Matrices, linear algebra, and eigenvalue problems

Read a matrix by rows and columns

A matrix is a rectangular array of entries. Its horizontal lists are rows; its vertical lists are columns. A matrix with m rows and n columns has shape m × n, and aᵢⱼ denotes the entry in row i, column j. These row/column counts are distinct from a physical dimension or unit.

Supply the exact real, dimensionless matrix A = [[2, 1]; [1, 2]]. In this compact notation a semicolon separates rows and commas separate entries within a row. A is 2 × 2, with a₁₂ = 1 and a₂₂ = 2. A square matrix has equal row and column counts. A column [q₁; q₂] is a 2 × 1 matrix.

For equal shapes, add corresponding entries; multiplying by a scalar multiplies each entry. Thus [u₁; u₂] + [w₁; w₂] = [u₁ + w₁; u₂ + w₂] and c[u₁; u₂] = [cu₁; cu₂]. In our length-valued cases c is a dimensionless real number, and additions require compatible physical quantities.

Fix the coordinate basis and units

Declare a fixed ideal plane, a reference origin and perpendicular axes, with right and up positive. The displacement components q₁ and q₂ both use metres. Their column q represents one displacement vector in this fixed basis. The canonical coordinate columns are e₁ = [1; 0] and e₂ = [0; 1], so q = q₁e₁ + q₂e₂. A basis supplies a unique such component representation. Here the e columns give dimensionless coordinate directions; the q coefficients carry length units.

Define an original hypothetical calculation p = A q for all real length components q₁ and q₂, using the same component basis and displacement quantity kind for input and output. All entries of A are dimensionless, so the output components have metre units. This equation supplies a mathematical map; it does not supply a force law, time rule or observed body's motion.

Matching shape alone is insufficient for a physical calculation. Every product term being added within an output component must have that output's compatible quantity kind and units. Coefficients can carry units in other models, possibly different units for different entries. For this unchanged dimensionless A, mixing a metre input with a second input in seconds would make expressions such as 2 m + 2 s invalid. Changing the basis or component ordering also requires the corresponding representation change, not silently reusing the old entries.

Multiply a row by a column

Matrix multiplication uses products across a row and down a column, then sums them. For our two-component input, A q = [2q₁ + q₂; q₁ + 2q₂]. With the exact supplied q = [1 m; −2 m], this gives [2(1 m) + (−2 m); 1 m + 2(−2 m)] = [0 m; −3 m].

In general, an m × n matrix can multiply an n × r matrix on its right, producing an m × r matrix. The inner counts n must match. The product of a 2 × 3 matrix with a 3 × 1 column therefore has shape 2 × 1. A shape check is done before arithmetic or physical interpretation.

The entry in row i, column j of a product is Σₖ aᵢₖ bₖⱼ, where Σₖ means add over the matching index k. This is not entry-by-entry multiplication of two whole matrices. For A acting on q, each sum has two terms; each term has metre units under our declared model.

Recognize linearity and order

The map T(q) = A q is linear: T(u + w) = T(u) + T(w) and T(cu) = cT(u) for compatible inputs and a dimensionless real scalar c. These equations define linearity here. They follow from distributing the row sums. In particular, zero maps to zero. Adding a fixed nonzero output displacement to every result would fail this zero-input requirement.

Now supply a second dimensionless map in the same basis, B = [[1, 2]; [0, 1]]. Applying B first and A second gives A(Bq) = (AB)q; the rightmost map acts first. For this pair, row–column multiplication gives AB = [[2, 5]; [1, 4]], while BA = [[4, 5]; [1, 2]].

The products differ even though both are 2 × 2. Matrix multiplication does not generally commute. Some pairs do commute; the statement is not that every pair differs. For our supplied q = [1 m; −2 m], (AB)q = [−8 m; −7 m] and (BA)q = [−6 m; −3 m]. The order changes this defined calculation without establishing an actual sequence of physical motions.

Separate a system from a map

Evaluating the map starts with q and calculates A q. Solving the linear system A q = b starts with a supplied target column b and seeks an unknown q. Its equations are 2q₁ + q₂ = b₁ and q₁ + 2q₂ = b₂. The unknowns occur only to the first power and are not multiplied together; these are linear equations. Target components b have metre units in our model.

The two-by-two identity is I = [[1, 0]; [0, 1]], which leaves a compatible column unchanged. An inverse A⁻¹ satisfies A⁻¹A = AA⁻¹ = I. It reverses the mathematical map. For a real two-by-two matrix K = [[a, b]; [c, d]], the determinant is det K = ad − bc; K is invertible exactly when this number is nonzero. We write det explicitly to distinguish it from a vector's magnitude.

For det K ≠ 0, the two-by-two inverse is K⁻¹ = (1/(ad − bc))[[d, −b]; [−c, a]]. Multiplying K by the numerator matrix in either order gives (ad − bc)I, which checks this formula. For A, det A = 4 − 1 = 3 and A⁻¹ = (1/3)[[2, −1]; [−1, 2]]. Thus the target b = [5 m; 4 m] has the unique input q = A⁻¹b = [2 m; 1 m]; substitution checks both equations.

Check what singular means

A square matrix with zero determinant is singular and has no two-sided inverse. That does not mean every associated system has no solution. The answer depends on the target: a singular system can be inconsistent, or it can admit more than one input.

Supply the separate exact dimensionless matrix S = [[0, 0]; [0, 2]] in the same fixed component basis. It acts as S q = [0 m; 2q₂] for length inputs. Its determinant is zero. The first input component disappears from the output, so the output cannot recover that component uniquely.

For a target whose first component is nonzero, the first equation is impossible. When that component is zero, the second equation fixes q₂ while q₁ remains any real length. These conclusions follow directly from this stipulated S; they illustrate why a determinant test must be paired with the actual equations and target. No division by a zero determinant is allowed.

Define an eigenpair with a nonzero vector

For a square matrix representing a map from a space back to itself, an eigenpair consists of a nonzero vector v and a scalar λ satisfying A v = λv. The Greek letter λ names the eigenvalue, and v is an eigenvector. Here we seek real λ and real columns, with input and output expressed in the same fixed basis. The eigenvector equation selects a component pattern whose output is a scalar multiple of itself; it is not a general statement that all inputs share this pattern.

The zero vector is excluded because A[0; 0] = λ[0; 0] holds for every λ and selects no eigenvalue. Zero can nevertheless be an eigenvalue. For our separate S, S e₁ = [0; 0] = 0e₁, while e₁ = [1; 0] is nonzero. A negative eigenvalue would reverse a nonzero pattern's sign; a zero one gives a zero output, so 'same direction' alone is an incomplete definition.

If v is an eigenvector, every nonzero dimensionless real multiple cv is also one with the same λ: A(cv) = cAv = cλv = λ(cv). A chosen eigenvector's magnitude and sign are therefore not unique. Our displayed v columns use dimensionless coordinate ratios. Multiplying such a column by a nonzero length gives a length-valued input with the same scaling property for our dimensionless A; λ remains dimensionless.

Find and check the symmetric example

The transpose Aᵀ swaps rows with columns. A real square matrix is symmetric when Aᵀ = A, equivalently aᵢⱼ = aⱼᵢ. Our A is symmetric: both off-diagonal entries equal 1. We will exhibit its two exact real eigenpairs, rather than assume every real matrix has this same result.

Rewrite the eigenvector equation as (A − λI)v = 0. A nonzero solution requires A − λI to be singular, so solve det(A − λI) = 0. Here this is (2 − λ)² − 1 = λ² − 4λ + 3 = (λ − 3)(λ − 1) = 0. A product of real numbers is zero only if at least one factor is zero, giving λ = 3 or λ = 1. Each candidate still needs a nonzero vector and a multiplication check.

At λ = 3, (A − 3I)v = 0 gives v₂ = v₁; choose v₊ =, and A v₊ = = 3v₊. At λ = 1 it gives v₂ = −v₁; choose v₋ = [1; −1], and A v₋ = [1; −1] = v₋. The plus/minus subscripts distinguish these chosen patterns. These are exact relations for the supplied static map. A physical oscillation mode, frequency or stability conclusion would require additional physical equations and conditions; no trajectory follows from these eigenpairs alone.

Worked examples

Try it yourself

1. For the dimensionless matrix R = [[4, −1, 0]; [2, 3, 5]], give its shape, number of entries and r₂₃. Does 2 × 3 here name a physical unit?

   Show the answer and reasoning

   R has shape 2 × 3 and six entries; r₂₃ = 5. The shape is a row/column count, not a physical unit.

   There are two rows and three entries in each row. The subscript 23 selects row 2, column 3. Counting the entries gives 2(3) = 6.

2. A dimensionless C has shape 2 × 3 and a dimensionless D has shape 3 × 1. Is CD defined, and what is its shape? Is DC defined?

   Show the answer and reasoning

   CD is defined with shape 2 × 1. DC is not defined under matrix multiplication.

   For CD the inner counts 3 and 3 match. For DC the inner counts are 1 and 2, which do not match. A reversed order must be checked separately.

3. Someone inserts q = [1 m; 2 s] into the unchanged dimensionless A and claims p₁ = 2 m + 2 s is a displacement. Is this physically valid?

   Show the answer and reasoning

   No. The terms have incompatible quantity kinds and dimensions, so this sum does not define a displacement in the declared model.

   A matched 2 × 1 shape does not repair the unit mismatch. The original map requires both inputs to be displacement components. Other models may use entries carrying units to produce compatible output terms, but those coefficients and quantity meanings must be explicitly supplied.

4. For the supplied S, solve Sq = [0 m; 6 m]. Then decide whether Sq = [1 m; 6 m] has any solution. All q components are real lengths.

   Show the answer and reasoning

   For [0 m; 6 m], q₂ = 3 m and q₁ can be any real length. The target [1 m; 6 m] has no solution.

   S gives [0 m; 2q₂]. The first target allows 0 m = 0 m and fixes only q₂. The second would require 0 m = 1 m, which is impossible. Thus singular does not by itself mean that every target lacks a solution.

5. For S, is e₁ = [1; 0] an eigenvector, and what is its eigenvalue? May [0; 0] be called an eigenvector? Use dimensionless coordinate columns.

   Show the answer and reasoning

   The nonzero e₁ is an eigenvector with eigenvalue 0. The zero column is not an eigenvector.

   Se₁ = [0; 0] = 0e₁ meets the equation with a nonzero input. The zero column would meet A v = λv for every λ, so the definition excludes it. Zero eigenvalue and zero eigenvector are different conditions.

6. For A, test whether w = is an eigenvector by comparing both components of Aw = λw. Also check the eigenvalue of [−4; −4]. These columns are dimensionless.

   Show the answer and reasoning

   [2; 1] is not an eigenvector. [−4; −4] is an eigenvector with eigenvalue 3.

   Aw =. Its first component would require λ = 5/2, while the second requires λ = 4; one scalar cannot satisfy both. A[−4; −4] = [−12; −12] = 3[−4; −4], and the input is nonzero.

What to carry forward

• A matrix's shape counts rows and columns; it does not name a physical unit.
• Check matching inner counts, multiplication order and compatible quantity terms before interpreting a product.
• A matrix represents a linear map in declared bases; evaluating the map and solving for an unknown input are different tasks.
• For a real two-by-two matrix, ad − bc is the determinant; nonzero determinant permits a two-sided inverse.
• A singular map can admit many inputs for one target and none for another; check the actual equations.
• An eigenvector must be nonzero, but its eigenvalue may be zero. Nonzero scalar multiples retain the same eigenvalue.
• The exact eigenpairs of the supplied symmetric matrix are mathematical results; a physical mode or stability interpretation needs additional equations and conditions.