Topic · Linear Algebra

Eigenvalues & Eigenvectors

Multiplying a vector by a matrix sends it somewhere new — typically pointing in a different direction and stretched to a different length. Picture a rotation: every arrow gets spun the same amount, no matter where it started. For most matrices, most input vectors come out aimed somewhere they weren't before. But a rare few hold their ground — they emerge pointing along the very same line they went in on, only scaled by some factor. Those special directions are the matrix's eigenvectors, the factor each one gets scaled by is its eigenvalue, and together they expose the hidden skeleton of any linear transformation.

12 min read Prereqs: Determinants Updated 2026·05·17

What you'll leave with

  • The defining equation $A\vec{v} = \lambda \vec{v}$ — and why that single line carries the whole subject.
  • A geometric picture: eigenvectors are the directions a transformation doesn't tilt.
  • The two-step recipe for computing them: characteristic equation, then null space.
  • Why a basis of eigenvectors turns matrix algebra into ordinary arithmetic — and where this shows up in the real world.

1. The defining equation

Eigenvector and eigenvalue

Given a square matrix $A$, a non-zero vector $\vec{v}$ is an eigenvector of $A$ if there is a scalar $\lambda$ such that

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

The scalar $\lambda$ is the corresponding eigenvalue. The pair $(\lambda, \vec{v})$ is called an eigenpair.

Read the equation slowly. The left side, $A\vec{v}$, is "apply the transformation $A$ to the vector $\vec{v}$." The right side, $\lambda \vec{v}$, is "stretch $\vec{v}$ by the number $\lambda$." The eigenvalue equation says: the complicated thing on the left and the simple thing on the right are the same vector. The matrix's action on $\vec{v}$, with all its potential mixing and shearing, collapses to a plain scalar multiplication.

That collapse is the whole point. Generic vectors get pushed around in complicated ways by $A$ — bent off their original direction, mixed with other components. Eigenvectors are the survivors. The transformation can stretch them, squash them, even flip them around (negative $\lambda$), but it never tilts them off the line they're on.

The eigenvalue $\lambda$ tells you how much the stretching is. The eigenvector $\vec{v}$ tells you where — which direction in space is invariant.

Etymology

"Eigen" is German for "own" or "characteristic" — these are the matrix's own directions, the ones intrinsic to it rather than to any choice of coordinate system. You'll sometimes see them called characteristic vectors and characteristic values.

2. Geometric intuition

Picture a 2-D plane with a few vectors scattered around it. Now apply some linear transformation $A$ — say, the matrix that horizontally stretches space by a factor of 2 and vertically squishes by a factor of $\tfrac{1}{2}$. Each input vector lands somewhere new.

For most vectors, the output points in a different direction than the input. The transformation rotated them. But two special directions are exceptions: anything purely horizontal stays horizontal (now twice as long), and anything purely vertical stays vertical (now half as long). Those two directions are the eigenvectors. The two stretch factors — $2$ and $\tfrac{1}{2}$ — are the eigenvalues.

The picture below shows three vectors under such a transformation. The eigenvector (orange, along the $x$-axis) is merely stretched — its image is on the same line. The other two vectors are rotated off their original directions.

Before: input vectors w u v After: A applied, A = [[2,0],[0,½]] Aw Au Av = 2v

Notice what happened. Vectors $\vec{u}$ and $\vec{w}$ both pointed somewhere diagonal on the left; after the transformation, they point somewhere different diagonal. They've been rotated. But the orange vector $\vec{v}$, which started along the $x$-axis, is still along the $x$-axis — just twice as long. That's the eigenvector signature: same line, different length.

Mental shortcut

If you can find the lines a matrix leaves invariant, you've found its eigenvectors. The stretching factor along each such line is the corresponding eigenvalue. Everything else about eigenanalysis is built on this picture.

3. Finding them

The defining equation $A\vec{v} = \lambda \vec{v}$ can be rearranged. Subtract $\lambda \vec{v}$ from both sides, and factor out $\vec{v}$ — being careful, because $A$ is a matrix and $\lambda$ is a scalar, so we need the identity $I$ to make the dimensions agree:

$$ A\vec{v} - \lambda \vec{v} = \vec{0} \quad\Longrightarrow\quad (A - \lambda I)\vec{v} = \vec{0}. $$

This is now a homogeneous linear system in $\vec{v}$. The trivial answer $\vec{v} = \vec{0}$ always works — but we excluded that from the definition. So we want a non-trivial solution. Such a solution exists exactly when the matrix $A - \lambda I$ is singular: that is, when its determinant is zero.

Characteristic equation

The eigenvalues of $A$ are the roots of

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

This is a polynomial equation in $\lambda$ — the characteristic polynomial of $A$. For an $n \times n$ matrix, it has degree $n$.

The recipe is therefore two stages:

  1. Find the eigenvalues. Compute $\det(A - \lambda I)$ as a polynomial in $\lambda$. Set it equal to zero. Solve.
  2. Find the eigenvectors. For each eigenvalue $\lambda_i$, plug back in and solve the homogeneous system $(A - \lambda_i I)\vec{v} = \vec{0}$. The null space of that matrix is the eigenspace for $\lambda_i$ — every non-zero vector in it is an eigenvector.

A 2×2 walkthrough

Let

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

Step 1. Form $A - \lambda I$ and take its determinant:

$$ A - \lambda I = \begin{pmatrix} 4 - \lambda & 1 \\ 2 & 3 - \lambda \end{pmatrix}, \quad \det(A - \lambda I) = (4 - \lambda)(3 - \lambda) - 2. $$

Expand:

$$ (4 - \lambda)(3 - \lambda) - 2 = \lambda^2 - 7\lambda + 12 - 2 = \lambda^2 - 7\lambda + 10. $$

Step 2. Solve $\lambda^2 - 7\lambda + 10 = 0$. This factors as $(\lambda - 5)(\lambda - 2) = 0$, giving

$$ \lambda_1 = 5, \qquad \lambda_2 = 2. $$

Step 3. For each eigenvalue, find the null space of $A - \lambda I$.

For $\lambda_1 = 5$:

$$ A - 5I = \begin{pmatrix} -1 & 1 \\ 2 & -2 \end{pmatrix}. $$

The rows are proportional, so the system reduces to $-v_1 + v_2 = 0$, i.e. $v_2 = v_1$. Any non-zero scalar multiple of $(1, 1)$ works. Pick the cleanest:

$$ \vec{v}_1 = \begin{pmatrix} 1 \\ 1 \end{pmatrix}. $$

For $\lambda_2 = 2$:

$$ A - 2I = \begin{pmatrix} 2 & 1 \\ 2 & 1 \end{pmatrix}. $$

Again rank-one, with $2v_1 + v_2 = 0$, so $v_2 = -2v_1$. Pick:

$$ \vec{v}_2 = \begin{pmatrix} 1 \\ -2 \end{pmatrix}. $$

That's everything: two eigenvalues, two eigenvectors. Verify with a quick check:

$$ A \vec{v}_1 = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}\begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 5 \\ 5 \end{pmatrix} = 5 \vec{v}_1 \quad \checkmark $$
Two useful shortcuts

For a $2 \times 2$ matrix, the characteristic polynomial is always $\lambda^2 - (\mathrm{tr}\,A)\lambda + \det A$. The sum of eigenvalues equals the trace; the product equals the determinant. Both facts generalize to $n \times n$ matrices and are an excellent sanity check on hand calculations.

4. Why they matter

Suppose you have an $n \times n$ matrix $A$ with $n$ linearly independent eigenvectors $\vec{v}_1, \dots, \vec{v}_n$ and corresponding eigenvalues $\lambda_1, \dots, \lambda_n$. Stack the eigenvectors as columns of a matrix $P$, and put the eigenvalues on the diagonal of $D$:

$$ P = \begin{pmatrix} | & | & & | \\ \vec{v}_1 & \vec{v}_2 & \cdots & \vec{v}_n \\ | & | & & | \end{pmatrix}, \qquad D = \begin{pmatrix} \lambda_1 & & & \\ & \lambda_2 & & \\ & & \ddots & \\ & & & \lambda_n \end{pmatrix}. $$

Then a beautiful identity holds:

$$ A = P D P^{-1}. $$

This is diagonalization. It says: in the coordinate system whose axes are the eigenvectors, $A$ is just $D$ — a diagonal matrix that does nothing more than scale each axis by its eigenvalue. The horrible-looking matrix $A$ in the standard basis turns into a simple stretching machine once you change perspective.

The payoff is enormous because diagonal matrices are trivial to compute with. For example, what is $A^{10}$?

$$ A^{10} = (PDP^{-1})^{10} = P D^{10} P^{-1}, $$

and $D^{10}$ is just the diagonal matrix with $\lambda_i^{10}$ on the diagonal — no matrix multiplications required. The same trick simplifies matrix exponentials, solutions to linear ODEs ($\frac{d\vec{x}}{dt} = A\vec{x}$ becomes $n$ uncoupled scalar equations, one per eigenvector), Markov chain long-run behavior, and almost any iterative process driven by a linear operator.

The takeaway: eigenvectors aren't just curiosities of a matrix. They're the right basis — the coordinate system in which $A$ shows its true, simple nature.

The diagonalization slogan

If $A = PDP^{-1}$, then anything you can do to a diagonal matrix, you can do to $A$: just sandwich the operation between $P$ and $P^{-1}$. Eigen-decomposition is a license to pretend $A$ is diagonal.

5. Applications, briefly

Eigenanalysis shows up wherever a linear operator drives a system, which turns out to be most of applied mathematics.

  • Google PageRank. The original algorithm models the web as a giant stochastic matrix $M$ of link-following probabilities. The PageRank of every page on the internet is just the entries of the dominant eigenvector of $M$ (the one with eigenvalue $1$). Find that eigenvector, you've ranked the web.
  • Principal Component Analysis (PCA). Given a cloud of data points, the directions of maximum variance are the eigenvectors of the data's covariance matrix; the eigenvalues give the variance along each. Keep the top few and you've compressed the data while preserving its dominant structure.
  • Vibrational modes. Springs, drum heads, molecules, bridges — any system with linear restoring forces vibrates in a superposition of normal modes, each of which is an eigenvector of the system's stiffness matrix, with its eigenvalue setting the squared frequency.
  • Quantum mechanics. Observable quantities are operators; their eigenvalues are the values you can measure; their eigenvectors are the states that produce a definite outcome. The entire formalism is eigenanalysis dressed in physics.
  • Stability analysis. Linearize any dynamical system near a fixed point. The eigenvalues of the Jacobian tell you whether small disturbances decay (stable) or blow up (unstable) — and how fast.

This is why eigenvalues earn the description "the single most influential idea in applied linear algebra": each item above is a billion-dollar industry, a Nobel-winning theory, or both, and each is at heart the same calculation you did by hand in section 3.

6. Playground: find the eigenvectors

Tune the four entries of a $2 \times 2$ matrix and watch what happens to the unit circle. Each thin grey arrow is an input vector $\vec{v}$ of length 1; each thin orange arrow is its image $A\vec{v}$. Where input and output land on the same line — that's an eigenvector, highlighted in green. The eigenvalue is how much longer (or shorter, or flipped) the green arrow got.

2.0 1.0 0.0 3.0
trace 5.00 det 6.00
λ₁ 3.00
λ₂ 2.00
v₁ (1.00, 1.00)
v₂ (1.00, 0.00)
note two real eigenvalues
input $\vec{v}$ output $A\vec{v}$ eigenvector direction
x y
2.0
1.0
0.0
3.0
Try it

Start with the default — a triangular matrix. Notice the two green eigen-directions; the orange arrows lie exactly on those lines (just longer). Now hit Rotation by 90°: every output arrow is perpendicular to its input, no green arrows appear, and the readout reports complex eigenvalues. That's the algebra and the picture telling you the same thing.

7. Common pitfalls

Pitfall · The zero vector is not an eigenvector

It's true that $A\vec{0} = \vec{0} = \lambda \vec{0}$ for every scalar $\lambda$ — which is exactly why we exclude it. If $\vec{0}$ counted, every number would be an eigenvalue of every matrix and the concept would be useless. The definition explicitly requires $\vec{v} \neq \vec{0}$. The eigenvalue $\lambda = 0$ is fine (it just means $A$ collapses some direction to zero), but the eigenvector itself must be non-zero.

Eigenvalues can be complex even for real matrices

The characteristic polynomial has real coefficients when $A$ is real, but its roots can still be complex. A 90° rotation in the plane has no real eigenvectors at all — every real direction gets rotated — and its eigenvalues are $\pm i$. Don't conclude "no solution" just because the discriminant goes negative; switch to $\mathbb{C}$ and keep going.

Eigenvectors are defined up to scaling

If $\vec{v}$ is an eigenvector with eigenvalue $\lambda$, then so is $c\vec{v}$ for every non-zero scalar $c$. There is no "the" eigenvector for a given eigenvalue — only an eigenspace, a whole line (or plane, if the eigenvalue is repeated). Pick a clean representative for hand work; for numerical work, normalize to unit length.

Algebraic vs. geometric multiplicity

An eigenvalue can be a repeated root of the characteristic polynomial (algebraic multiplicity $k$) but still have only one independent eigenvector (geometric multiplicity $1$). When the two disagree the matrix is defective — it can't be diagonalized, and you need Jordan form instead. Always check that you found as many independent eigenvectors as the eigenvalue's algebraic multiplicity before claiming a basis.

8. Worked examples

Try each one before opening the solution. The mechanics are mechanical — but mistakes love to hide in sign errors and bookkeeping.

Example 1 · Find the eigenvalues of $A = \begin{pmatrix} 3 & 1 \\ 0 & 2 \end{pmatrix}$

Step 1. Form $A - \lambda I$:

$$ A - \lambda I = \begin{pmatrix} 3 - \lambda & 1 \\ 0 & 2 - \lambda \end{pmatrix}. $$

Step 2. Take the determinant. Because the lower-left entry is zero, this is just the product of the diagonal:

$$ \det(A - \lambda I) = (3 - \lambda)(2 - \lambda). $$

Step 3. Set equal to zero:

$$ (3 - \lambda)(2 - \lambda) = 0 \quad\Longrightarrow\quad \lambda_1 = 3, \;\; \lambda_2 = 2. $$

Sanity check. Trace is $3 + 2 = 5 = \lambda_1 + \lambda_2$. Determinant is $6 = \lambda_1 \lambda_2$. ✓

For any triangular matrix, the eigenvalues are exactly the diagonal entries — a fact worth remembering.

Example 2 · Find eigenvectors of $A = \begin{pmatrix} 3 & 1 \\ 0 & 2 \end{pmatrix}$

Eigenvalues from Example 1 are $\lambda_1 = 3$ and $\lambda_2 = 2$.

For $\lambda_1 = 3$:

$$ A - 3I = \begin{pmatrix} 0 & 1 \\ 0 & -1 \end{pmatrix}. $$

This says $v_2 = 0$ and $v_1$ is free. Take

$$ \vec{v}_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}. $$

For $\lambda_2 = 2$:

$$ A - 2I = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}. $$

This says $v_1 + v_2 = 0$, so $v_2 = -v_1$. Take

$$ \vec{v}_2 = \begin{pmatrix} 1 \\ -1 \end{pmatrix}. $$
Example 3 · Verify $A\vec{v} = \lambda \vec{v}$ for the pair from Example 2

Take $\lambda_2 = 2$, $\vec{v}_2 = (1, -1)$. Compute $A \vec{v}_2$:

$$ A \vec{v}_2 = \begin{pmatrix} 3 & 1 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} 1 \\ -1 \end{pmatrix} = \begin{pmatrix} 3 \cdot 1 + 1 \cdot (-1) \\ 0 \cdot 1 + 2 \cdot (-1) \end{pmatrix} = \begin{pmatrix} 2 \\ -2 \end{pmatrix}. $$

And $\lambda_2 \vec{v}_2 = 2 \cdot (1, -1) = (2, -2)$. They match. ✓

This kind of check is cheap and catches a huge fraction of arithmetic mistakes — always run it.

Example 4 · A rotation has no real eigenvalues (unless trivial)

The matrix for a $90°$ counter-clockwise rotation in the plane is

$$ R = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}. $$

The characteristic polynomial:

$$ \det(R - \lambda I) = \det \begin{pmatrix} -\lambda & -1 \\ 1 & -\lambda \end{pmatrix} = \lambda^2 + 1. $$

Setting $\lambda^2 + 1 = 0$ gives $\lambda = \pm i$ — no real solutions.

Why this makes geometric sense. A real eigenvector would be a real direction that the rotation leaves invariant. But every direction in the plane gets rotated by $90°$. So there is no real eigenvector, and the algebra confirms the picture.

(General rotation matrices have real eigenvalues only when the angle is $0$ or $180°$ — the trivial cases where every vector or its negation is fixed.)

Example 5 · Diagonalize $A = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix}$

From section 3, the eigenvalues are $\lambda_1 = 5, \lambda_2 = 2$ with eigenvectors $\vec{v}_1 = (1, 1)$ and $\vec{v}_2 = (1, -2)$.

Step 1. Build $P$ from the eigenvectors and $D$ from the eigenvalues, in matching order:

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

Step 2. Compute $P^{-1}$. For a $2 \times 2$, $P^{-1} = \frac{1}{\det P} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$. Here $\det P = (1)(-2) - (1)(1) = -3$:

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

Step 3. Verify $A = PDP^{-1}$. Compute $DP^{-1}$ first:

$$ D P^{-1} = \begin{pmatrix} 5 & 0 \\ 0 & 2 \end{pmatrix} \cdot \tfrac{1}{3}\begin{pmatrix} 2 & 1 \\ 1 & -1 \end{pmatrix} = \tfrac{1}{3}\begin{pmatrix} 10 & 5 \\ 2 & -2 \end{pmatrix}. $$

Then $P D P^{-1}$:

$$ \tfrac{1}{3}\begin{pmatrix} 1 & 1 \\ 1 & -2 \end{pmatrix}\begin{pmatrix} 10 & 5 \\ 2 & -2 \end{pmatrix} = \tfrac{1}{3}\begin{pmatrix} 12 & 3 \\ 6 & 9 \end{pmatrix} = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix} = A. \;\;\checkmark $$

And now, the payoff: $A^{10} = P D^{10} P^{-1}$ with $D^{10} = \mathrm{diag}(5^{10}, 2^{10}) = \mathrm{diag}(9{,}765{,}625, \,1024)$ — no ten-fold matrix multiplication required.

Example 6 · A 3×3 with a known triangular shortcut

Let

$$ A = \begin{pmatrix} 2 & 1 & 4 \\ 0 & 3 & 5 \\ 0 & 0 & 6 \end{pmatrix}. $$

The matrix is upper-triangular. For any triangular matrix, $\det(A - \lambda I)$ is just the product of the diagonal entries of $A - \lambda I$:

$$ \det(A - \lambda I) = (2 - \lambda)(3 - \lambda)(6 - \lambda). $$

So the eigenvalues are read directly off the diagonal: $\lambda_1 = 2, \lambda_2 = 3, \lambda_3 = 6$.

Sanity check. Trace $= 2 + 3 + 6 = 11 = \lambda_1 + \lambda_2 + \lambda_3$. Determinant $= 2 \cdot 3 \cdot 6 = 36 = \lambda_1 \lambda_2 \lambda_3$. ✓

For a general $3 \times 3$, expanding $\det(A - \lambda I)$ produces a cubic in $\lambda$ — the characteristic polynomial. Factor it (or use the cubic formula in extremis) to find the three eigenvalues, then solve $(A - \lambda_i I)\vec{v} = \vec{0}$ for each.

Example 7 · The eigenspace of a repeated eigenvalue

Consider

$$ A = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}. $$

Its characteristic polynomial is $(2 - \lambda)^2$, so $\lambda = 2$ is a repeated root with algebraic multiplicity $2$.

Now solve $(A - 2I)\vec{v} = \vec{0}$. But $A - 2I$ is the zero matrix, so every vector in $\mathbb{R}^2$ is a solution. The eigenspace for $\lambda = 2$ is all of $\mathbb{R}^2$ — a two-dimensional space. The geometric multiplicity is also $2$, matching the algebraic multiplicity, so $A$ is diagonalizable (trivially: it's already diagonal).

Contrast with the defective case

$$ B = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}. $$

Same characteristic polynomial $(2 - \lambda)^2$, algebraic multiplicity $2$. But $B - 2I = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ forces $v_2 = 0$ with $v_1$ free, so the eigenspace is one-dimensional — geometric multiplicity is only $1$. The two multiplicities disagree; $B$ cannot be diagonalized.

Sources & further reading

The treatment above is synthesized from the standard references. If anything reads ambiguously here, the primary sources are the ground truth — and the deeper links are where to turn when this page has done its job.

Test your understanding

A quiz that builds from easy to hard. Pick an answer to get instant feedback and a worked explanation. Your progress is saved in this browser — come back anytime to continue.

Question 1 of 28
0 correct