer
Av = (0, -2) = -v, so λ = -1 and A^2v = v = (0, 2).
Matrix-vector multiplication gives Av = (3·0 + 0·2, 0·0 + (-1)·2) = (0, -2). Since v = (0, 2), the output is -1 times v. The second application uses (-1)^2 = 1, so A^2v = (0, 2).
Card 5 / transfer
Front / question
Let A = [[2, 0], [0, 1/2]] and w = (1, 1). Compute Aw and A^2w. Why can you not choose one λ and write A^2w = λ^2w here?
Back / answer
Aw = (2, 1/2) and A^2w = (4, 1/4). The first coordinate doubles while the second halves, so w is not an eigenvector and has no single λ for this rule.
One application gives Aw = (2·1, (1/2)·1) = (2, 1/2). A second application gives A(Aw) = (2·2, (1/2)·(1/2)) = (4, 1/4). If Aw were λw, both coordinates would equal the same λ. They are 2 and 1/2, so the eigenvector shortcut does not apply to w.
Card 6 / transfer
Front / question
For M = [[1, 1], [0, 1]], compare the true matrix square M^2 with the array obtained by squaring each entry. Are they equal?
Back / answer
M^2 = MM = [[1, 2], [0, 1]], while entry-by-entry squaring gives [[1, 1], [0, 1]]. They are different.
For MM, the top-right entry is 1·1 + 1·1 = 2; the other entries are 1, 0, and 1. Squaring the four entries separately gives 1, 1, 0, and 1. A matrix power represents composition through matrix multiplication, which combines rows and columns.
06 / Take it forward
Where this appears again
Discrete linear models repeat a rule of the form x_(k+1) = Ax_k. If the initial state is an eigenvector, this lesson gives its exact state after k steps. For a general state, more work is needed to account for all its directions; a single eigenvalue is not a verdict about the whole system.
One-minute reference
- Check Av = λv with v ≠ 0 before using the eigenvector shortcut.
- A^k means k applications of A; A^0 is the identity. Step zero is v, even for λ = 0, because no multiplication has happened yet.
- For whole-number steps k ≥ 1 on an eigenvector, A^k v = λ^k v. This follows by repeatedly applying Av = λv; step zero is v.
- For real λ, |λ| controls size: above 1 grows, between 0 and 1 shrinks, equal to 1 keeps the same size. A negative λ also reverses direction each step.
- λ = 0 sends an eigenvector to zero in one step; a nonzero λ with |λ| < 1 approaches zero without reaching it in a finite number of steps.