Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How can we physically interpret an eigenvalue or an eigenvector in linear algebra?

share|cite|improve this question
Amazing -- the questioner's profile says he or she hasn't been seen since 16 minutes after posting this question, before any of the answers were written -- so many good answers, and the questioner never read them... – joriki Jul 27 '11 at 13:30

To add yet another answer: I'm surprised that normal modes haven't been mentioned yet. Whenever you expand the behaviour of a mechanical system around a point of equilibrium, you get coupled equations of motion of the form


where $\vec{x}$ is a vector containing all coordinates of the system and $A$ is a square matrix. Such a system is solved by diagonalizing $A$, i.e. transforming to a basis of eigenvectors of $A$ in which $A$ becomes diagonal; this decouples the equations of motion, which are transformed into a set of simple harmonic oscillator equations that can be solved individually. The eigenvalues of $A$ correspond to the eigenfrequencies of the system (they are their negative squares), and the eigenvectors are the so-called normal modes -- patterns of displacement that allow the system to oscillate at a single frequency.

share|cite|improve this answer

In classical mechanics, the eigenvectors of the inertia tensor correspond to the principal axes of the object; these are directions about which the angular momentum will be point in the same direction as the direction of rotation, so that the object can rotate without precession in this direction (without any torque).

In quantum mechanics, to say that a state is an eigenfunction of an observable (i.e. Hermitian operator) equates to saying that measurement of that observable on that state is certain to return a definite value. In the special case of a (time-independent) Hamiltonian, the Schrodinger equation implies that the eigenstate in question is a stationary one---it evolves in time by multiplication by an exponential (of frequency proportional to the energy), and the expectation value of any measurement on the state remains constant in time.

share|cite|improve this answer
Nitpick: An object can rotate without precession about any of the three principal axes, but rotation about the intermediate one is not stable. – Rahul Mar 13 '11 at 7:27
@Rahul: Dear Rahul, thanks for the correction. – Akhil Mathew Mar 13 '11 at 14:14

The clearest interpretation I've found of the eigenvalues and eigenvectors of an operator in physics is in the wave equation

$$\frac{\partial^2 u}{\partial t^2} = \Delta u$$

where $\Delta$ denotes the Laplacian. The eigenvectors of $\Delta$ with negative eigenvalues satisfy $\Delta v = -\lambda^2 v$, and from any such eigenvector one can construct a solution

$$u = A \cos (\lambda t + B) v$$

of the wave equation. The solutions one gets in this way are precisely the standing waves of frequency $\frac{\lambda}{2\pi}$, and under nice conditions every solution is a linear combination of standing waves.

In common physical situations, the Hamiltonian one uses in the Schrödinger equation in quantum mechanics is closely related to the Laplacian, which is one reason solutions to the Schrödinger equation are referred to as wave functions.

share|cite|improve this answer
Let me mention that you can interpret a wide class of real symmetric matrices as Laplacians of suitable weighted finite graphs, giving a direct physical interpretation of their eigenvectors and eigenvalues: see… . – Qiaochu Yuan Mar 13 '11 at 18:04

Mathematically an eigenvector of a linear transformation is just a vector whose direction is fixed under the transformation, while its eigenvalue is the amount its magnitude changes by. To understand the physical meaning of it requires first understanding the physical meaning of a linear transformation.

Akhil already gave a very good answer detailing two common examples. Let me here give a slightly more esoteric one.

Consider electromagnetism as a relativistic theory (Maxwell's theory is inherently compatible with special relativity). We can unite the electric field and magnetic field, which are each represented as spatial vectors (3 components), into a space-time quantity, a 4-by-4 matrix known as the Faraday tensor denoted by $F_{ab}$. It is an antisymmetric matrix $F_{ab} = - F_{ba}$, and transforms covariantly under coordinate transformations of Minkowski space. (In other words, geometrically it is a two-form.) Expressed in terms of the components of the electric and magnetic fields, you have

$$ F = \begin{pmatrix} 0 & -E_x & -E_y & - E_z \\ E_x & 0 & -B_z & B_y \\ E_y & B_z & 0 & -B_x \\ E_z & -B_y & B_x & 0 \end{pmatrix} $$

Written as such we cannot consider the eigenvectors of $F_{ab}$. Geometrically it sends vectors into co-vectors, and eigenvectors are only meaningfully defined if the domain and codomain represent the same vector space. However, as we are working geometrically, we have a canonical map of co-vectors into vectors using the "raising operation" associated to the Minkowski metric.

In other words, you multiply $F_{ab}$ on the right by the matrix $\eta^{ca}$ which has $(-1,1,1,1)$ along its diagonal, and 0 elsewhere. If you don't follow the whole shpiel above, don't worry too much. We can summarise it in the following: associated to an electromagnetic field on space-time, there is a natural linear transformation $\tilde{F}^a{}_b$ which can be expressed as $\sum_c \eta^{ac}F_{cb}$ and as a matrix

$$ \tilde{F} = \begin{pmatrix} 0 & E_x & E_y & E_z \\ E_x & 0 & -B_z & B_y \\ E_y & B_z & 0 & -B_x \\ E_z & -B_y & B_x & 0 \end{pmatrix} $$

Let us consider its eigenvectors! Suppose $v$ is a 4-vector such that $\tilde{F}v = \lambda v$. Now we take the Minkowski inner product against $v$ itself. You have that

$$ \lambda \eta(v,v) = \eta( \tilde{F} v,v) = \eta( \eta\circ F v,v) = F(v,v) = 0$$

The third equality uses the fact that $\eta = \eta^{-1}$, and the last inequality uses that $F$ is antisymmetric. (Notice that the above is also true with respect to the Euclidean inner product if $\tilde{F}$ were defined that way. And in particular it says that relative to a Euclidean inner product there are no real eigenvectors with non-zero eigenvalue...) So we conclude that $v$ either lies in the null-space of $\tilde{F}$ (from $\lambda = 0$; and hence in the null-space of $F$), or, $\eta(v,v) = 0$.

The latter condition says that $v$ is a light-like vector in Minkowski space.

A somewhat involved linear algebra computation using the symmetries of the problem gives us the following theorem:

Theorem (Classification of electromagnetic fields) Given an arbitrary electromagnetic field, one of the following three must hold:

  1. $\tilde{F}$ is the zero linear transformation: it sends all vectors to zero.
  2. $\tilde{F}$ has only one real eigenvector. It is a repeated eigenvector of multiplicity two, and its eigenvalue is 0.
  3. $\tilde{F}$ has exactly two real eigenvectors. They each have multiplicity one, and their eigenvalues agree in magnitude but differs in sign.

(There's also a bit more you can say about the 3rd case, and I'll refer you to the wikipedia article. )

In the case where $F$ is non-zero, the eigenvectors are called the principal null directions of the electromagnetic field. The precise physical interpretation of the object is tricky in general, but in the second case, where $\tilde{F}$ has only one real eigenvector, the interpretation is clear: the unique principal null direction represents the direction of travel for the electromagnetic wave (recall that light always travels at the speed of light and on a light-like curve). (Recall that the eigenvalue is zero here, so there isn't much to interpret in this case.) (And you can think of the more complicated case 3 as situation where at the point the electromagnetic field is not described by the propagation of just one wave alone, but the superposition of many.)

As a last side remark: the same procedure can be used to describe, analogously, the propagation of gravitational disturbances, and is extremely useful in general relativity.

share|cite|improve this answer

They are characteristic of oscillating systems. There can be certain "modes" which always oscillate at the same frequencies, these are the eigenvalues and the frequencies the eigenvectos.

Have a look at the video here.

At resonance frequencies, the rice forms a pattern of gaps , which maps out the nodes of the standing wave pattern. The pattern you see is an eigenstate of that shape; the frequency squared is its eigenvalue.

share|cite|improve this answer

In terms of transformations of $E^{n}$ ($n$-dimensional Euclidean space), the eigenvectors of a transformation are axes along which the transformation does not rotate the space (but may or may not stretch or compress it). For each eigenvector there is a corresponding eigenvalue, which measures how much stretching or compression occurs along the axis defined by the eigenvector.

share|cite|improve this answer
This is a geometrical interpretation, not a physical one. – Rasmus Mar 13 '11 at 11:08
In many physics problems, such as those concerning changing reference frame in classical or relativistic mechanics, this is the correct interpretation. I gave this broad explanation because he made no mention of quantum mechanics or eigenfunctions of linear operators, which I associate more closely with quantum phenomena. – Alex Becker Mar 14 '11 at 1:44

in quantum mechanics, you can only measure the stable state of a system, which is defined to be the eigen state of some operator. For example, $p=-i\frac{h}{2\pi}\nabla$. Hence eigen vectors and eigen values play a fundamental role in quantum mechanics. You may read Chapter 1-2 in Dirac's book Principle of Quantum Mechanics to get a better understanding. I am also an undergraduate, I expect a much detailed answer from others as eigenvectors just appeared in everywhere. A not bad place for start is here:

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.