Eigenvalues and Eigenvectors

When you multiply a vector by a matrix, the vector usually changes direction. But for most matrices there are a few special directions that stay put: vectors along them only get stretched or shrunk. Those vectors are eigenvectors, and the stretch factors are eigenvalues. They make it easy to find high powers of a matrix, which is exactly what you need to predict where a population, a market share or a system ends up in the long run.

Key ideas

What eigenvectors and eigenvalues are

A non-zero vector $\vec{v}$ is an eigenvector of a square matrix $A$ if multiplying by $A$ just scales it:

$$A\vec{v} = \lambda\vec{v}$$

The number $\lambda$ (Greek “lambda”) is the eigenvalue that goes with $\vec{v}$. For example, with $A = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}$:

$$A\begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 5 \\ 5 \end{pmatrix} = 5\begin{pmatrix} 1 \\ 1 \end{pmatrix}, \qquad A\begin{pmatrix} 1 \\ -2 \end{pmatrix} = \begin{pmatrix} 2 \\ -4 \end{pmatrix} = 2\begin{pmatrix} 1 \\ -2 \end{pmatrix}$$

So $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$ is an eigenvector with eigenvalue $5$, and $\begin{pmatrix} 1 \\ -2 \end{pmatrix}$ is an eigenvector with eigenvalue $2$. But $A\begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 4 \\ 2 \end{pmatrix}$ points in a new direction, so $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ is not an eigenvector.

Eigenvectors of $A$ (blue, with their images in orange) stay on their own lines; $(1, 0)$ (green) gets turned to $(4, 2)$.

Any non-zero multiple of an eigenvector is also an eigenvector with the same eigenvalue: $\begin{pmatrix} 3 \\ 3 \end{pmatrix}$ works just as well as $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$. So an eigenvector is really a direction, and you can choose any convenient one.

The characteristic polynomial

Rewrite $A\vec{v} = \lambda\vec{v}$ as $(A - \lambda I)\vec{v} = \vec{0}$. For a non-zero $\vec{v}$ to satisfy this, $A - \lambda I$ must be singular, so

$$\det(A - \lambda I) = 0$$

For $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$:

$$\det\begin{pmatrix} a - \lambda & b \\ c & d - \lambda \end{pmatrix} = (a - \lambda)(d - \lambda) - bc$$

This quadratic in $\lambda$ is the characteristic polynomial. Setting it to $0$ gives the characteristic equation, and its roots are the eigenvalues. Expanding gives a handy shortcut:

$$\lambda^2 - (a + d)\lambda + (ad - bc) = 0$$

that is, $\lambda^2 - (\text{sum of the leading diagonal})\lambda + \det A = 0$.

Finding eigenvectors

For each eigenvalue $\lambda$, solve $(A - \lambda I)\vec{v} = \vec{0}$. Because $A - \lambda I$ is singular, its two rows give the same equation, so you get a direction rather than a single vector. Pick a simple vector in that direction.

This is the one place in the course where you work with a matrix that has no inverse: you can’t solve $(A - \lambda I)\vec{v} = \vec{0}$ with an inverse matrix.

Diagonalization and powers

If a $2 \times 2$ matrix $A$ has two distinct real eigenvalues $\lambda_1$ and $\lambda_2$ with eigenvectors $\vec{v}_1$ and $\vec{v}_2$, put the eigenvectors in the columns of $P$ and the eigenvalues, in the same order, on the diagonal of $D$:

$$P = \begin{pmatrix} \vec{v}_1 & \vec{v}_2 \end{pmatrix}, \qquad D = \begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix}, \qquad A = PDP^{-1}$$

Writing $A$ like this is diagonalization. It makes powers easy, because the $P^{-1}P$ pairs in the middle cancel:

$$A^n = (PDP^{-1})(PDP^{-1})\cdots(PDP^{-1}) = PD^nP^{-1}, \qquad D^n = \begin{pmatrix} \lambda_1^n & 0 \\ 0 & \lambda_2^n \end{pmatrix}$$

Long-term behaviour

If $\vec{x}_n = A^n\vec{x}_0$ describes a system after $n$ steps, the eigenvalues control what happens:

• An eigenvalue with $|\lambda| \lt 1$ gives a term $\lambda^n \to 0$: that part dies away.
• An eigenvalue $\lambda = 1$ gives a term that stays the same: that’s the steady state.
• An eigenvalue with $|\lambda| \gt 1$ gives a term that grows.

For a population moving between two places (a transition matrix whose columns add to $1$), one eigenvalue is always $1$, and its eigenvector gives the long-term split. This links to Markov chains, and eigenvalues also describe the solutions of coupled differential equations in phase portraits.

Your GDC can find eigenvalues and eigenvectors for you, and can check $PDP^{-1}$ and $A^n$. You should also be able to do $2 \times 2$ matrices by hand.

Worked examples

Example 1: Checking for an eigenvector

Let $B = \begin{pmatrix} 3 & 2 \\ 1 & 4 \end{pmatrix}$. Decide whether each vector is an eigenvector of $B$, and if so, give its eigenvalue.

$$\text{(a) } \begin{pmatrix} 2 \\ -1 \end{pmatrix} \qquad \text{(b) } \begin{pmatrix} 1 \\ 2 \end{pmatrix}$$

Solution.

(a) $B\begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 6 - 2 \\ 2 - 4 \end{pmatrix} = \begin{pmatrix} 4 \\ -2 \end{pmatrix} = 2\begin{pmatrix} 2 \\ -1 \end{pmatrix}$. Yes: it’s an eigenvector with eigenvalue $2$.

(b) $B\begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 3 + 4 \\ 1 + 8 \end{pmatrix} = \begin{pmatrix} 7 \\ 9 \end{pmatrix}$. Is this a multiple of $\begin{pmatrix} 1 \\ 2 \end{pmatrix}$? It would need $7 = \lambda$ and $9 = 2\lambda$, which can’t both be true. So no, it’s not an eigenvector.

Example 2: Eigenvalues and eigenvectors by hand

Find the eigenvalues and corresponding eigenvectors of $A = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}$.

Solution. The characteristic equation:

$$\begin{aligned} \det(A - \lambda I) = (4 - \lambda)(3 - \lambda) - 1(2) &= 0 \\ \lambda^2 - 7\lambda + 10 &= 0 \\ (\lambda - 5)(\lambda - 2) &= 0 \end{aligned}$$

The eigenvalues are $\lambda = 5$ and $\lambda = 2$. (Shortcut check: sum of the diagonal $= 7$, $\det A = 12 - 2 = 10$.)

For $\lambda = 5$: $A - 5I = \begin{pmatrix} -1 & 1 \\ 2 & -2 \end{pmatrix}$. With $\vec{v} = \begin{pmatrix} x \\ y \end{pmatrix}$, the first row gives $-x + y = 0$, so $y = x$. (The second row, $2x - 2y = 0$, says the same thing.) Take $\vec{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$.

For $\lambda = 2$: $A - 2I = \begin{pmatrix} 2 & 1 \\ 2 & 1 \end{pmatrix}$. Both rows give $2x + y = 0$, so $y = -2x$. Take $\vec{v}_2 = \begin{pmatrix} 1 \\ -2 \end{pmatrix}$.

Check: $A\begin{pmatrix} 1 \\ -2 \end{pmatrix} = \begin{pmatrix} 4 - 2 \\ 2 - 6 \end{pmatrix} = \begin{pmatrix} 2 \\ -4 \end{pmatrix} = 2\begin{pmatrix} 1 \\ -2 \end{pmatrix}$ ✓.

Example 3: Diagonalizing and finding a power

For $A = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}$ from Example 2:

• (a) Write $A$ in the form $PDP^{-1}$.
• (b) Find an expression for $A^n$, and use it to find $A^3$.

Solution.

(a) Put the eigenvectors in the columns of $P$, and the eigenvalues in the same order in $D$:

$$P = \begin{pmatrix} 1 & 1 \\ 1 & -2 \end{pmatrix}, \qquad D = \begin{pmatrix} 5 & 0 \\ 0 & 2 \end{pmatrix}$$

$\det P = -2 - 1 = -3$, so

$$P^{-1} = \frac{1}{-3}\begin{pmatrix} -2 & -1 \\ -1 & 1 \end{pmatrix} = \frac{1}{3}\begin{pmatrix} 2 & 1 \\ 1 & -1 \end{pmatrix}$$

(b)

$$\begin{aligned} A^n = PD^nP^{-1} &= \begin{pmatrix} 1 & 1 \\ 1 & -2 \end{pmatrix}\begin{pmatrix} 5^n & 0 \\ 0 & 2^n \end{pmatrix}\frac{1}{3}\begin{pmatrix} 2 & 1 \\ 1 & -1 \end{pmatrix} \\ &= \frac{1}{3}\begin{pmatrix} 5^n & 2^n \\ 5^n & -2 \cdot 2^n \end{pmatrix}\begin{pmatrix} 2 & 1 \\ 1 & -1 \end{pmatrix} \\ &= \frac{1}{3}\begin{pmatrix} 2 \cdot 5^n + 2^n & 5^n - 2^n \\ 2 \cdot 5^n - 2 \cdot 2^n & 5^n + 2 \cdot 2^n \end{pmatrix} \end{aligned}$$

For $n = 3$, with $5^3 = 125$ and $2^3 = 8$:

$$A^3 = \frac{1}{3}\begin{pmatrix} 258 & 117 \\ 234 & 141 \end{pmatrix} = \begin{pmatrix} 86 & 39 \\ 78 & 47 \end{pmatrix}$$

Check: calculating $A^3$ directly on a GDC gives the same matrix. ✓

Example 4: Two towns in the long run

Each year, $10\%$ of the people in town $A$ move to town $B$, and $20\%$ of the people in town $B$ move to town $A$. Everyone else stays. At the start, $5000$ people live in $A$ and $7000$ in $B$. Assume the total stays the same.

• (a) Write down the matrix $M$ with $\begin{pmatrix} a_{n+1} \\ b_{n+1} \end{pmatrix} = M\begin{pmatrix} a_n \\ b_n \end{pmatrix}$.
• (b) Find the eigenvalues and eigenvectors of $M$.
• (c) Find expressions for $a_n$ and $b_n$, and describe the long-term populations.

Solution.

(a) Next year, town $A$ keeps $90\%$ of its people and gains $20\%$ of $B$‘s: $a_{n+1} = 0.9a_n + 0.2b_n$. Similarly $b_{n+1} = 0.1a_n + 0.8b_n$.

$$M = \begin{pmatrix} 0.9 & 0.2 \\ 0.1 & 0.8 \end{pmatrix}$$

(b) $\lambda^2 - 1.7\lambda + (0.72 - 0.02) = 0$, that is, $\lambda^2 - 1.7\lambda + 0.7 = 0$, which factors as $(\lambda - 1)(\lambda - 0.7) = 0$. So $\lambda = 1$ or $\lambda = 0.7$.

For $\lambda = 1$: $M - I = \begin{pmatrix} -0.1 & 0.2 \\ 0.1 & -0.2 \end{pmatrix}$ gives $-0.1x + 0.2y = 0$, so $x = 2y$: $\vec{v}_1 = \begin{pmatrix} 2 \\ 1 \end{pmatrix}$.

For $\lambda = 0.7$: $M - 0.7I = \begin{pmatrix} 0.2 & 0.2 \\ 0.1 & 0.1 \end{pmatrix}$ gives $x + y = 0$: $\vec{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$.

(c) With $P = \begin{pmatrix} 2 & 1 \\ 1 & -1 \end{pmatrix}$, $D = \begin{pmatrix} 1 & 0 \\ 0 & 0.7 \end{pmatrix}$ and $P^{-1} = \dfrac{1}{3}\begin{pmatrix} 1 & 1 \\ 1 & -2 \end{pmatrix}$:

$$\begin{aligned} \begin{pmatrix} a_n \\ b_n \end{pmatrix} = PD^nP^{-1}\begin{pmatrix} 5000 \\ 7000 \end{pmatrix} &= \begin{pmatrix} 2 & 1 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & 0.7^n \end{pmatrix}\begin{pmatrix} 4000 \\ -3000 \end{pmatrix} \\ &= \begin{pmatrix} 2 & 1 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} 4000 \\ -3000(0.7)^n \end{pmatrix} = \begin{pmatrix} 8000 - 3000(0.7)^n \\ 4000 + 3000(0.7)^n \end{pmatrix} \end{aligned}$$

As $n \to \infty$, $0.7^n \to 0$, so the populations approach $8000$ in town $A$ and $4000$ in town $B$. That’s the eigenvector $\begin{pmatrix} 2 \\ 1 \end{pmatrix}$ scaled to the total of $12\,000$.

Check after one year: $a_1 = 8000 - 3000(0.7) = 5900$, and directly $0.9(5000) + 0.2(7000) = 5900$ ✓.

Common mistakes

Accepting the zero vector as an eigenvector. $A\vec{0} = \lambda\vec{0}$ for every $\lambda$, so it tells you nothing. Eigenvectors must be non-zero.

Subtracting λ from every entry. $A - \lambda I$ subtracts $\lambda$ only from the leading diagonal. The other entries stay the same.

Expecting a single answer for an eigenvector. The equations from $(A - \lambda I)\vec{v} = \vec{0}$ always reduce to one relation like $y = -2x$. Any non-zero vector satisfying it is correct, so $\begin{pmatrix} 1 \\ -2 \end{pmatrix}$ and $\begin{pmatrix} -3 \\ 6 \end{pmatrix}$ are both fine.

Mismatching the orders in P and D. The eigenvalue in column $1$ of $D$ must belong to the eigenvector in column $1$ of $P$. Swap one without the other and $PDP^{-1}$ is no longer $A$.

Writing Aⁿ = PⁿDⁿ(P⁻¹)ⁿ or PⁿDⁿP⁻¹. Only $D$ gets the power: $A^n = PD^nP^{-1}$. That’s the whole point of diagonalizing.

Raising a non-diagonal matrix to a power entry by entry. $D^n$ is found by raising each diagonal entry to the power $n$, but this only works because $D$ is diagonal. For any other matrix, $A^n$ is not found entry by entry.

Practice

1. (Warm-up) Show that $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$ is an eigenvector of $\begin{pmatrix} 3 & 2 \\ 1 & 4 \end{pmatrix}$, and state its eigenvalue.

Solution

$$\begin{pmatrix} 3 & 2 \\ 1 & 4 \end{pmatrix}\begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 5 \\ 5 \end{pmatrix} = 5\begin{pmatrix} 1 \\ 1 \end{pmatrix}$$

It’s an eigenvector with eigenvalue $5$.

2. (Warm-up) Find the characteristic polynomial and the eigenvalues of $C = \begin{pmatrix} 5 & -2 \\ 1 & 2 \end{pmatrix}$.

Solution

$$\det(C - \lambda I) = (5 - \lambda)(2 - \lambda) - (-2)(1) = \lambda^2 - 7\lambda + 10 + 2 = \lambda^2 - 7\lambda + 12$$

$\lambda^2 - 7\lambda + 12 = (\lambda - 3)(\lambda - 4) = 0$, so the eigenvalues are $3$ and $4$.

3. (Warm-up) Find the eigenvalues of $T = \begin{pmatrix} 3 & 0 \\ 5 & -1 \end{pmatrix}$. What do you notice?

Solution

$$\det(T - \lambda I) = (3 - \lambda)(-1 - \lambda) - 0(5) = 0$$

So $\lambda = 3$ or $\lambda = -1$. These are just the entries on the leading diagonal: when one of the off-diagonal entries is $0$, the eigenvalues are the diagonal entries.

4. (Core) Find an eigenvector for each eigenvalue of $C = \begin{pmatrix} 5 & -2 \\ 1 & 2 \end{pmatrix}$ from question 2.

Solution

$\lambda = 3$: $C - 3I = \begin{pmatrix} 2 & -2 \\ 1 & -1 \end{pmatrix}$, giving $x - y = 0$. Eigenvector $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$.

$\lambda = 4$: $C - 4I = \begin{pmatrix} 1 & -2 \\ 1 & -2 \end{pmatrix}$, giving $x = 2y$. Eigenvector $\begin{pmatrix} 2 \\ 1 \end{pmatrix}$.

Check: $C\begin{pmatrix} 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 10 - 2 \\ 2 + 2 \end{pmatrix} = \begin{pmatrix} 8 \\ 4 \end{pmatrix} = 4\begin{pmatrix} 2 \\ 1 \end{pmatrix}$ ✓.

5. (Core) For $C = \begin{pmatrix} 5 & -2 \\ 1 & 2 \end{pmatrix}$:

• (a) Write $C = PDP^{-1}$, giving $P$, $D$ and $P^{-1}$.
• (b) Use $C^5 = PD^5P^{-1}$ to find $C^5$.

Solution

(a) Using question 4:

$$P = \begin{pmatrix} 1 & 2 \\ 1 & 1 \end{pmatrix}, \qquad D = \begin{pmatrix} 3 & 0 \\ 0 & 4 \end{pmatrix}, \qquad P^{-1} = \frac{1}{-1}\begin{pmatrix} 1 & -2 \\ -1 & 1 \end{pmatrix} = \begin{pmatrix} -1 & 2 \\ 1 & -1 \end{pmatrix}$$

(b) $3^5 = 243$ and $4^5 = 1024$.

$$\begin{aligned} C^5 &= \begin{pmatrix} 1 & 2 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} 243 & 0 \\ 0 & 1024 \end{pmatrix}\begin{pmatrix} -1 & 2 \\ 1 & -1 \end{pmatrix} \\ &= \begin{pmatrix} 243 & 2048 \\ 243 & 1024 \end{pmatrix}\begin{pmatrix} -1 & 2 \\ 1 & -1 \end{pmatrix} \\ &= \begin{pmatrix} -243 + 2048 & 486 - 2048 \\ -243 + 1024 & 486 - 1024 \end{pmatrix} = \begin{pmatrix} 1805 & -1562 \\ 781 & -538 \end{pmatrix} \end{aligned}$$

A GDC calculation of $C^5$ agrees.

6. (Core) A matrix $A$ has eigenvalue $3$ with eigenvector $\begin{pmatrix} 2 \\ 1 \end{pmatrix}$ and eigenvalue $-1$ with eigenvector $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$. Find $A$.

Solution

$A = PDP^{-1}$ with $P = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}$ and $D = \begin{pmatrix} 3 & 0 \\ 0 & -1 \end{pmatrix}$. $\det P = 2 - 1 = 1$, so $P^{-1} = \begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix}$.

$$A = \begin{pmatrix} 6 & -1 \\ 3 & -1 \end{pmatrix}\begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix} = \begin{pmatrix} 6 + 1 & -6 - 2 \\ 3 + 1 & -3 - 2 \end{pmatrix} = \begin{pmatrix} 7 & -8 \\ 4 & -5 \end{pmatrix}$$

(Here $PD = \begin{pmatrix} 6 & -1 \\ 3 & -1 \end{pmatrix}$.) Check: $A\begin{pmatrix} 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 14 - 8 \\ 8 - 5 \end{pmatrix} = \begin{pmatrix} 6 \\ 3 \end{pmatrix} = 3\begin{pmatrix} 2 \\ 1 \end{pmatrix}$ ✓.

7. (Core) A town has two gyms with $1000$ members between them. Each month, $15\%$ of Gym $A$‘s members switch to Gym $B$, and $5\%$ of Gym $B$‘s members switch to Gym $A$. At the start, Gym $A$ has $600$ members and Gym $B$ has $400$.

• (a) Write down the matrix $G$ such that $\begin{pmatrix} a_{n+1} \\ b_{n+1} \end{pmatrix} = G\begin{pmatrix} a_n \\ b_n \end{pmatrix}$.
• (b) Find the eigenvalues of $G$ and an eigenvector for each.
• (c) Find expressions for $a_n$ and $b_n$, and the long-term number of members at each gym.

Solution

(a)

$$G = \begin{pmatrix} 0.85 & 0.05 \\ 0.15 & 0.95 \end{pmatrix}$$

(b) $\lambda^2 - 1.8\lambda + (0.8075 - 0.0075) = \lambda^2 - 1.8\lambda + 0.8 = (\lambda - 1)(\lambda - 0.8) = 0$, so $\lambda = 1$ or $0.8$.

$\lambda = 1$: $-0.15x + 0.05y = 0$, so $y = 3x$: eigenvector $\begin{pmatrix} 1 \\ 3 \end{pmatrix}$.

$\lambda = 0.8$: $0.05x + 0.05y = 0$, so $y = -x$: eigenvector $\begin{pmatrix} 1 \\ -1 \end{pmatrix}$.

(c) $P = \begin{pmatrix} 1 & 1 \\ 3 & -1 \end{pmatrix}$, $\det P = -4$, $P^{-1} = \dfrac{1}{4}\begin{pmatrix} 1 & 1 \\ 3 & -1 \end{pmatrix}$, and $P^{-1}\begin{pmatrix} 600 \\ 400 \end{pmatrix} = \begin{pmatrix} 250 \\ 350 \end{pmatrix}$.

$$\begin{pmatrix} a_n \\ b_n \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 3 & -1 \end{pmatrix}\begin{pmatrix} 250 \\ 350(0.8)^n \end{pmatrix} = \begin{pmatrix} 250 + 350(0.8)^n \\ 750 - 350(0.8)^n \end{pmatrix}$$

As $n \to \infty$, $0.8^n \to 0$: in the long run Gym $A$ has about $250$ members and Gym $B$ about $750$.

Check: $a_1 = 250 + 280 = 530$, and directly $0.85(600) + 0.05(400) = 510 + 20 = 530$ ✓.

8. (Challenge) Let $M = \begin{pmatrix} 0.7 & 0.6 \\ 0.3 & 0.4 \end{pmatrix}$.

• (a) Show that the eigenvalues of $M$ are $1$ and $0.1$, and find an eigenvector for each.
• (b) Find $M^n$ in terms of $n$.
• (c) Find the matrix that $M^n$ approaches as $n \to \infty$, and explain what its columns tell you.

Solution

(a) $\lambda^2 - 1.1\lambda + (0.28 - 0.18) = \lambda^2 - 1.1\lambda + 0.1 = (\lambda - 1)(\lambda - 0.1) = 0$. ✓

$\lambda = 1$: $-0.3x + 0.6y = 0$, so $x = 2y$: $\begin{pmatrix} 2 \\ 1 \end{pmatrix}$. $\lambda = 0.1$: $0.6x + 0.6y = 0$: $\begin{pmatrix} 1 \\ -1 \end{pmatrix}$.

(b) $P = \begin{pmatrix} 2 & 1 \\ 1 & -1 \end{pmatrix}$, $P^{-1} = \dfrac{1}{3}\begin{pmatrix} 1 & 1 \\ 1 & -2 \end{pmatrix}$.

$$\begin{aligned} M^n &= \begin{pmatrix} 2 & 1 \\ 1 & -1 \end{pmatrix}\begin{pmatrix} 1 & 0 \\ 0 & 0.1^n \end{pmatrix}\frac{1}{3}\begin{pmatrix} 1 & 1 \\ 1 & -2 \end{pmatrix} \\ &= \frac{1}{3}\begin{pmatrix} 2 & 0.1^n \\ 1 & -0.1^n \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 1 & -2 \end{pmatrix} = \frac{1}{3}\begin{pmatrix} 2 + 0.1^n & 2 - 2(0.1)^n \\ 1 - 0.1^n & 1 + 2(0.1)^n \end{pmatrix} \end{aligned}$$

Check $n = 1$: $\dfrac{1}{3}\begin{pmatrix} 2.1 & 1.8 \\ 0.9 & 1.2 \end{pmatrix} = \begin{pmatrix} 0.7 & 0.6 \\ 0.3 & 0.4 \end{pmatrix}$ ✓.

(c) As $n \to \infty$, $0.1^n \to 0$:

$$M^n \to \frac{1}{3}\begin{pmatrix} 2 & 2 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} \frac{2}{3} & \frac{2}{3} \\ \frac{1}{3} & \frac{1}{3} \end{pmatrix}$$

Both columns are the same: whatever the starting split, the long-term split is $\dfrac{2}{3}$ and $\dfrac{1}{3}$, the eigenvector $\begin{pmatrix} 2 \\ 1 \end{pmatrix}$ scaled so its entries add to $1$.

9. (Challenge) The matrix $A = \begin{pmatrix} 1 & k \\ 2 & 3 \end{pmatrix}$ has an eigenvalue of $5$.

• (a) Find $k$.
• (b) Find the other eigenvalue, and an eigenvector for each eigenvalue.

Solution

(a) $5$ is a root of $\det(A - \lambda I) = 0$:

$$\det\begin{pmatrix} 1 - 5 & k \\ 2 & 3 - 5 \end{pmatrix} = (-4)(-2) - 2k = 8 - 2k = 0 \quad\Rightarrow\quad k = 4$$

(b) With $k = 4$: $\lambda^2 - 4\lambda + (3 - 8) = \lambda^2 - 4\lambda - 5 = (\lambda - 5)(\lambda + 1) = 0$. The other eigenvalue is $-1$. (The two eigenvalues add to the sum of the diagonal, $1 + 3 = 4$.)

$\lambda = 5$: $A - 5I = \begin{pmatrix} -4 & 4 \\ 2 & -2 \end{pmatrix}$, so $y = x$: eigenvector $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$.

$\lambda = -1$: $A + I = \begin{pmatrix} 2 & 4 \\ 2 & 4 \end{pmatrix}$, so $x = -2y$: eigenvector $\begin{pmatrix} -2 \\ 1 \end{pmatrix}$.

Check: $A\begin{pmatrix} -2 \\ 1 \end{pmatrix} = \begin{pmatrix} -2 + 4 \\ -4 + 3 \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \end{pmatrix} = -1\begin{pmatrix} -2 \\ 1 \end{pmatrix}$ ✓.