Why does requiring a non-zero eigenvector force the determinant to be zero?

The equation $(A-\lambda I)\mathbf{v}=\mathbf{0}$ always has the trivial solution $\mathbf{v}=\mathbf{0}$. We want a non-zero solution, and a square linear system has non-zero solutions only when its matrix is singular, that is, when its determinant is zero. So demanding a genuine eigenvector is exactly the condition $\det(A-\lambda I)=0$, which is why that becomes the equation for $\lambda$.

My eigenvector looks different from the answer key. Am I wrong?

Probably not. An eigenvector is only defined up to a non-zero scalar, so $\begin{pmatrix}1\\1\end{pmatrix}$, $\begin{pmatrix}2\\2\end{pmatrix}$ and $\begin{pmatrix}-1\\-1\end{pmatrix}$ all describe the same eigen-direction. Check whether your vector is a scalar multiple of the expected one; if it is, it is correct. IB mark schemes and our engine accept any non-zero scalar multiple.

When can a 2x2 matrix actually be diagonalized?

Whenever it has two linearly independent eigenvectors. If the two eigenvalues are distinct (the usual case in AI HL), their eigenvectors are automatically independent, so $A=PDP^{-1}$ works. Trouble only arises with a repeated eigenvalue that supplies just one independent eigenvector; such a matrix is not diagonalizable, but you will rarely meet one at this level.

Why is $A^n=PD^nP^{-1}$ so much easier than multiplying $A$ by itself $n$ times?

Because $A^n=(PDP^{-1})(PDP^{-1})\cdots(PDP^{-1})$, and each interior $P^{-1}P$ collapses to the identity, leaving $PD^nP^{-1}$. Raising the diagonal matrix $D$ to the power $n$ just means raising each eigenvalue to that power, which is trivial. So one diagonalization replaces $n$ full matrix multiplications, and it even lets you write a closed formula for any $n$.

How do eigenvalues connect to Markov chains and coupled differential equations?

Both are systems driven by a matrix. In a Markov chain the state updates as $\mathbf{x}_{n+1}=A\mathbf{x}_n$, so $\mathbf{x}_n=A^n\mathbf{x}_0$; the dominant eigenvalue (always 1 for a Markov transition matrix) and its eigenvector give the steady state the chain settles into. For coupled linear differential equations $\dot{\mathbf{x}}=A\mathbf{x}$, the solution is built from $e^{\lambda t}$ terms along the eigenvectors, so the signs and sizes of the eigenvalues decide whether the system grows, decays or oscillates. In both cases the eigenvalues are the DNA of the long-term behaviour.

Eigenvalues and eigenvectors — IB Maths AI HL

Revise Eigenvalues and eigenvectors for IB Mathematics: Applications and Interpretation (1.7)

Looking for free IB Mathematics: Applications and Interpretation notes on Eigenvalues and eigenvectors? This Higher Level lesson covers official syllabus point 1.7 on Maths AI HL. Use the visual steps, key terms, and quick checks below — then practise with IB criterion marking on Eigenvalues and eigenvectors. Built for students searching "IB Maths AI 1.7", "Eigenvalues and eigenvectors IB revision", or a free alternative to paid IB note sites.

Eigenvalues and eigenvectors

Multiply a random vector by a matrix and, in general, two things change at once: its length and its direction. That double change is what makes matrices hard to reason about over many steps. Eigenvalues and eigenvectors cut through it. An eigenvector is a direction the matrix refuses to twist, so along it the transformation reduces to a single number, the eigenvalue, that simply scales. Find those invariant directions and a complicated matrix falls apart into a few clean stretches. This HL lesson takes 1.7 in full for the 2x2 case: the defining relation $A\mathbf{v}=\lambda\mathbf{v}$ , the characteristic equation that delivers the eigenvalues, the eigenvectors that go with them, the diagonalization $A=PDP^{-1}$ that repackages the matrix, the power formula $A^n=PD^nP^{-1}$ , and the pay-off, reading the long-term fate of a system straight from its dominant eigenvalue. As in every calculation topic, the worked examples are laid out exactly the way our engine marks your own practice.

BY THE END, YOU CAN…

1State and use the defining relation $A\mathbf{v}=\lambda\mathbf{v}$ and explain what makes a vector an eigenvector and a scalar an eigenvalue.
2Form the characteristic equation $\det(A-\lambda I)=0$ for a 2x2 matrix and solve it to find the eigenvalues.
3Solve $(A-\lambda I)\mathbf{v}=\mathbf{0}$ to find a corresponding eigenvector for each eigenvalue, giving a representative of the family up to scale.
4Diagonalize a 2x2 matrix as $A=PDP^{-1}$ and use $A^n=PD^nP^{-1}$ to compute matrix powers.
5Interpret the dominant eigenvalue and its eigenvector as the driver of long-term behaviour, linking to Markov chains and coupled differential equations.

In simple terms

A friendly intro before the formal notes — no formulas yet.

The Directions a Transformation Leaves Alone

Most vectors get knocked off course when you multiply them by a matrix: they change both length and direction. An eigenvector is a rare vector that keeps its direction, so the matrix only stretches or flips it. The stretch factor is the eigenvalue. Find those special directions and you understand what the whole transformation really does.

THINK OF IT LIKE…

Picture a spinning globe. Almost every point on the surface sweeps out a circle as the globe turns, so its direction from the centre keeps changing. But the two points on the axis, the poles, never move: their direction is fixed. Those poles are the eigenvectors of the rotation, and because they are not moved at all, their eigenvalue is 1. Every transformation has its own private axes like this, and once you know them you know how the transformation behaves on everything else.

1
Write the transformation as a square matrix $A$ and the target relation as $A\mathbf{v}=\lambda\mathbf{v}$ : a matrix times a special vector equals a number times the same vector.
2
Rearrange to $(A-\lambda I)\mathbf{v}=\mathbf{0}$ and demand a non-zero $\mathbf{v}$ , which forces $\det(A-\lambda I)=0$ . Solve this characteristic equation for the eigenvalues $\lambda$ .
3
For each eigenvalue, substitute back into $(A-\lambda I)\mathbf{v}=\mathbf{0}$ and solve the resulting equations for the eigenvector, choosing the simplest representative of its scale family.
4
Assemble $P$ from the eigenvectors and $D$ from the eigenvalues to get $A=PDP^{-1}$ , then read off powers and long-term behaviour from $A^n=PD^nP^{-1}$ .

Explore the concept

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Key formulas

Tap any symbol to reveal exactly what it means and its units.

$A\mathbf{v} = \lambda\mathbf{v}, \qquad \mathbf{v}\neq\mathbf{0}$

$A\mathbf{v}=\lambda\mathbf{v} \;\Longrightarrow\; A\mathbf{v}-\lambda\mathbf{v}=\mathbf{0} \;\Longrightarrow\; A\mathbf{v}-\lambda I\mathbf{v}=\mathbf{0} \;\Longrightarrow\; (A-\lambda I)\mathbf{v}=\mathbf{0}$

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

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

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

Full topic notes

Formal explanation with the rigour you need for the exam.

The defining relation

Everything starts from one equation. A square matrix $A$ acts on a column vector $\mathbf{v}$ ; usually $A\mathbf{v}$ points somewhere new. But for certain special non-zero vectors the output points along the same line as the input, so it is just a scalar multiple of $\mathbf{v}$ . Those vectors are the eigenvectors, and the scalar is the eigenvalue.

$A\mathbf{v} = \lambda\mathbf{v}, \qquad \mathbf{v}\neq\mathbf{0}$

Reading the equation geometrically: if $\lambda>1$ the eigenvector is stretched, if $0<\lambda<1$ it is shrunk, if $\lambda<0$ it is flipped to the opposite direction (and scaled), if $\lambda=1$ it is left exactly where it was, and if $\lambda=0$ it is collapsed onto the origin. The whole behaviour of $A$ is encoded in these few numbers and directions.

$A$ is an $n\times n$ square matrix; in this lesson $2\times 2$ .
$\mathbf{v}$ is a non-zero column vector, the eigenvector, giving an invariant direction.
$\lambda$ is a scalar, the eigenvalue, giving the stretch/flip factor along that direction.
The zero vector is excluded on purpose: $A\mathbf{0}=\lambda\mathbf{0}$ holds for every $\lambda$ , so it would tell us nothing.

The characteristic equation

To find the eigenvalues we turn $A\mathbf{v}=\lambda\mathbf{v}$ into a condition on $\lambda$ alone. Move everything to one side, insert the identity so the vector can be factored out, and gather terms.

This is a homogeneous linear system. It always has the useless solution $\mathbf{v}=\mathbf{0}$ ; we want a genuine, non-zero eigenvector. A square system has non-zero solutions only when its matrix is singular, so we require the determinant of $(A-\lambda I)$ to vanish. That single condition is the characteristic equation.

\det(A-\lambda I)=0

For a $2\times 2$ matrix $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ this expands to $(a-\lambda)(d-\lambda)-bc=0$ , a quadratic in $\lambda$ . Written out, $\lambda^2-(a+d)\lambda+(ad-bc)=0$ , i.e. $\lambda^2-\operatorname{tr}(A)\,\lambda+\det(A)=0$ . So the two eigenvalues sum to the trace and multiply to the determinant, a check worth remembering.

EXAM TIP

The command term matters. On Paper 1 a 2x2 eigenvalue question expects you to actually form and solve $\det(A-\lambda I)=0$ by hand; writing the characteristic equation is itself a marked line. Reserve the GDC's eigenvalue function for checking your answer or for anything larger than 2x2.

Finding the eigenvectors

Each eigenvalue opens a door back to its eigenvectors. Substitute the value of $\lambda$ into $(A-\lambda I)\mathbf{v}=\mathbf{0}$ and solve the resulting equations for $\mathbf{v}=\begin{pmatrix}x\\y\end{pmatrix}$ . Because $\lambda$ was chosen to make the matrix singular, the two rows are always proportional, so they collapse to a single relation between $x$ and $y$ . That relation defines a whole line of eigenvectors; you quote one convenient representative.

Set up $(A-\lambda I)\mathbf{v}=\mathbf{0}$ with the specific eigenvalue substituted in.
The two rows give the same relation; take either one, e.g. $px+qy=0$ .
Choose a simple non-zero solution, such as setting $y=1$ (or clearing fractions) to get whole-number components.
Any non-zero scalar multiple of your answer is equally correct: the eigenvector is a direction, not a fixed arrow.

Diagonalization: $A=PDP^{-1}$

Once you have the eigenpairs you can rebuild $A$ in a form that exposes its stretches directly. Place the eigenvectors as the columns of a matrix $P$ , and the matching eigenvalues on the diagonal of a matrix $D$ in the same order. Then, provided the eigenvectors are linearly independent (guaranteed here whenever the eigenvalues are distinct), $A=PDP^{-1}$ .

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

The order is the whole game: the first column of $P$ must be the eigenvector for the first diagonal entry $\lambda_1$ , and the second column for $\lambda_2$ . You may swap both together, but never one without the other. Geometrically, $P^{-1}$ rewrites a vector in the eigenvector basis, $D$ scales along each eigen-direction, and $P$ translates back; the messy mixing that $A$ seemed to do is really just independent stretching in disguise.

Powers of a matrix: $A^n=PD^nP^{-1}$

Diagonalization pays off spectacularly for powers. Because the interior factors cancel, $A^n=(PDP^{-1})^n=PD^nP^{-1}$ , and raising a diagonal matrix to a power just raises each diagonal entry to that power. A repeated matrix multiplication becomes a couple of exponentiations of numbers.

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

Long-term behaviour: the dominant eigenvalue

The power formula also tells you the destiny of a system. Write the starting state in the eigenvector basis, $\mathbf{x}_0=c_1\mathbf{v}_1+c_2\mathbf{v}_2$ . Then $\mathbf{x}_n=A^n\mathbf{x}_0=c_1\lambda_1^n\mathbf{v}_1+c_2\lambda_2^n\mathbf{v}_2$ . As $n$ grows, the term with the eigenvalue of largest magnitude, the dominant eigenvalue, swamps the other. The system lines up along that eigenvector, and its size is governed by that eigenvalue.

This is why eigenvalues are the natural language of two headline HL applications. In a Markov chain the state updates as $\mathbf{x}_{n+1}=A\mathbf{x}_n$ with a transition matrix whose columns are probabilities; such a matrix always has $\lambda=1$ as its dominant eigenvalue, and the corresponding eigenvector, scaled so its entries sum to 1, is the steady-state distribution the chain converges to. In coupled linear differential equations $\dot{\mathbf{x}}=A\mathbf{x}$ , the solution is a combination of $e^{\lambda t}$ terms directed along the eigenvectors, so real positive eigenvalues give growth, real negative ones give decay, and complex eigenvalues give oscillation. In every case the eigenvalues decide the long-run story before you compute a single trajectory.

Dominant eigenvalue = the one with the largest $|\lambda|$ . It dictates the long-term growth rate.
$|\lambda_{\text{dom}}|>1$ : the system grows without bound; $|\lambda_{\text{dom}}|<1$ : it decays to zero; $|\lambda_{\text{dom}}|=1$ : it settles to a steady state.
The eigenvector of the dominant eigenvalue gives the limiting ratio of the components (e.g. the stable age structure of a population).
A percentage growth rate follows from the dominant eigenvalue: a factor of $\lambda$ per step means a $(\lambda-1)\times 100\%$ change per step.

Common mistakes examiners penalise

Writing the characteristic equation as $\det(A)-\lambda I=0$ or $\det(A)-\lambda=0$ — it is $\det(A-\lambda I)=0$ : subtract $\lambda$ from the diagonal FIRST, then take the determinant.
Solving $A\mathbf{v}=\mathbf{0}$ instead of $(A-\lambda I)\mathbf{v}=\mathbf{0}$ — the eigenvector satisfies $(A-\lambda I)\mathbf{v}=\mathbf{0}$ for the specific eigenvalue, not $A\mathbf{v}=\mathbf{0}$ .
Giving $\mathbf{v}=\mathbf{0}$ as an eigenvector — the zero vector is excluded by definition; you must supply a non-zero representative.
Marking a scalar multiple 'wrong' — eigenvectors are non-unique up to scale, so $\begin{pmatrix}1\\1\end{pmatrix}$ and $\begin{pmatrix}3\\3\end{pmatrix}$ describe the same eigenvector; do not discard a correct multiple.
Mismatching the order of $P$ and $D$ — the $k$ th column of $P$ must belong to the $k$ th diagonal entry of $D$ ; a mismatched pairing breaks $A=PDP^{-1}$ .
Writing $A^n=P^nD^nP^{-1}$ or $PDP^{-n}$ — the correct power formula is $A^n=PD^nP^{-1}$ ; only the diagonal matrix is raised to the power.
Reaching for the GDC on a 2x2 by-hand question — Paper 1 expects the characteristic equation shown and solved; missing that method line forfeits the M and A marks even if the eigenvalues are right.
Picking the wrong dominant eigenvalue — dominance is by MAGNITUDE $|\lambda|$ , so $-0.6$ does not beat $1.5$ , but $-2$ would beat $1.5$ .

Model answer — marked the way our engine marks it

On Paper 1 the marks are analytic: each is pinned to a specific line of working, a method mark (M) or an accuracy mark (A), where an accuracy mark depends on the method mark it follows. Follow-through (FT) means an earlier slip need not cost the marks that depend on it, provided the later step is done correctly on your own figures. The engine also ignores subsequent working (ISW) once a correct answer appears, and accepts any correct equivalent, including any non-zero scalar multiple of an eigenvector. That protection exists only if the method is on the page, so study how each mark below is earned by a specific line.

Where this leads

Eigen-thinking is the bridge from static matrices to systems that move. The diagonalization you built here is exactly what makes Markov chains solvable: raise the transition matrix to a power with $A^n=PD^nP^{-1}$ and the steady state falls out of the $\lambda=1$ eigenvector. The same eigenvalues reappear when you solve coupled differential equations $\dot{\mathbf{x}}=A\mathbf{x}$ , where their signs decide growth, decay or oscillation, and their eigenvectors trace the natural axes of the flow. Even phase-portrait sketching and stability analysis are, at heart, reading off the dominant eigenvalue. Master the 2x2 machinery, characteristic equation, eigenvectors, $A=PDP^{-1}$ and its powers, and the dynamic-systems topics that follow become variations on a calculation you already control.

Worked examples

See the formulas applied — reveal one step at a time, like the exam.

EXAMPLE 1

Let $A=\begin{pmatrix}4&-2\\1&1\end{pmatrix}$ . \n (a) Find the eigenvalues of $A$ . \n (b) For each eigenvalue, find a corresponding eigenvector. [6]

1
(a) Eigenvalues. Form $A-\lambda I$ and set its determinant to zero.\n $A-\lambda I=\begin{pmatrix}4-\lambda&-2\\1&1-\lambda\end{pmatrix}$ , so $\det(A-\lambda I)=(4-\lambda)(1-\lambda)-(-2)(1)=0$ . [M1]\nExpanding: $4-4\lambda-\lambda+\lambda^2+2=0\Rightarrow \lambda^2-5\lambda+6=0$ . [A1]\nFactorising: $(\lambda-2)(\lambda-3)=0$ , so $\lambda_1=2$ and $\lambda_2=3$ . [A1]\n(Check: $\lambda_1+\lambda_2=5=\operatorname{tr}(A)$ and $\lambda_1\lambda_2=6=\det(A)$ .)\n\n**(b) Eigenvectors.**\nFor $\lambda_1=2$ : solve $(A-2I)\mathbf{v}=\mathbf{0}$ , i.e. $\begin{pmatrix}2&-2\\1&-1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$ . [M1]\nBoth rows give $x=y$ , so a representative eigenvector is $\mathbf{v}_1=\begin{pmatrix}1\\1\end{pmatrix}$ . [A1]\n\nFor $\lambda_2=3$ : solve $(A-3I)\mathbf{v}=\mathbf{0}$ , i.e. $\begin{pmatrix}1&-2\\1&-2\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$ .\nBoth rows give $x=2y$ , so a representative eigenvector is $\mathbf{v}_2=\begin{pmatrix}2\\1\end{pmatrix}$ . [A1]\n\nAny non-zero scalar multiples, such as $\begin{pmatrix}-1\\-1\end{pmatrix}$ or $\begin{pmatrix}4\\2\end{pmatrix}$ , are equally acceptable.

✓ FULLY WORKED

EXAMPLE 2

For the matrix $A=\begin{pmatrix}4&-2\\1&1\end{pmatrix}$ from the previous example, whose eigenvalues are $2$ and $3$ with eigenvectors $\begin{pmatrix}1\\1\end{pmatrix}$ and $\begin{pmatrix}2\\1\end{pmatrix}$ , write down matrices $P$ and $D$ such that $A=PDP^{-1}$ , and verify the inverse used. [4]

1
Take the eigenvectors as the columns of $P$ and the matching eigenvalues on the diagonal of $D$ , keeping the same order (column 1 with $\lambda=2$ , column 2 with $\lambda=3$ ):\n $P=\begin{pmatrix}1&2\\1&1\end{pmatrix}, \qquad D=\begin{pmatrix}2&0\\0&3\end{pmatrix}$ . [A1 for $P$ ] [A1 for $D$ in matching order]\n\nInverse of $P$ . $\det(P)=(1)(1)-(2)(1)=-1$ , so\n $P^{-1}=\dfrac{1}{-1}\begin{pmatrix}1&-2\\-1&1\end{pmatrix}=\begin{pmatrix}-1&2\\1&-1\end{pmatrix}$ . [M1]\n\nVerify. $PD=\begin{pmatrix}1&2\\1&1\end{pmatrix}\begin{pmatrix}2&0\\0&3\end{pmatrix}=\begin{pmatrix}2&6\\2&3\end{pmatrix}$ , and $PDP^{-1}=\begin{pmatrix}2&6\\2&3\end{pmatrix}\begin{pmatrix}-1&2\\1&-1\end{pmatrix}=\begin{pmatrix}4&-2\\1&1\end{pmatrix}=A$ . [A1]\n\nThe reconstruction returns $A$ exactly, confirming the factorisation.

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EXAMPLE 3

A system evolves by $\mathbf{x}_{n}=M^{n}\mathbf{x}_0$ , where $M=\begin{pmatrix}3&0\\1&2\end{pmatrix}$ . \n (a) Find the eigenvalues and corresponding eigenvectors of $M$ . \n (b) Hence find a formula for $M^n$ in the form $PD^nP^{-1}$ , and use it to evaluate $M^{3}$ . [7]

1
(a) Eigenpairs. $\det(M-\lambda I)=(3-\lambda)(2-\lambda)-0=0$ (the matrix is lower-triangular). [M1]\nSo $(3-\lambda)(2-\lambda)=0$ , giving $\lambda_1=3$ , $\lambda_2=2$ . [A1]\nFor $\lambda_1=3$ : $(M-3I)\mathbf{v}=\begin{pmatrix}0&0\\1&-1\end{pmatrix}\mathbf{v}=\mathbf{0}\Rightarrow x=y$ , so $\mathbf{v}_1=\begin{pmatrix}1\\1\end{pmatrix}$ . [A1]\nFor $\lambda_2=2$ : $(M-2I)\mathbf{v}=\begin{pmatrix}1&0\\1&0\end{pmatrix}\mathbf{v}=\mathbf{0}\Rightarrow x=0$ , so $\mathbf{v}_2=\begin{pmatrix}0\\1\end{pmatrix}$ . [A1]\n\n**(b) Power formula.** With $P=\begin{pmatrix}1&0\\1&1\end{pmatrix}$ and $D=\begin{pmatrix}3&0\\0&2\end{pmatrix}$ , we have $\det(P)=1$ and $P^{-1}=\begin{pmatrix}1&0\\-1&1\end{pmatrix}$ . [M1]\n $M^n=PD^nP^{-1}=\begin{pmatrix}1&0\\1&1\end{pmatrix}\begin{pmatrix}3^n&0\\0&2^n\end{pmatrix}\begin{pmatrix}1&0\\-1&1\end{pmatrix}=\begin{pmatrix}3^n&0\\3^n-2^n&2^n\end{pmatrix}$ . [A1]\n\nEvaluate $M^3$ . Put $n=3$ : $3^3=27$ , $2^3=8$ , so $M^3=\begin{pmatrix}27&0\\27-8&8\end{pmatrix}=\begin{pmatrix}27&0\\19&8\end{pmatrix}$ . [A1]\n(You can confirm by multiplying $M\cdot M\cdot M$ directly.)

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EXAMPLE 4

A population of juveniles and adults is modelled by $\mathbf{p}_{n+1}=L\mathbf{p}_n$ with $L=\begin{pmatrix}0&1.5\\0.6&0.8\end{pmatrix}$ . Using your GDC where appropriate: \n (a) find the eigenvalues of $L$ ; \n (b) state the long-term annual percentage growth rate of the population; \n (c) find the long-term ratio of juveniles to adults. [6]

1
(a) Eigenvalues. The characteristic equation is $\det(L-\lambda I)=(0-\lambda)(0.8-\lambda)-(1.5)(0.6)=0$ . [M1]\nSo $\lambda^2-0.8\lambda-0.9=0$ , and solving (GDC or formula) gives $\lambda_1=1.5$ , $\lambda_2=-0.6$ . [A1]\n\n**(b) Growth rate.** The dominant eigenvalue is the one of largest magnitude, $\lambda_1=1.5$ . [A1]\nIn the long run the population is multiplied by $1.5$ each year, a growth of $1.5-1=0.5$ , i.e. $50\%$ per year. [A1]\n\n**(c) Long-term ratio.** The limiting ratio is the eigenvector for the dominant eigenvalue. Solve $(L-1.5I)\mathbf{v}=\mathbf{0}$ : $\begin{pmatrix}-1.5&1.5\\0.6&-0.7\end{pmatrix}\mathbf{v}=\mathbf{0}$ . The top row gives $-1.5x+1.5y=0\Rightarrow x=y$ . [M1]\nSo the long-term ratio of juveniles to adults is $1:1$ . [A1]\n(A negative eigenvalue like $-0.6$ contributes a fading, alternating term that dies away, leaving the dominant eigenvector to set the stable structure.)

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EXAMPLE 5

Find the eigenvalues and corresponding eigenvectors of $A=\begin{pmatrix}2&1\\1&2\end{pmatrix}$ . [5]

1
Model answer — full working.\n\nCharacteristic equation. Form $A-\lambda I=\begin{pmatrix}2-\lambda&1\\1&2-\lambda\end{pmatrix}$ and set its determinant to zero:\n $\det(A-\lambda I)=(2-\lambda)^2-(1)(1)=0.$ \nExpanding, $(2-\lambda)^2-1=0\Rightarrow \lambda^2-4\lambda+3=0\Rightarrow(\lambda-1)(\lambda-3)=0.$ \nSo the eigenvalues are $\lambda_1=1$ and $\lambda_2=3$ .\n(Check: sum $=4=\operatorname{tr}(A)$ , product $=3=\det(A)$ .)\n\nEigenvector for $\lambda=3$ . Solve $(A-3I)\mathbf{v}=\mathbf{0}$ : $\begin{pmatrix}-1&1\\1&-1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$ . Both rows give $x=y$ , so $\mathbf{v}=\begin{pmatrix}1\\1\end{pmatrix}$ .\n\nEigenvector for $\lambda=1$ . Solve $(A-I)\mathbf{v}=\mathbf{0}$ : $\begin{pmatrix}1&1\\1&1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$ . Both rows give $x=-y$ , so $\mathbf{v}=\begin{pmatrix}1\\-1\end{pmatrix}$ .\n\nAnswer: $\lambda=1$ with $\begin{pmatrix}1\\-1\end{pmatrix}$ , and $\lambda=3$ with $\begin{pmatrix}1\\1\end{pmatrix}$ .\n\n---\nHow our marking engine awards the 5 marks:\n\n- M1 — set up the characteristic equation. Awarded for forming $\det(A-\lambda I)=0$ , i.e. $(2-\lambda)^2-1=0$ , subtracting $\lambda$ from the diagonal before taking the determinant. It is the method that is rewarded, so it survives an arithmetic slip further down.\n- A1 — characteristic equation. Awarded for the correct quadratic $\lambda^2-4\lambda+3=0$ (or the equivalent factored form). This accuracy mark stands on the M1 above.\n- A1 — eigenvalues. Awarded for $\lambda=1$ and $\lambda=3$ . FT applies: a candidate who reaches a slightly different quadratic but solves it correctly earns this on their own equation.\n- M1 — solve for the eigenvectors. A method mark for substituting an eigenvalue into $(A-\lambda I)\mathbf{v}=\mathbf{0}$ and reducing to a relation between the components. The engine checks that $(A-\lambda I)\mathbf{v}=\mathbf{0}$ is used, not $A\mathbf{v}=\mathbf{0}$ .\n- A1 — both eigenvectors. Awarded for a correct non-zero eigenvector for each eigenvalue, paired correctly. Any non-zero scalar multiple is accepted, so $\begin{pmatrix}2\\2\end{pmatrix}$ for $\lambda=3$ or $\begin{pmatrix}-1\\1\end{pmatrix}$ for $\lambda=1$ scores. This A-mark depends on the M1 and is FT on the candidate's own eigenvalues.\n\n**'Accept equivalent forms and any scalar multiple.'** The engine accepts the characteristic equation as $(2-\lambda)^2-1=0$ , $\lambda^2-4\lambda+3=0$ or $(\lambda-1)(\lambda-3)=0$ , and accepts each eigenvector up to a non-zero scalar. Once correct eigenpairs appear, ISW means later restatements do not lose marks.\n\nBottom line: of the 5 marks, two are method marks that survive an arithmetic slip, and the accuracy marks are shielded by follow-through, but only if the characteristic equation and the $(A-\lambda I)\mathbf{v}=\mathbf{0}$ step are actually written down. A student who jumps to ' $\lambda=1,3$ ' with no equation, or who leaves the eigenvectors out, risks losing 2-3 marks that the method would have protected.

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How it all connects

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MAIN IDEAEigenvalues and eigenvectors

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Glossary

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Revision flashcards

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QUESTION · 1 / 11

Eigenvector

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ANSWER

A non-zero vector $\mathbf{v}$ whose direction is unchanged by the matrix $A$ : $A\mathbf{v}=\lambda\mathbf{v}$ for some scalar $\lambda$ . The matrix only scales it (and may flip it), never rotates it off its line.

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Key takeaways

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✓
$A$ is an $n\times n$ square matrix; in this lesson $2\times 2$ .
✓
$\mathbf{v}$ is a non-zero column vector, the eigenvector, giving an invariant direction.
✓
$\lambda$ is a scalar, the eigenvalue, giving the stretch/flip factor along that direction.
✓
The zero vector is excluded on purpose: $A\mathbf{0}=\lambda\mathbf{0}$ holds for every $\lambda$ , so it would tell us nothing.

Practice — then mark it

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Real past paper[0]

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Before you move on: do Get a Paper 2 eigenvalue question marked: form the characteristic equation, solve for the eigenvalues and give the eigenvectors with full working on paper, snap a photo, and get examiner-style feedback on exactly where you win and lose marks.

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Eigenvalues and eigenvectors

BY THE END, YOU CAN…

1State and use the defining relation $A\mathbf{v}=\lambda\mathbf{v}$ and explain what makes a vector an eigenvector and a scalar an eigenvalue.
2Form the characteristic equation $\det(A-\lambda I)=0$ for a 2x2 matrix and solve it to find the eigenvalues.
3Solve $(A-\lambda I)\mathbf{v}=\mathbf{0}$ to find a corresponding eigenvector for each eigenvalue, giving a representative of the family up to scale.
4Diagonalize a 2x2 matrix as $A=PDP^{-1}$ and use $A^n=PD^nP^{-1}$ to compute matrix powers.
5Interpret the dominant eigenvalue and its eigenvector as the driver of long-term behaviour, linking to Markov chains and coupled differential equations.

In simple terms

A friendly intro before the formal notes — no formulas yet.

The Directions a Transformation Leaves Alone

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1
Write the transformation as a square matrix $A$ and the target relation as $A\mathbf{v}=\lambda\mathbf{v}$ : a matrix times a special vector equals a number times the same vector.
2
Rearrange to $(A-\lambda I)\mathbf{v}=\mathbf{0}$ and demand a non-zero $\mathbf{v}$ , which forces $\det(A-\lambda I)=0$ . Solve this characteristic equation for the eigenvalues $\lambda$ .
3
For each eigenvalue, substitute back into $(A-\lambda I)\mathbf{v}=\mathbf{0}$ and solve the resulting equations for the eigenvector, choosing the simplest representative of its scale family.
4
Assemble $P$ from the eigenvectors and $D$ from the eigenvalues to get $A=PDP^{-1}$ , then read off powers and long-term behaviour from $A^n=PD^nP^{-1}$ .

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Key formulas

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$A\mathbf{v} = \lambda\mathbf{v}, \qquad \mathbf{v}\neq\mathbf{0}$

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

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

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

Full topic notes

Formal explanation with the rigour you need for the exam.

The defining relation

$A\mathbf{v} = \lambda\mathbf{v}, \qquad \mathbf{v}\neq\mathbf{0}$

$A$ is an $n\times n$ square matrix; in this lesson $2\times 2$ .
$\mathbf{v}$ is a non-zero column vector, the eigenvector, giving an invariant direction.
$\lambda$ is a scalar, the eigenvalue, giving the stretch/flip factor along that direction.
The zero vector is excluded on purpose: $A\mathbf{0}=\lambda\mathbf{0}$ holds for every $\lambda$ , so it would tell us nothing.

The characteristic equation

\det(A-\lambda I)=0

EXAM TIP

Finding the eigenvectors

Set up $(A-\lambda I)\mathbf{v}=\mathbf{0}$ with the specific eigenvalue substituted in.
The two rows give the same relation; take either one, e.g. $px+qy=0$ .
Choose a simple non-zero solution, such as setting $y=1$ (or clearing fractions) to get whole-number components.
Any non-zero scalar multiple of your answer is equally correct: the eigenvector is a direction, not a fixed arrow.

Diagonalization: $A=PDP^{-1}$

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

Powers of a matrix: $A^n=PD^nP^{-1}$

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

Long-term behaviour: the dominant eigenvalue

Dominant eigenvalue = the one with the largest $|\lambda|$ . It dictates the long-term growth rate.
$|\lambda_{\text{dom}}|>1$ : the system grows without bound; $|\lambda_{\text{dom}}|<1$ : it decays to zero; $|\lambda_{\text{dom}}|=1$ : it settles to a steady state.
The eigenvector of the dominant eigenvalue gives the limiting ratio of the components (e.g. the stable age structure of a population).
A percentage growth rate follows from the dominant eigenvalue: a factor of $\lambda$ per step means a $(\lambda-1)\times 100\%$ change per step.

Common mistakes examiners penalise

Writing the characteristic equation as $\det(A)-\lambda I=0$ or $\det(A)-\lambda=0$ — it is $\det(A-\lambda I)=0$ : subtract $\lambda$ from the diagonal FIRST, then take the determinant.
Solving $A\mathbf{v}=\mathbf{0}$ instead of $(A-\lambda I)\mathbf{v}=\mathbf{0}$ — the eigenvector satisfies $(A-\lambda I)\mathbf{v}=\mathbf{0}$ for the specific eigenvalue, not $A\mathbf{v}=\mathbf{0}$ .
Giving $\mathbf{v}=\mathbf{0}$ as an eigenvector — the zero vector is excluded by definition; you must supply a non-zero representative.
Marking a scalar multiple 'wrong' — eigenvectors are non-unique up to scale, so $\begin{pmatrix}1\\1\end{pmatrix}$ and $\begin{pmatrix}3\\3\end{pmatrix}$ describe the same eigenvector; do not discard a correct multiple.
Mismatching the order of $P$ and $D$ — the $k$ th column of $P$ must belong to the $k$ th diagonal entry of $D$ ; a mismatched pairing breaks $A=PDP^{-1}$ .
Writing $A^n=P^nD^nP^{-1}$ or $PDP^{-n}$ — the correct power formula is $A^n=PD^nP^{-1}$ ; only the diagonal matrix is raised to the power.
Reaching for the GDC on a 2x2 by-hand question — Paper 1 expects the characteristic equation shown and solved; missing that method line forfeits the M and A marks even if the eigenvalues are right.
Picking the wrong dominant eigenvalue — dominance is by MAGNITUDE $|\lambda|$ , so $-0.6$ does not beat $1.5$ , but $-2$ would beat $1.5$ .

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Where this leads

Worked examples

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EXAMPLE 1

Let $A=\begin{pmatrix}4&-2\\1&1\end{pmatrix}$ . \n (a) Find the eigenvalues of $A$ . \n (b) For each eigenvalue, find a corresponding eigenvector. [6]

1
(a) Eigenvalues. Form $A-\lambda I$ and set its determinant to zero.\n $A-\lambda I=\begin{pmatrix}4-\lambda&-2\\1&1-\lambda\end{pmatrix}$ , so $\det(A-\lambda I)=(4-\lambda)(1-\lambda)-(-2)(1)=0$ . [M1]\nExpanding: $4-4\lambda-\lambda+\lambda^2+2=0\Rightarrow \lambda^2-5\lambda+6=0$ . [A1]\nFactorising: $(\lambda-2)(\lambda-3)=0$ , so $\lambda_1=2$ and $\lambda_2=3$ . [A1]\n(Check: $\lambda_1+\lambda_2=5=\operatorname{tr}(A)$ and $\lambda_1\lambda_2=6=\det(A)$ .)\n\n**(b) Eigenvectors.**\nFor $\lambda_1=2$ : solve $(A-2I)\mathbf{v}=\mathbf{0}$ , i.e. $\begin{pmatrix}2&-2\\1&-1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$ . [M1]\nBoth rows give $x=y$ , so a representative eigenvector is $\mathbf{v}_1=\begin{pmatrix}1\\1\end{pmatrix}$ . [A1]\n\nFor $\lambda_2=3$ : solve $(A-3I)\mathbf{v}=\mathbf{0}$ , i.e. $\begin{pmatrix}1&-2\\1&-2\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$ .\nBoth rows give $x=2y$ , so a representative eigenvector is $\mathbf{v}_2=\begin{pmatrix}2\\1\end{pmatrix}$ . [A1]\n\nAny non-zero scalar multiples, such as $\begin{pmatrix}-1\\-1\end{pmatrix}$ or $\begin{pmatrix}4\\2\end{pmatrix}$ , are equally acceptable.

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EXAMPLE 2

1
Take the eigenvectors as the columns of $P$ and the matching eigenvalues on the diagonal of $D$ , keeping the same order (column 1 with $\lambda=2$ , column 2 with $\lambda=3$ ):\n $P=\begin{pmatrix}1&2\\1&1\end{pmatrix}, \qquad D=\begin{pmatrix}2&0\\0&3\end{pmatrix}$ . [A1 for $P$ ] [A1 for $D$ in matching order]\n\nInverse of $P$ . $\det(P)=(1)(1)-(2)(1)=-1$ , so\n $P^{-1}=\dfrac{1}{-1}\begin{pmatrix}1&-2\\-1&1\end{pmatrix}=\begin{pmatrix}-1&2\\1&-1\end{pmatrix}$ . [M1]\n\nVerify. $PD=\begin{pmatrix}1&2\\1&1\end{pmatrix}\begin{pmatrix}2&0\\0&3\end{pmatrix}=\begin{pmatrix}2&6\\2&3\end{pmatrix}$ , and $PDP^{-1}=\begin{pmatrix}2&6\\2&3\end{pmatrix}\begin{pmatrix}-1&2\\1&-1\end{pmatrix}=\begin{pmatrix}4&-2\\1&1\end{pmatrix}=A$ . [A1]\n\nThe reconstruction returns $A$ exactly, confirming the factorisation.

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EXAMPLE 3

1
(a) Eigenpairs. $\det(M-\lambda I)=(3-\lambda)(2-\lambda)-0=0$ (the matrix is lower-triangular). [M1]\nSo $(3-\lambda)(2-\lambda)=0$ , giving $\lambda_1=3$ , $\lambda_2=2$ . [A1]\nFor $\lambda_1=3$ : $(M-3I)\mathbf{v}=\begin{pmatrix}0&0\\1&-1\end{pmatrix}\mathbf{v}=\mathbf{0}\Rightarrow x=y$ , so $\mathbf{v}_1=\begin{pmatrix}1\\1\end{pmatrix}$ . [A1]\nFor $\lambda_2=2$ : $(M-2I)\mathbf{v}=\begin{pmatrix}1&0\\1&0\end{pmatrix}\mathbf{v}=\mathbf{0}\Rightarrow x=0$ , so $\mathbf{v}_2=\begin{pmatrix}0\\1\end{pmatrix}$ . [A1]\n\n**(b) Power formula.** With $P=\begin{pmatrix}1&0\\1&1\end{pmatrix}$ and $D=\begin{pmatrix}3&0\\0&2\end{pmatrix}$ , we have $\det(P)=1$ and $P^{-1}=\begin{pmatrix}1&0\\-1&1\end{pmatrix}$ . [M1]\n $M^n=PD^nP^{-1}=\begin{pmatrix}1&0\\1&1\end{pmatrix}\begin{pmatrix}3^n&0\\0&2^n\end{pmatrix}\begin{pmatrix}1&0\\-1&1\end{pmatrix}=\begin{pmatrix}3^n&0\\3^n-2^n&2^n\end{pmatrix}$ . [A1]\n\nEvaluate $M^3$ . Put $n=3$ : $3^3=27$ , $2^3=8$ , so $M^3=\begin{pmatrix}27&0\\27-8&8\end{pmatrix}=\begin{pmatrix}27&0\\19&8\end{pmatrix}$ . [A1]\n(You can confirm by multiplying $M\cdot M\cdot M$ directly.)

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EXAMPLE 4

1
(a) Eigenvalues. The characteristic equation is $\det(L-\lambda I)=(0-\lambda)(0.8-\lambda)-(1.5)(0.6)=0$ . [M1]\nSo $\lambda^2-0.8\lambda-0.9=0$ , and solving (GDC or formula) gives $\lambda_1=1.5$ , $\lambda_2=-0.6$ . [A1]\n\n**(b) Growth rate.** The dominant eigenvalue is the one of largest magnitude, $\lambda_1=1.5$ . [A1]\nIn the long run the population is multiplied by $1.5$ each year, a growth of $1.5-1=0.5$ , i.e. $50\%$ per year. [A1]\n\n**(c) Long-term ratio.** The limiting ratio is the eigenvector for the dominant eigenvalue. Solve $(L-1.5I)\mathbf{v}=\mathbf{0}$ : $\begin{pmatrix}-1.5&1.5\\0.6&-0.7\end{pmatrix}\mathbf{v}=\mathbf{0}$ . The top row gives $-1.5x+1.5y=0\Rightarrow x=y$ . [M1]\nSo the long-term ratio of juveniles to adults is $1:1$ . [A1]\n(A negative eigenvalue like $-0.6$ contributes a fading, alternating term that dies away, leaving the dominant eigenvector to set the stable structure.)

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EXAMPLE 5

Find the eigenvalues and corresponding eigenvectors of $A=\begin{pmatrix}2&1\\1&2\end{pmatrix}$ . [5]

1
Model answer — full working.\n\nCharacteristic equation. Form $A-\lambda I=\begin{pmatrix}2-\lambda&1\\1&2-\lambda\end{pmatrix}$ and set its determinant to zero:\n $\det(A-\lambda I)=(2-\lambda)^2-(1)(1)=0.$ \nExpanding, $(2-\lambda)^2-1=0\Rightarrow \lambda^2-4\lambda+3=0\Rightarrow(\lambda-1)(\lambda-3)=0.$ \nSo the eigenvalues are $\lambda_1=1$ and $\lambda_2=3$ .\n(Check: sum $=4=\operatorname{tr}(A)$ , product $=3=\det(A)$ .)\n\nEigenvector for $\lambda=3$ . Solve $(A-3I)\mathbf{v}=\mathbf{0}$ : $\begin{pmatrix}-1&1\\1&-1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$ . Both rows give $x=y$ , so $\mathbf{v}=\begin{pmatrix}1\\1\end{pmatrix}$ .\n\nEigenvector for $\lambda=1$ . Solve $(A-I)\mathbf{v}=\mathbf{0}$ : $\begin{pmatrix}1&1\\1&1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}$ . Both rows give $x=-y$ , so $\mathbf{v}=\begin{pmatrix}1\\-1\end{pmatrix}$ .\n\nAnswer: $\lambda=1$ with $\begin{pmatrix}1\\-1\end{pmatrix}$ , and $\lambda=3$ with $\begin{pmatrix}1\\1\end{pmatrix}$ .\n\n---\nHow our marking engine awards the 5 marks:\n\n- M1 — set up the characteristic equation. Awarded for forming $\det(A-\lambda I)=0$ , i.e. $(2-\lambda)^2-1=0$ , subtracting $\lambda$ from the diagonal before taking the determinant. It is the method that is rewarded, so it survives an arithmetic slip further down.\n- A1 — characteristic equation. Awarded for the correct quadratic $\lambda^2-4\lambda+3=0$ (or the equivalent factored form). This accuracy mark stands on the M1 above.\n- A1 — eigenvalues. Awarded for $\lambda=1$ and $\lambda=3$ . FT applies: a candidate who reaches a slightly different quadratic but solves it correctly earns this on their own equation.\n- M1 — solve for the eigenvectors. A method mark for substituting an eigenvalue into $(A-\lambda I)\mathbf{v}=\mathbf{0}$ and reducing to a relation between the components. The engine checks that $(A-\lambda I)\mathbf{v}=\mathbf{0}$ is used, not $A\mathbf{v}=\mathbf{0}$ .\n- A1 — both eigenvectors. Awarded for a correct non-zero eigenvector for each eigenvalue, paired correctly. Any non-zero scalar multiple is accepted, so $\begin{pmatrix}2\\2\end{pmatrix}$ for $\lambda=3$ or $\begin{pmatrix}-1\\1\end{pmatrix}$ for $\lambda=1$ scores. This A-mark depends on the M1 and is FT on the candidate's own eigenvalues.\n\n**'Accept equivalent forms and any scalar multiple.'** The engine accepts the characteristic equation as $(2-\lambda)^2-1=0$ , $\lambda^2-4\lambda+3=0$ or $(\lambda-1)(\lambda-3)=0$ , and accepts each eigenvector up to a non-zero scalar. Once correct eigenpairs appear, ISW means later restatements do not lose marks.\n\nBottom line: of the 5 marks, two are method marks that survive an arithmetic slip, and the accuracy marks are shielded by follow-through, but only if the characteristic equation and the $(A-\lambda I)\mathbf{v}=\mathbf{0}$ step are actually written down. A student who jumps to ' $\lambda=1,3$ ' with no equation, or who leaves the eigenvectors out, risks losing 2-3 marks that the method would have protected.

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How it all connects

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MAIN IDEAEigenvalues and eigenvectors

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Glossary

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QUESTION · 1 / 11

Eigenvector

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Key takeaways

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✓
$A$ is an $n\times n$ square matrix; in this lesson $2\times 2$ .
✓
$\mathbf{v}$ is a non-zero column vector, the eigenvector, giving an invariant direction.
✓
$\lambda$ is a scalar, the eigenvalue, giving the stretch/flip factor along that direction.
✓
The zero vector is excluded on purpose: $A\mathbf{0}=\lambda\mathbf{0}$ holds for every $\lambda$ , so it would tell us nothing.

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Revise Eigenvalues and eigenvectors for IB Mathematics: Applications and Interpretation (1.7)

In simple terms

Explore the concept

Key formulas

Full topic notes

The defining relation

The characteristic equation

Finding the eigenvectors

Diagonalization: $A=PDP^{-1}$

Powers of a matrix: $A^n=PD^nP^{-1}$

Long-term behaviour: the dominant eigenvalue

Common mistakes examiners penalise

Model answer — marked the way our engine marks it

Where this leads

Worked examples

How it all connects

Glossary

Quick check

Revision flashcards

Key takeaways

Practice — then mark it

Get a Paper 2 eigenvalue question marked: form the characteristic equation, solve for the eigenvalues and give the eigenvectors with full working

Extra simulations & links

Frequently asked

Checkpoint

Reading it isn’t knowing it — prove it.

In simple terms

Explore the concept

Key formulas

Full topic notes

The defining relation

The characteristic equation

Finding the eigenvectors

Diagonalization: $A=PDP^{-1}$

Powers of a matrix: $A^n=PD^nP^{-1}$

Long-term behaviour: the dominant eigenvalue

Common mistakes examiners penalise

Model answer — marked the way our engine marks it

Where this leads

Worked examples

How it all connects

Glossary

Quick check

Revision flashcards

Key takeaways

Practice — then mark it

Get a Paper 2 eigenvalue question marked: form the characteristic equation, solve for the eigenvalues and give the eigenvectors with full working

Extra simulations & links

Frequently asked

Checkpoint

Reading it isn’t knowing it — prove it.

Discuss Eigenvalues and eigenvectors