IB Mathematics Applications & Interpretation Topic…

Finding The Directions A Matrix Does Not Turn

When a matrix multiplies a vector, it usually stretches it and swings it into a new direction.
For most matrices, though, there are a few special directions that only get stretched, never turned.
A vector along such a direction is an eigenvector, and the factor it is stretched by is the eigenvalue.

Note

These special directions are the secret to computing huge matrix powers quickly, which is exactly what the long-run Markov chains and population models in later topics rely on.

What Eigenvalues And Eigenvectors Mean

An eigenvector $\mathbf{v}$ of a matrix $M$ is a non-zero vector whose direction is unchanged when multiplied by $M$.
Multiplying by $M$ only scales it, by a number $\lambda$ called the eigenvalue. $$M\mathbf{v} = \lambda\mathbf{v}, \quad \mathbf{v} \neq \mathbf{0}$$
Here is what that means with actual numbers. Take $M = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$ and the vector $\mathbf{v} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$. $$M\mathbf{v} = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}\begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \end{pmatrix} = 3\mathbf{v}$$
The output points the same way as $\mathbf{v}$, just three times as long, so $\mathbf{v}$ is an eigenvector with eigenvalue $\lambda = 3$.
By contrast, $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ maps to $\begin{pmatrix} 2 \\ 1 \end{pmatrix}$, which points in a new direction, so it is not an eigenvector.

Note

Any non-zero multiple of an eigenvector is also an eigenvector, so eigenvectors are really directions, not single points.
A $2 \times 2$ matrix has two eigenvalues, which in this course are real and distinct.

Finding Eigenvalues: The Characteristic Equation

Start from the definition $M\mathbf{v} = \lambda\mathbf{v}$ and move everything to one side.
$M\mathbf{v} - \lambda\mathbf{v} = \mathbf{0}$, which factorises as $(M - \lambda I)\mathbf{v} = \mathbf{0}$.
For a non-zero $\mathbf{v}$ to exist, the matrix $M - \lambda I$ must be singular.
A singular matrix has determinant 0, so $\det(M - \lambda I) = 0$.
This determinant gives the characteristic equation, which is a quadratic in $\lambda$ for a $2 \times 2$ matrix. $$\det(M - \lambda I) = 0$$
For a general $2 \times 2$ matrix, expanding $\det\begin{pmatrix} a-\lambda & b \\ c & d-\lambda \end{pmatrix}$ gives $(a-\lambda)(d-\lambda) - bc = 0$.
1. This quadratic is the characteristic equation, and its two solutions are the eigenvalues.

Example question

Find the eigenvalues and eigenvectors of $M = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$.

Solution

Form $M - \lambda I = \begin{pmatrix} 2-\lambda & 1 \\ 1 & 2-\lambda \end{pmatrix}$.
Set the determinant to 0: $(2-\lambda)(2-\lambda) - (1)(1) = 0$, that is $(2-\lambda)^2 - 1 = 0$.
Solve: $(2-\lambda)^2 = 1$, so $2 - \lambda = \pm 1$, giving $\lambda = 1$ or $\lambda = 3$.
For $\lambda = 3$: $(M - 3I)\mathbf{v} = \mathbf{0}$ gives $-v_1 + v_2 = 0$, so $\mathbf{v} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$.
For $\lambda = 1$: $(M - I)\mathbf{v} = \mathbf{0}$ gives $v_1 + v_2 = 0$, so $\mathbf{v} = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$.

Common Mistake

Form $M - \lambda I$ carefully: subtract $\lambda$ only from the diagonal entries.
Each eigenvalue has a whole line of eigenvectors, so any non-zero multiple of your answer is correct.

Diagonalization: Writing $M = PDP^{-1}$

Once you have the eigenvalues and eigenvectors, you can rebuild $M$ in a form that is easy to work with.
Put the eigenvectors as the columns of a matrix $P$, and the matching eigenvalues on the diagonal of a matrix $D$.
The order must match: column 1 of $P$ pairs with the first eigenvalue in $D$.
Then $M = PDP^{-1}$, which separates the stretching (in $D$) from the change of direction (in $P$). $$M = PDP^{-1}$$
Use the eigenpairs we found for $M = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$ to build $P$ and $D$.
The eigenvectors become the columns of $P$, and the matching eigenvalues go on the diagonal of $D$. $$P = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}, \qquad D = \begin{pmatrix} 3 & 0 \\ 0 & 1 \end{pmatrix}$$
1. Column 1 of $P$ is the $\lambda = 3$ eigenvector, so 3 sits in the first diagonal slot of $D$.

Note

This works only when the eigenvalues are real and distinct, which is all this course requires.
Keep the eigenvector columns of $P$ in the same order as the eigenvalues in $D$.

Matrix Powers Made Easy

Diagonalization pays off when you need a high power of $M$, such as many steps of a model.
Because the $P$ and $P^{-1}$ factors cancel in the middle, only the diagonal matrix is raised to the power.
$M^n = (PDP^{-1})(PDP^{-1})\cdots$ collapses to $PD^nP^{-1}$, because $P^{-1}P = I$ each time.
Raising a diagonal matrix to a power just raises each diagonal entry to that power, which is easy.
This power formula is given in your formula booklet under AHL 1.15.

$$M^n = PD^nP^{-1}, \qquad D^n = \begin{pmatrix} \lambda_1^{\,n} & 0 \\ 0 & \lambda_2^{\,n} \end{pmatrix}$$

Example question

Using $P = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$ and $D = \begin{pmatrix} 3 & 0 \\ 0 & 1 \end{pmatrix}$ for $M = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$, find $M^2$ with $M^2 = PD^2P^{-1}$ and check it directly.

Solution

Square the diagonal: $D^2 = \begin{pmatrix} 9 & 0 \\ 0 & 1 \end{pmatrix}$.
Find $P^{-1}$: $\det P = (1)(-1) - (1)(1) = -2$, so $P^{-1} = -\dfrac{1}{2}\begin{pmatrix} -1 & -1 \\ -1 & 1 \end{pmatrix} = \begin{pmatrix} \tfrac{1}{2} & \tfrac{1}{2} \\ \tfrac{1}{2} & -\tfrac{1}{2} \end{pmatrix}$.
Multiply out: $PD^2 = \begin{pmatrix} 9 & 1 \\ 9 & -1 \end{pmatrix}$, then $PD^2P^{-1} = \begin{pmatrix} 5 & 4 \\ 4 & 5 \end{pmatrix}$.
Check directly: $M^2 = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} = \begin{pmatrix} 5 & 4 \\ 4 & 5 \end{pmatrix}$, which matches.

Example question

A matrix $M$ has eigenvalues $\lambda_1 = 1$ and $\lambda_2 = 0.5$. Describe what happens to $M^n$ as $n$ becomes very large.

Solution

In $M^n = PD^nP^{-1}$, only $D^n$ changes with $n$.
The diagonal entries are $\lambda_1^{\,n} = 1^n = 1$ and $\lambda_2^{\,n} = 0.5^n$.
As $n \to \infty$, $0.5^n \to 0$, so $D^n \to \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$.
So the long-run behaviour is controlled entirely by the eigenvalue 1 and its eigenvector, which becomes the steady state.

Common Mistake

Only $D$ is raised to the power: $M^n = PD^nP^{-1}$, never $P^nD^nP^{-1}$.
Raise each diagonal entry separately; do not add or mix them.

Why It Matters: Steady States And Long-Run Behaviour

The eigenvalues reveal the long-run behaviour of any repeated process modelled by a matrix.
An eigenvalue of 1 marks the steady state of a Markov chain, which you meet in AHL 4.19.
So this section is the key that unlocks the long-term predictions those later topics ask for.

Largest eigenvalue	What the model does over time
Equal to 1	Settles to a steady state
Between 0 and 1	Decays towards zero
Greater than 1	Grows without bound

Active recall

Show that $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$ is an eigenvector of $\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$ and state its eigenvalue.
How do you set up the characteristic equation for a $2 \times 2$ matrix?
In $M = PDP^{-1}$, what goes into $P$ and what goes into $D$?
Why is $M^n = PD^nP^{-1}$ so much easier to compute than multiplying $M$ by itself $n$ times?
What does an eigenvalue of 1 tell you about a long-run process?

Finding The Directions A Matrix Does Not Turn

When a matrix multiplies a vector, it usually stretches it and swings it into a new direction.
For most matrices, though, there are a few special directions that only get stretched, never turned.
A vector along such a direction is an eigenvector, and the factor it is stretched by is the eigenvalue.

Note

These special directions are the secret to computing huge matrix powers quickly, which is exactly what the long-run Markov chains and population models in later topics rely on.

What Eigenvalues And Eigenvectors Mean

An eigenvector $\mathbf{v}$ of a matrix $M$ is a non-zero vector whose direction is unchanged when multiplied by $M$.
Multiplying by $M$ only scales it, by a number $\lambda$ called the eigenvalue. $$M\mathbf{v} = \lambda\mathbf{v}, \quad \mathbf{v} \neq \mathbf{0}$$
Here is what that means with actual numbers. Take $M = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$ and the vector $\mathbf{v} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$. $$M\mathbf{v} = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}\begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \end{pmatrix} = 3\mathbf{v}$$
The output points the same way as $\mathbf{v}$, just three times as long, so $\mathbf{v}$ is an eigenvector with eigenvalue $\lambda = 3$.
By contrast, $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ maps to $\begin{pmatrix} 2 \\ 1 \end{pmatrix}$, which points in a new direction, so it is not an eigenvector.

Note

Any non-zero multiple of an eigenvector is also an eigenvector, so eigenvectors are really directions, not single points.
A $2 \times 2$ matrix has two eigenvalues, which in this course are real and distinct.

Finding Eigenvalues: The Characteristic Equation

Start from the definition $M\mathbf{v} = \lambda\mathbf{v}$ and move everything to one side.
$M\mathbf{v} - \lambda\mathbf{v} = \mathbf{0}$, which factorises as $(M - \lambda I)\mathbf{v} = \mathbf{0}$.
For a non-zero $\mathbf{v}$ to exist, the matrix $M - \lambda I$ must be singular.
A singular matrix has determinant 0, so $\det(M - \lambda I) = 0$.
This determinant gives the characteristic equation, which is a quadratic in $\lambda$ for a $2 \times 2$ matrix. $$\det(M - \lambda I) = 0$$
For a general $2 \times 2$ matrix, expanding $\det\begin{pmatrix} a-\lambda & b \\ c & d-\lambda \end{pmatrix}$ gives $(a-\lambda)(d-\lambda) - bc = 0$.
1. This quadratic is the characteristic equation, and its two solutions are the eigenvalues.

Example question

Find the eigenvalues and eigenvectors of $M = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$.

Solution

Form $M - \lambda I = \begin{pmatrix} 2-\lambda & 1 \\ 1 & 2-\lambda \end{pmatrix}$.
Set the determinant to 0: $(2-\lambda)(2-\lambda) - (1)(1) = 0$, that is $(2-\lambda)^2 - 1 = 0$.
Solve: $(2-\lambda)^2 = 1$, so $2 - \lambda = \pm 1$, giving $\lambda = 1$ or $\lambda = 3$.
For $\lambda = 3$: $(M - 3I)\mathbf{v} = \mathbf{0}$ gives $-v_1 + v_2 = 0$, so $\mathbf{v} = \begin{pmatrix} 1 \\ 1 \end{pmatrix}$.
For $\lambda = 1$: $(M - I)\mathbf{v} = \mathbf{0}$ gives $v_1 + v_2 = 0$, so $\mathbf{v} = \begin{pmatrix} 1 \\ -1 \end{pmatrix}$.

Common Mistake

Form $M - \lambda I$ carefully: subtract $\lambda$ only from the diagonal entries.
Each eigenvalue has a whole line of eigenvectors, so any non-zero multiple of your answer is correct.

Diagonalization: Writing $M = PDP^{-1}$

Once you have the eigenvalues and eigenvectors, you can rebuild $M$ in a form that is easy to work with.
Put the eigenvectors as the columns of a matrix $P$, and the matching eigenvalues on the diagonal of a matrix $D$.
The order must match: column 1 of $P$ pairs with the first eigenvalue in $D$.
Then $M = PDP^{-1}$, which separates the stretching (in $D$) from the change of direction (in $P$). $$M = PDP^{-1}$$
Use the eigenpairs we found for $M = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$ to build $P$ and $D$.
The eigenvectors become the columns of $P$, and the matching eigenvalues go on the diagonal of $D$. $$P = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}, \qquad D = \begin{pmatrix} 3 & 0 \\ 0 & 1 \end{pmatrix}$$
1. Column 1 of $P$ is the $\lambda = 3$ eigenvector, so 3 sits in the first diagonal slot of $D$.

Note

This works only when the eigenvalues are real and distinct, which is all this course requires.
Keep the eigenvector columns of $P$ in the same order as the eigenvalues in $D$.

Matrix Powers Made Easy

Diagonalization pays off when you need a high power of $M$, such as many steps of a model.
Because the $P$ and $P^{-1}$ factors cancel in the middle, only the diagonal matrix is raised to the power.
$M^n = (PDP^{-1})(PDP^{-1})\cdots$ collapses to $PD^nP^{-1}$, because $P^{-1}P = I$ each time.
Raising a diagonal matrix to a power just raises each diagonal entry to that power, which is easy.
This power formula is given in your formula booklet under AHL 1.15.

$$M^n = PD^nP^{-1}, \qquad D^n = \begin{pmatrix} \lambda_1^{\,n} & 0 \\ 0 & \lambda_2^{\,n} \end{pmatrix}$$

Example question

Solution

Square the diagonal: $D^2 = \begin{pmatrix} 9 & 0 \\ 0 & 1 \end{pmatrix}$.
Find $P^{-1}$: $\det P = (1)(-1) - (1)(1) = -2$, so $P^{-1} = -\dfrac{1}{2}\begin{pmatrix} -1 & -1 \\ -1 & 1 \end{pmatrix} = \begin{pmatrix} \tfrac{1}{2} & \tfrac{1}{2} \\ \tfrac{1}{2} & -\tfrac{1}{2} \end{pmatrix}$.
Multiply out: $PD^2 = \begin{pmatrix} 9 & 1 \\ 9 & -1 \end{pmatrix}$, then $PD^2P^{-1} = \begin{pmatrix} 5 & 4 \\ 4 & 5 \end{pmatrix}$.
Check directly: $M^2 = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} = \begin{pmatrix} 5 & 4 \\ 4 & 5 \end{pmatrix}$, which matches.

Example question

A matrix $M$ has eigenvalues $\lambda_1 = 1$ and $\lambda_2 = 0.5$. Describe what happens to $M^n$ as $n$ becomes very large.

Solution

In $M^n = PD^nP^{-1}$, only $D^n$ changes with $n$.
The diagonal entries are $\lambda_1^{\,n} = 1^n = 1$ and $\lambda_2^{\,n} = 0.5^n$.
As $n \to \infty$, $0.5^n \to 0$, so $D^n \to \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$.
So the long-run behaviour is controlled entirely by the eigenvalue 1 and its eigenvector, which becomes the steady state.

Common Mistake

Only $D$ is raised to the power: $M^n = PD^nP^{-1}$, never $P^nD^nP^{-1}$.
Raise each diagonal entry separately; do not add or mix them.

Why It Matters: Steady States And Long-Run Behaviour

The eigenvalues reveal the long-run behaviour of any repeated process modelled by a matrix.
An eigenvalue of 1 marks the steady state of a Markov chain, which you meet in AHL 4.19.
So this section is the key that unlocks the long-term predictions those later topics ask for.

Largest eigenvalue	What the model does over time
Equal to 1	Settles to a steady state
Between 0 and 1	Decays towards zero
Greater than 1	Grows without bound

Active recall

Show that $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$ is an eigenvector of $\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$ and state its eigenvalue.
How do you set up the characteristic equation for a $2 \times 2$ matrix?
In $M = PDP^{-1}$, what goes into $P$ and what goes into $D$?
Why is $M^n = PD^nP^{-1}$ so much easier to compute than multiplying $M$ by itself $n$ times?
What does an eigenvalue of 1 tell you about a long-run process?

IB Mathematics Applications & Interpretation AHL 1.15—eigenvalues and Eigenvectors Notes

IB Mathematics Applications & Interpretation AHL 1.15—eigenvalues and Eigenvectors Notes

Finding The Directions A Matrix Does Not Turn

What Eigenvalues And Eigenvectors Mean

Finding Eigenvalues: The Characteristic Equation

Diagonalization: Writing $M = PDP^{-1}$

Matrix Powers Made Easy

Why It Matters: Steady States And Long-Run Behaviour

Finding The Directions A Matrix Does Not Turn

What Eigenvalues And Eigenvectors Mean

Finding Eigenvalues: The Characteristic Equation

Diagonalization: Writing $M = PDP^{-1}$

Matrix Powers Made Easy

Why It Matters: Steady States And Long-Run Behaviour

Number and Algebra

Functions

Geometry and Trigonometry

Statistics and Probability

Calculus

IB Mathematics Applications & Interpretation AHL 1.15—eigenvalues and Eigenvectors Notes

Number and Algebra

Functions

Geometry and Trigonometry

Statistics and Probability

Calculus

IB Mathematics Applications & Interpretation AHL 1.15—eigenvalues and Eigenvectors Notes

Finding The Directions A Matrix Does Not Turn

What Eigenvalues And Eigenvectors Mean

Finding Eigenvalues: The Characteristic Equation

Diagonalization: Writing $M = PDP^{-1}$

Matrix Powers Made Easy

Why It Matters: Steady States And Long-Run Behaviour

Finding The Directions A Matrix Does Not Turn

What Eigenvalues And Eigenvectors Mean

Finding Eigenvalues: The Characteristic Equation

Diagonalization: Writing $M = PDP^{-1}$

Matrix Powers Made Easy

Why It Matters: Steady States And Long-Run Behaviour