# Matrix Diagonalisation

Business I · Mathematics II · https://tryals.app/learn/business-i/matrix-diagonalisation

## Decomposing a Matrix

A matrix $A$ is **diagonalisable** if it can be written as $A = PDP^{-1}$, where $D$ is a **diagonal matrix** of eigenvalues and $P$ has eigenvectors as columns.

**Why this matters**: matrix powers become trivial via $A^n = PD^nP^{-1}$. Raising $D$ to the $n$-th power requires zero matrix multiplications.

**The Recipe**:
1. Solve $\det(A - \lambda I) = 0$ for eigenvalues.
2. Solve $(A - \lambda I)\mathbf{v} = \mathbf{0}$ for eigenvectors.
3. Form $P$ and $D$.

| Problem | Without $PDP^{-1}$ | With it |
| --- | --- | --- |
| $A^{100}$ | 99 multiplications | Raise diagonal numbers |
| Long run | Simulate | Read dominant eigenvalue |

> **Pitfall**: Distinct eigenvalues guarantee diagonalisability, but repeated ones may lack independent eigenvectors.

## Matrix Powers and Steady States

Diagonalisation yields the closed-form solution: $\mathbf{x}_n = c_1 \lambda_1^n \mathbf{v}_1 + c_2 \lambda_2^n \mathbf{v}_2$.

**Dominant eigenvalue controls long-run behaviour**:

| $|\lambda_{\text{max}}|$ | System Behaviour |
| --- | --- |
| $< 1$ | Decays to zero |
| $= 1$ | Converges to steady state |
| $> 1$ | Explodes |

**Example**: For Markov matrices, $\lambda_1 = 1$ gives the **stationary distribution**. As $n \to \infty$, $0.3^n \to 0$, leaving only the dominant eigenvector.

> **Tip**: In Markov matrices, the dominant eigenvalue is 1. Its normalised eigenvector _is_ the long-run market share.

## Practice questions

7 of this lesson's 14 practice questions, with answers. The full set is in the app.

### 1. Order the diagonalisation procedure.

**Answer:**

1. Find eigenvalues: solve $\det(A - \lambda I) = 0$
2. Find eigenvectors: solve $(A - \lambda I)\mathbf{v} = \mathbf{0}$
3. Form $P$ from eigenvector columns
4. Form $D$ with eigenvalues on diagonal
5. Verify $A = PDP^{-1}$

**Why:** Eigenvalues first (characteristic equation), then eigenvectors (null space), then assemble the decomposition.

Page: https://tryals.app/practice/business-i/matrix-diagonalisation/order-the-diagonalisation-procedure

### 2. Diagonalisation writes A as PDP⁻¹. What is the change-of-basis story that makes A¹⁰⁰ = PD¹⁰⁰P⁻¹ obvious?

A. P and P⁻¹ commute with D, which allows them to be moved outside the power
B. The powers of P grow slowly enough to be neglected for large exponents
C. P⁻¹ translates in, D scales, P translates back; the rest cancels
D. Diagonal matrices commute with everything, so the order can be rearranged freely

**Answer:** C. P⁻¹ translates in, D scales, P translates back; the rest cancels

**Why:** Each adjacent P⁻¹P collapses to the identity, so a hundred sandwiches telescope into one, translate, scale a hundred times, translate back. Option 2 is false in general, and mistaking a telescoping cancellation for commutativity is the standard slip here.

Page: https://tryals.app/practice/business-i/matrix-diagonalisation/diagonalisation-writes-a-as-pdp-what-is-the-change-of-basis-story

### 3. A 2x2 matrix has eigenvalues $\lambda_1 = 3$ and $\lambda_2 = 5$. What is $\det(A)$?

A. -15
B. 15
C. 8
D. 2

**Answer:** B. 15

**Why:** $\det(A) = \lambda_1\lambda_2 = 3 \times 5 = 15$. The determinant is the PRODUCT of the eigenvalues and the trace is their sum, so 8 is the trace and 2 is their difference, neither is the determinant. The product rule is why a zero eigenvalue forces a zero determinant: one factor kills it, which is singularity seen through the eigenvalues.

Page: https://tryals.app/practice/business-i/matrix-diagonalisation/a-2x2-matrix-has-eigenvalues-1-3-and-2-5-what-is-det-a

### 4. Sort: which properties guarantee diagonalisability, and which do not?

**Answer:**

- Guarantees diagonalisability: All eigenvalues are distinct, Matrix is symmetric, Enough independent eigenvectors exist
- Does NOT guarantee: Matrix has a repeated eigenvalue, Matrix is invertible, Matrix is real

**Why:** Distinct eigenvalues or symmetry guarantee it. Repeated eigenvalues, invertibility, or being real are not sufficient conditions.

Page: https://tryals.app/practice/business-i/matrix-diagonalisation/sort-which-properties-guarantee-diagonalisability-and-which-do-not

### 5. In the decomposition $A = PDP^{-1}$, the columns of $P$ are the eigenvectors of $A$.

**Answer:** True

**Why:** Each column of $P$ is an eigenvector; the corresponding diagonal entry of $D$ is its eigenvalue. The decomposition encodes $A\mathbf{v}_i = \lambda_i \mathbf{v}_i$ for all eigenvectors simultaneously.

Page: https://tryals.app/practice/business-i/matrix-diagonalisation/in-the-decomposition-a-pdp-the-columns-of-p-are-the-eigenvectors

### 6. After 10 periods, a Markov component with $\lambda = 0.5$ has magnitude $0.5^{10}$. **Estimate this value** (in thousandths).

**Answer:** 0.977 (within ±0.5)

**Why:** $0.5^{10} = 1/1024 \approx 0.00098$, essentially zero. After 10 doublings of halving, the component has almost completely died out.

Page: https://tryals.app/practice/business-i/matrix-diagonalisation/after-10-periods-a-markov-component-with-0-5-has-magnitude

### 7. Diagonalising $A = PDP^{-1}$ makes $A^{n}$ easy because $D^{n}$ raises the diagonal entries to the $n$th power. Why does that also reveal the long-run behaviour of the system?

A. Because the determinant of $A^{n}$ approaches zero
B. Because all eigenvalues converge to the same value as $n$ grows
C. Because $P$ and $P^{-1}$ cancel, leaving only the initial condition
D. Because the largest eigenvalue's power eventually dominates

**Answer:** D. Because the largest eigenvalue's power eventually dominates

**Why:** Powers separate the eigenvalues by magnitude. This is exactly why a Markov chain settles into its stationary distribution: the $\lambda = 1$ eigenvector survives while everything smaller decays.

Page: https://tryals.app/practice/business-i/matrix-diagonalisation/diagonalising-a-pdp-makes-a-easy-because-d-raises-the-diagonal