LINEAR-ALGEBRA · Interactive Practice | Unit 23 · Video 1

Why Every Markov Matrix Hides a 1

IKey Formulas

Formula Name Description
∑iaij=1\sum_i a_{ij} = 1 Markov column rule Each column of AA sums to 1
Ax=xA\mathbf{x} = \mathbf{x} Steady-state equation Eigenvector for λ=1\lambda = 1
(1,1,…,1) (A−I)=0(1, 1, \ldots, 1)\,(A - I) = \mathbf{0} Left null space identity Row of ones kills A−IA - I
det⁡(A−λI)=det⁡(AT−λI)\det(A - \lambda I) = \det(A^T - \lambda I) Transpose principle AA and ATA^T share eigenvalues

IIVisualization 1 — Columns of A−IA - I Sum to Zero

Subtracting II drops each diagonal entry by one, pulling every Markov column sum from 11 straight down to 00.

💡 Because every column of A−IA - I sums to 00, the row (1,1,1)(1, 1, 1) times A−IA - I is all zeros — the row of ones sits in the left null space of A−IA - I.

IIIVisualization 2 — Watching the Steady State Emerge

Start anywhere in the probability simplex and keep applying AA; every path is drawn into one fixed point.

💡 That shared limit is the steady state — the right null-space eigenvector of A−IA - I. It is not (1,1,1)(1, 1, 1); the row of ones lives in the left null space.

IVVisualization 3 — The Logical Chain at a Glance

Seven structural steps carry "columns sum to 11" all the way to "λ=1\lambda = 1 is an eigenvalue."

💡 The chain reaches λ=1\lambda = 1 by pure structural reasoning about column sums — no characteristic polynomial and no determinant expansion.

VQuiz Questions

Question 1

Consider the matrix A=(0.40.20.60.8).A = \begin{pmatrix} 0.4 & 0.2 \\ 0.6 & 0.8 \end{pmatrix}.

Without computing any determinant or characteristic polynomial, what is one eigenvalue you can identify immediately?

✅ Correct! Every Markov matrix has λ=1\lambda = 1 as an eigenvalue.

❌ Not quite. Check that this is a Markov matrix (non-negative entries, columns summing to 1) and recall the guaranteed eigenvalue.

Show solution

Solution:

Check that AA is a Markov matrix:

  • All entries are non-negative ✓
  • Column 1 sum: 0.4+0.6=10.4 + 0.6 = 1 ✓
  • Column 2 sum: 0.2+0.8=10.2 + 0.8 = 1 ✓

Since AA is Markov, λ=1\lambda = 1 is automatically an eigenvalue. The structural reason: columns of A−IA - I now sum to 00, so (1,1) (A−I)=(0,0)(1, 1)\,(A - I) = (0, 0), which means A−IA - I is singular, which means det⁡(A−I)=0\det(A - I) = 0, which means λ=1\lambda = 1 solves the characteristic equation.

Question 2

True or False: For any Markov matrix AA, the row vector (1,1,…,1)(1, 1, \ldots, 1) is the steady-state eigenvector — that is, A(1,1,…,1)T=(1,1,…,1)TA(1,1,\ldots,1)^T = (1,1,\ldots,1)^T.

✅ Correct! (1,1,…,1)(1,1,\ldots,1) is in the left null space of A−IA - I, not the right. The right null space gives the steady state.

❌ Not quite. Recall that the row of ones kills A−IA - I from the left, while eigenvectors satisfy Ax=xA\mathbf{x} = \mathbf{x} on the right.

Show solution

Solution:

This is False. The row of ones (1,1,…,1)(1, 1, \ldots, 1) lives in the left null space of A−IA - I:

(1,1,…,1) (A−I)=0(1, 1, \ldots, 1)\,(A - I) = \mathbf{0}

That works because each entry of the product is a column sum of A−IA - I, and Markov columns sum to 0 after subtracting II.

The steady-state eigenvector x\mathbf{x} lives in the right null space: (A−I)x=0(A - I)\mathbf{x} = \mathbf{0}, equivalently Ax=xA\mathbf{x} = \mathbf{x}. In general x≠(1,1,…,1)T\mathbf{x} \neq (1,1,\ldots,1)^T.

For example, with A=(0.10.010.30.20.990.30.700.4),A = \begin{pmatrix} 0.1 & 0.01 & 0.3 \\ 0.2 & 0.99 & 0.3 \\ 0.7 & 0 & 0.4 \end{pmatrix}, the steady-state vector is approximately (0.017,0.962,0.020)(0.017, 0.962, 0.020), not (1,1,1)(1, 1, 1).

Question 3

The proof that λ=1\lambda = 1 is a Markov eigenvalue uses the chain:

  1. Columns of AA sum to 1
  2. Columns of A−IA - I sum to 0
  3. (1,1,…,1)(1, 1, \ldots, 1) is in the left null space of A−IA - I
  4. ???
  5. Columns of A−IA - I are linearly dependent
  6. A−IA - I is singular
  7. λ=1\lambda = 1 is an eigenvalue of AA

Which statement fills step 4, and what fact justifies the jump from step 4 to step 5?

✅ Correct! The transpose principle is the linchpin — without it, finding a left null vector wouldn't help us prove the matrix is singular.

❌ Not quite. Think about what the row of ones in the left null space tells you about the rows, and what fact lets you transfer that to the columns.

Show solution

Solution:

The missing step is "rows of A−IA - I are linearly dependent." We get this directly from step 3: a nonzero left null vector (1,1,…,1)(1, 1, \ldots, 1) produces a linear combination of the rows that equals zero, so the rows are dependent.

The bridge from "rows dependent" to "columns dependent" uses the transpose principle:

det⁡(A−λI)=det⁡((A−λI)T)=det⁡(AT−λI).\det(A - \lambda I) = \det\bigl((A - \lambda I)^T\bigr) = \det(A^T - \lambda I).

Equivalently, a matrix and its transpose have the same rank, so row-rank equals column-rank. If the rows are dependent, the rank is less than full, so the columns must be dependent too.

The other options are wrong:

  • ❌ A Markov matrix does not generally have a zero row.
  • ❌ det⁡(A−I)=0\det(A - I) = 0 (not 1) — that's actually what we're trying to prove.
  • ❌ Markov matrices are not generally symmetric.

Question 4

Suppose AA is a 4×44 \times 4 Markov matrix and x\mathbf{x} is its steady-state probability vector (so Ax=xA\mathbf{x} = \mathbf{x} and the entries of x\mathbf{x} are non-negative and sum to 1). Which statement is true?

✅ Correct! The steady state lives in the right null space of A−IA - I and is fixed by every power of AA.

❌ Not quite. Rearrange Ax=xA\mathbf{x} = \mathbf{x} into a null-space statement, then ask what happens when you apply AA repeatedly.

Show solution

Solution:

Ax=xA\mathbf{x} = \mathbf{x} rearranges to (A−I)x=0(A - I)\mathbf{x} = \mathbf{0}, which means x\mathbf{x} lies in the right null space of A−IA - I.

Because Ax=xA\mathbf{x} = \mathbf{x}, applying AA leaves x\mathbf{x} unchanged: A2x=A(Ax)=Ax=x,…,A100x=x.A^2 \mathbf{x} = A(A\mathbf{x}) = A\mathbf{x} = \mathbf{x}, \quad \ldots, \quad A^{100}\mathbf{x} = \mathbf{x}.

This is exactly what "steady state" means.

The other options:

  • ❌ The left null space contains (1,1,1,1)(1,1,1,1), not the steady-state vector.
  • ❌ x\mathbf{x} is uniform only for very special (doubly stochastic) Markov matrices.
  • ❌ The second Markov theorem says all eigenvalues satisfy ∣λ∣≤1|\lambda| \leq 1, so nothing blows up; λ=1\lambda = 1 exactly preserves x\mathbf{x}.

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