When a Mathematical System Stops Changing

01 / Visual model

A fixed horizontal part and a shrinking, reversing vertical part

02 / Full working

Walk through one example

Let F(x) = Ax with A = [[1, 0], [0, -1/2]]. Start at x_0 = (1, 2). Is x* = (1, 0) fixed? Where do the next three states go?

1. Test the candidateF(x*) = A(1, 0) = (1, 0) = x*

   The whole vector is unchanged. Since x* is nonzero, it is an eigenvector with eigenvalue 1.

2. Take one stepx_1 = A(1, 2) = (1, -1)

   The first coordinate stays 1; the second becomes -1. The state is not fixed because x_1 differs from x_0.

3. Take two more stepsx_2 = A(1, -1) = (1, 1/2); x_3 = A(1, 1/2) = (1, -1/4)

   The vertical part reverses sign and halves each time. Successive states lie on alternating sides of the fixed line y = 0, but none of these states is on it.

4. Separate the fixed part from the changing partx_k = (1, 0) + (0, 2(-1/2)^k)

   The first part is fixed; the second is a vertical eigenvector contribution. Its magnitude tends to zero, while its sign alternates. For any finite k, 2(-1/2)^k is nonzero.

ResultThe states approach x* = (1, 0), but this particular path never lands exactly on it at a finite step. 'Approaching a fixed state' and 'already fixed' describe different things.

03 / Name the rule

Now name the rule we used

A fixed state x* satisfies F(x*) = x*. For a linear update F(x) = Ax, a nonzero fixed state is an eigenvector with λ = 1.

Read aloudA star on x marks the state we are testing. Apply the rule F to that state. If the output is the same whole state, it is fixed. For a linear matrix rule, Ax star equals x star means one application scales this nonzero vector by one. A state can also approach a fixed state while still changing at every finite step.

Objects in the expression

F and x_(k+1) = F(x_k)

F is an update rule. Starting from x_0, each new state is the output of applying F to the previous state. This notation describes repeated steps, not a continuous motion between them.

x*

A candidate fixed state. The star labels a special value; it is neither multiplication nor an exponent. Check it by substituting into F.

fixed versus period two

Fixed means one step returns the same state: F(x) = x. A genuine two-step cycle means F(F(x)) = x but F(x) ≠ x. Returning after two steps does not by itself make a state fixed.

approach

The distance between x_k and a fixed state gets arbitrarily small as k grows. A sequence may approach its destination without equaling it at any finite step.

04 / A tempting wrong turn

Where a reading can go astray

See one unchanged direction and declare the entire system stable, or call an alternating path fixed because it returns after two steps.

A = [[1, 0], [0, 2]] fixes every vector on the horizontal axis but sends (1, 1) to (1, 2), then (1, 4), so other states move away from that axis. A factor of -1 sends a nonzero eigenvector v to -v and back to v: that is period two, not fixed, because the first step changes it.

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

   For F(x) = 3x - 4 on real numbers, test whether x* = 2 is fixed. Is x = 1 fixed?

   Show a hint

   Substitute each proposed value into F and compare output with input.

   Reveal worked answer

   2 is fixed because F(2) = 2. 1 is not fixed because F(1) = -1.

   F(2) = 3·2 - 4 = 2, so applying the rule leaves 2 unchanged.

   F(1) = 3·1 - 4 = -1, which differs from 1.

   The fixed-point test asks for equality after one application, not just a nearby result.

2. recognize

   Let F(x) = 2 - x. Starting at 5 gives 5 → -3 → 5. Is 5 fixed or on a genuine two-step cycle? Find one fixed state of this F.

   Show a hint

   A fixed state must survive the first arrow, not merely the second.

   Reveal worked answer

   5 is on a genuine two-step cycle, not fixed. The fixed state is x* = 1.

   F(5) = 2 - 5 = -3, so 5 changes after one step.

   F(-3) = 2 - (-3) = 5, so it returns after exactly two steps.

   Solving F(x) = x gives 2 - x = x, hence x = 1; indeed F(1) = 1.

3. apply

   Let x_(k+1) = A x_k: each new state is A times the current state. For A = [[1, 0], [0, -1/2]] and x_0 = (2, 4), find x_1, x_2, x_3 and the fixed state this path approaches. Does it ever reach that state in a finite number of steps?

   Show a hint

   Keep the first coordinate; multiply the second by -1/2 at each step.

   Reveal worked answer

   x_1 = (2, -2), x_2 = (2, 1), x_3 = (2, -1/2). They approach (2, 0) but never reach it at any finite step.

   The matrix leaves the first coordinate at 2 and multiplies the second by -1/2 each time.

   The second coordinates are 4, -2, 1, -1/2, and in general 4(-1/2)^k.

   Their magnitudes tend to zero, but 4(-1/2)^k is nonzero for every finite k. Also A(2, 0) = (2, 0), so (2, 0) is fixed.

4. apply

   A nonzero vector v satisfies Av = -v. Classify the repeated path v → Av → A^2v as fixed, a genuine two-step cycle, or neither.

   Show a hint

   Apply A once to -v, using linearity.

   Reveal worked answer

   It is a genuine two-step cycle: v → -v → v.

   The first step is Av = -v, which differs from v because v is nonzero.

   The second is A(-v) = -Av = -(-v) = v.

   Returning after two steps, with a change after one, is precisely a period-two cycle.

5. transfer

   A = [[1, 0], [0, 2]] fixes v = (1, 0). Starting instead at w = (1, 1), find Aw and A^2w. Does the fixed vector v prove that every state approaches a fixed state?

   Show a hint

   Compare the horizontal and vertical coordinates of w after each step.

   Reveal worked answer

   Aw = (1, 2), A^2w = (1, 4). No: the vertical coordinate grows, so this starting state does not approach the fixed horizontal axis.

   Av = (1, 0) verifies v is fixed.

   But Aw = (1, 2) and A(Aw) = (1, 4); the second coordinate doubles each time.

   One fixed eigendirection describes what happens along that direction only. It cannot establish stability for all states.

6. transfer

   Compare two rules with the same fixed state 7. Rule P(x) = 7; rule Q(x) = 7 + (x - 7)/2. Starting at x_0 = 2, which rule reaches 7 in a finite step, and which merely approaches it?

   Show a hint

   Track the error x_k - 7 for each rule.

   Reveal worked answer

   P reaches 7 after one step and stays there. Q approaches 7 but never reaches it from 2 in a finite number of steps.

   Both have fixed state 7 because P(7) = Q(7) = 7.

   P(2) = 7, so its path is 2 → 7 → 7 → ... .

   For Q, the initial error is 2 - 7 = -5 and each step halves it: x_k - 7 = -5/2^k. This is nonzero for finite k but tends to zero.

06 / Take it forward

Where this appears again

A fixed point is a useful question for any repeated rule F, including nonlinear ones: solve F(x*) = x*. A matrix is a special case. In linear discrete systems, an eigenvalue of 1 marks a nonzero fixed direction, while other directions can shrink, grow, or cycle. Before making a statement about an entire system, inspect the other directions and the states under discussion. No analogy to a physical system is implied by this abstract rule alone.