Eigenvalues and Eigenvectors

Eigenvalues answer a deceptively simple question: when a matrix acts on a vector, is there a direction that the matrix only stretches and never turns? Those special directions are the eigenvectors, and the stretch factors are the eigenvalues. They are the hidden skeleton of a linear system, and on the CELE they surface both directly and indirectly: the critical buckling load of a column, the natural frequencies of a structure, and the principal stresses at a point are all eigenvalue problems, while decoupling a system of differential equations quietly rides on the same idea. This lesson builds the machine from the characteristic equation up and carries every worked example all the way to a finished number.

What an eigenvalue is

A nonzero vector v is an eigenvector of a square matrix A when A leaves its direction unchanged and only scales it. The scale factor lambda is the eigenvalue, and the defining relation is:

A v = lambda v, with v not equal to the zero vector.

Move everything to one side, factoring out v with the identity matrix I so the dimensions match:

(A - lambda I) v = 0

This is a homogeneous system. It always has the trivial answer v = 0, but an eigenvector must be nonzero, and a homogeneous system has a nonzero solution only when its coefficient matrix is singular. So the eigenvalues are exactly the values of lambda that make A - lambda I singular, which is to say that make its determinant vanish:

det(A - lambda I) = 0

That single equation is the characteristic equation, and it is where every eigenvalue problem begins.

The characteristic equation of a 2 by 2 matrix

For a 2 by 2 matrix A = [[a, b], [c, d]], form A - lambda I by subtracting lambda from each main-diagonal entry:

A - lambda I = [[a - lambda, b], [c, d - lambda]]

Its determinant is (a - lambda)(d - lambda) - bc. Expanding and collecting terms gives a tidy quadratic that every 2 by 2 problem reduces to:

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

The coefficient of lambda is the negative of the trace (the sum of the main-diagonal entries), and the constant term is the determinant. In shorthand:

lambda^2 - (trace) lambda + (det) = 0

Worked example: Find the eigenvalues of A = [[4, 1], [2, 3]]. The trace is 4 + 3 = 7 and the determinant is (4)(3) - (1)(2) = 12 - 2 = 10, so the characteristic equation is lambda^2 - 7 lambda + 10 = 0. Factor: (lambda - 5)(lambda - 2) = 0, giving lambda = 5 and lambda = 2. The two eigenvalues are 5 and 2. Notice the built-in check: 5 + 2 = 7 matches the trace and (5)(2) = 10 matches the determinant.

Finding the eigenvectors

Once an eigenvalue is known, its eigenvectors are the nonzero solutions of (A - lambda I) v = 0. Because A - lambda I is singular by construction, its two rows are proportional, so a single equation fixes the direction and any nonzero scalar multiple of the answer is also an eigenvector. We usually report the simplest integer vector.

Worked example: Find eigenvectors of A = [[4, 1], [2, 3]] for the eigenvalues 5 and 2 found above.

For lambda = 5, A - 5I = [[-1, 1], [2, -2]]. The top row gives -v1 + v2 = 0, so v2 = v1, and choosing v1 = 1 gives the eigenvector v = (1, 1). For lambda = 2, A - 2I = [[2, 1], [2, 1]]. The top row gives 2 v1 + v2 = 0, so v2 = -2 v1, and choosing v1 = 1 gives the eigenvector v = (1, -2). A quick sanity check on the first: A v = [[4, 1], [2, 3]] (1, 1) = (4 + 1, 2 + 3) = (5, 5) = 5 (1, 1) = lambda v, exactly as required.

Trace and determinant as instant checks

Two facts turn into free answer-checks and sometimes shortcut a whole problem. For any n by n matrix, the sum of the eigenvalues equals the trace, and the product of the eigenvalues equals the determinant.

Relation	What it equals	Why it is useful
Sum of eigenvalues	the trace (sum of main-diagonal entries)	verifies a pair of roots in one addition
Product of eigenvalues	the determinant	a zero eigenvalue forces det = 0, so the matrix is singular
Characteristic equation	det(A - lambda I) = 0	its roots are the eigenvalues, real or complex
Number of eigenvalues	n, counted with multiplicity	an n by n matrix has n roots in the characteristic equation

A useful consequence: a matrix is singular exactly when 0 is one of its eigenvalues, because the product of the eigenvalues is the determinant. For a symmetric matrix, whose CELE role is stiffness, mass, and stress matrices, all eigenvalues are guaranteed real, and a matrix whose eigenvalues are all greater than 0 is positive definite.

Eigenvalues of a 3 by 3 matrix

For a 3 by 3 matrix the characteristic equation det(A - lambda I) = 0 is a cubic in lambda, so there are three eigenvalues counted with multiplicity. Cofactor expansion produces the cubic; a clever substitution often keeps the algebra clean.

Worked example: Find the eigenvalues of A = [[4, 1, 1], [1, 4, 1], [1, 1, 4]]. Every diagonal entry is 4, so let m = 4 - lambda and write A - lambda I = [[m, 1, 1], [1, m, 1], [1, 1, m]]. Expanding along the first row:

det = m(m^2 - 1) - 1(m - 1) + 1(1 - m) = m^3 - m - m + 1 + 1 - m = m^3 - 3m + 2

Set m^3 - 3m + 2 = 0. It factors as (m - 1)^2 (m + 2) = 0, so m = 1 (a double root) or m = -2. Undo the substitution with lambda = 4 - m: m = 1 gives lambda = 3, and m = -2 gives lambda = 6. The eigenvalues are lambda = 6, 3, 3, where 3 is a repeated (double) eigenvalue. Check: the sum 6 + 3 + 3 = 12 equals the trace 4 + 4 + 4, and the product (6)(3)(3) = 54 equals det(A).

A special case worth memorizing: for a triangular matrix (all entries above or all below the main diagonal are zero), the eigenvalues are just the main-diagonal entries, so no cubic is needed. For A = [[2, 7, 9], [0, 3, 5], [0, 0, 8]] the eigenvalues are 2, 3, and 8 read straight off the diagonal.

Diagonalization

When an n by n matrix has n independent eigenvectors, it can be factored into a diagonal core. Place the eigenvectors as the columns of a matrix P and the matching eigenvalues on the diagonal of D. Then:

A = P D P^-1, and equivalently P^-1 A P = D

This is diagonalization, and it is what makes repeated matrix action cheap. Because D is diagonal, its powers are found entry by entry, so:

A^k = P D^k P^-1

Worked example: For A = [[4, 1], [2, 3]] we found eigenvalues 5 and 2 with eigenvectors (1, 1) and (1, -2). Then P = [[1, 1], [1, -2]] and D = [[5, 0], [0, 2]]. To get the eigenvalues of A^2 you do not multiply the matrix out: the eigenvalues of A^2 are 5^2 and 2^2, that is 25 and 4. As a direct check, A^2 = [[4, 1], [2, 3]] times [[4, 1], [2, 3]] = [[18, 7], [14, 11]], whose trace 18 + 11 = 29 equals 25 + 4 and whose determinant (18)(11) - (7)(14) = 198 - 98 = 100 equals (25)(4). Diagonalization also decouples a system of differential equations dx/dt = A x: in the eigenvector coordinates each mode evolves on its own as e^(lambda t), so the eigenvalues govern growth, decay, and stability.

Applications in civil engineering

Three board-standard problems are eigenvalue problems in disguise. Column buckling gives the critical load as the smallest eigenvalue of a stiffness relation; structural vibration gives the natural frequencies from the eigenvalues of the dynamic matrix (stiffness divided by mass); and the state of stress at a point gives the principal stresses as the eigenvalues of the stress matrix, with the principal directions as its eigenvectors.

Worked example (principal stresses): The plane stress at a point is described by the symmetric matrix [[60, 20], [20, 30]] MPa (the diagonal holds the normal stresses and the off-diagonal the shear stress). The principal stresses are its eigenvalues. Trace = 60 + 30 = 90 and det = (60)(30) - (20)(20) = 1800 - 400 = 1400, so lambda^2 - 90 lambda + 1400 = 0. The discriminant is 90^2 - 4(1400) = 8100 - 5600 = 2500, and sqrt(2500) = 50, so lambda = (90 +/- 50) / 2, giving sigma1 = 70 MPa and sigma2 = 20 MPa. For the principal direction, lambda = 70 gives (A - 70I) = [[-10, 20], [20, -40]], whose first row -10 v1 + 20 v2 = 0 yields v1 = 2 v2, so the eigenvector is (2, 1). Its angle from the x-axis is theta = arctan(1/2) = 26.57 degrees. So the major principal stress is 70 MPa acting along a direction 26.57 degrees from the x-axis, and the minor principal stress is 20 MPa on the perpendicular direction.

Worked example (natural frequency): A two-story shear building reduces to the eigenvalue problem for its dynamic matrix [[250, -150], [-150, 250]] in units of (rad/s)^2, where the eigenvalues are the squared natural frequencies omega^2. Trace = 500 and det = (250)(250) - (-150)(-150) = 62500 - 22500 = 40000, so lambda^2 - 500 lambda + 40000 = 0. The discriminant is 500^2 - 4(40000) = 250000 - 160000 = 90000, and sqrt(90000) = 300, so lambda = (500 +/- 300) / 2, giving eigenvalues 400 and 100 (rad/s)^2. Take square roots for the frequencies: omega1 = sqrt(100) = 10 rad/s and omega2 = sqrt(400) = 20 rad/s. Converting to cycles, f1 = omega1 / (2 pi) = 10 / 6.2832 = 1.59 Hz and f2 = omega2 / (2 pi) = 20 / 6.2832 = 3.18 Hz. The fundamental (lowest) natural frequency of the building is 1.59 Hz.

Exam-day strategy

• Always start from det(A - lambda I) = 0; subtract lambda from each main-diagonal entry only, never from the off-diagonal entries, before taking the determinant.
• For a 2 by 2 matrix, jump straight to lambda^2 - (trace) lambda + (det) = 0 rather than expanding the determinant by hand; it is faster and the sign on the trace term is the usual slip.
• Check any eigenvalue answer in two additions: the roots must sum to the trace and multiply to the determinant. If either fails, you have an arithmetic error.
• Read the eigenvalues straight off the diagonal of a triangular matrix and skip the cubic entirely.
• To get an eigenvector, plug the eigenvalue back in, use one row of A - lambda I, and pick the simplest integer vector; remember any nonzero multiple is equally valid, so do not panic if your vector differs from the key by a scale factor.
• Translate the words: principal stresses are eigenvalues of the stress matrix (report in MPa), natural frequencies are square roots of the eigenvalues of the dynamic matrix (report omega in rad/s, then f in Hz), and the smallest eigenvalue of a buckling stiffness relation sets the critical load.
• A zero eigenvalue means a zero determinant, so a singular matrix has no inverse and its system has no unique solution; that single fact often answers a multiple-choice item outright.