Normalize a fundamental matrix and change coordinates

Try

Two independent solutions of a linear system can become columns of a matrix. If their initial columns are (1,0) and (1,1), does multiplying that matrix by the desired initial state automatically produce the correct solution? Test the proposed rule with initial state (0,1): it would start at (1,1), which is wrong. The columns form a basis of solutions, but their initial basis is not the identity.

Predict

Predict which side of the solution-column matrix should be multiplied by the inverse initial matrix. Then predict whether shearing or reflecting coordinate axes changes the actual physical trajectory when the initial coordinates are transformed consistently.

Experiment

Change the parameter and coordinate basis

Temporary analytic teaching values. Controls do not save evidence or award understanding. Record a prediction and explanation in the lesson practice.

Parameter case a = −4a = −3 (zero determinant)a = −2.7a = 1 − 2√3 (exact repeated root)a = −2a = −1 (zero trace)a = 0a = 1 + 2√3 (exact repeated root)a = 5 Coordinate basis Original coordinatesSheared basis: (1,0), (1,1)Reflected basis: (1,0), (0,−1) Elapsed time 00.10.250.5 Reset inspector

A(a) = [a, −3; 1, 1]. Center; stable, not attracting. Full real-parameter region: a = −1. Named radical cases use the exact repeated-root formula; their displayed decimals are approximations.

Trace 0.00000, determinant 2.00000. Fixed physical initial state x(0) = (1,−1). Time 0.25: x = (1.42780, -0.93815).

Physical solution matrix E(t), with E(0)=I

[ 0.69332, -0.73447; 0.24482, 1.18297 ]

Basis matrix S, with x = S z

[ 1.00000, 0.00000; 0.00000, 1.00000 ]

Transformed generator S⁻¹ A S

[ -1.00000, -3.00000; 1.00000, 1.00000 ]

Transformed solution matrix S⁻¹ E(t) S

[ 0.69332, -0.73447; 0.24482, 1.18297 ]

z(0) = (1.00000, -1.00000); z(t) = (1.42780, -0.93815). Reconstructing S z(t) gives (1.42780, -0.93815). Changing basis preserves the physical trajectory and changes its coordinates.

The displayed trajectory stays inside the fixed −4 to 4 window. It has 65 analytic samples and is not a time-integrator output or a proof of stability. The portrait always uses physical x coordinates; a reflected z portrait would reverse orientation.

Keep the parameter at the center case and time at 0.25. Change from original to sheared and reflected bases. The generator and z coordinates change; the physical x coordinates and portrait remain fixed. Set time zero to check both normalized solution matrices are the identity. Choose a repeated-root case and explain why a formula requiring two independent eigenvectors would fail even though the inspector still has a solution matrix.

Understand

A basis of solutions needs an initial normalization

Let Φ(t) have independent solution columns for x′ = A(t)x, where A is continuous on an interval. At a chosen initial time t₀, solve Φ(t₀)c = x₀ for the constant coefficient vector c. The resulting solution is:

U(t₀,t₀) is the identity. Differentiation gives U′ = A(t)U. Independence at one time implies independence throughout the interval: if a nonzero column combination vanished at another time, uniqueness of the zero IVP would make it vanish at t₀ too, contradicting independence. The existing existence and uniqueness lesson supplies this local uniqueness argument, extended along the common interval. For time-varying A, do not replace U by exp(tA(t)); differentiation then introduces extra terms.

For constant A, the normalized matrix at t₀ = 0 is E(t) = exp(tA). Its power series converges absolutely and can be differentiated on bounded time intervals. This justification already appears in the skew-adjoint exponential construction; it applies to a general finite matrix using the same norm bound. Directly checking E′ = AE and E(0) = I is another sufficient verification by uniqueness.

Worked construction with a nonidentity initial basis

Take eigenvectors (1,0), (1,1) with eigenvalues −1, 2. The basis matrix is S = [1,1;0,1]. Its inverse is [1,−1;0,1]. Thus A = S diag(−1,2) S⁻¹ = [−1,3;0,2]. The raw solution columns are Φ = S diag(exp(−t), exp(2t)); Φ(0) = S. Normalizing on the right gives:

$$E(t)=\Phi(t)S^{-1}=\begin{pmatrix}e^{-t}&e^{2t}-e^{-t}\\0&e^{2t}\end{pmatrix}.$$

For initial state (0,1), the second column gives x(t) = (exp(2t) − exp(−t), exp(2t)). Both coordinates start correctly. Differentiating the first gives 2exp(2t) + exp(−t), which agrees with −x₁ + 3x₂. This check would expose a missing normalization factor.

Coordinate conventions determine multiplication order

For a constant invertible basis matrix S, define physical coordinates by x = S z. Then S z′ = A S z, so the coordinate generator and flow are:

Verify both the derivative and initial identity of the conjugated flow. Similarity preserves eigenvalues and stability. It need not preserve visual angles, Euclidean lengths or the label “counterclockwise” under a reflected coordinate system. The inspector keeps its portrait in physical x coordinates so changing S does not redraw a different physical experiment. For a time-varying basis the transformed generator also contains −S⁻¹S′; the displayed rule assumes S is constant.

A construction that survives repeated eigenvalues

For a real 2×2 matrix, write T = trace(A), Δ = det(A), B = A − (T/2)I and d = T²/4 − Δ. Cayley–Hamilton gives A² − TA + ΔI = 0 and hence B² = dI. Every even power of B is a scalar multiple of I and every odd power a multiple of B. Grouping the absolutely convergent exponential series yields:

At d = 0, B may be zero, producing a scalar exponential, or nonzero with B² = 0, producing a defective matrix and a necessary t factor. For A = [−1,2;0,−1], E(t) = exp(−t)[1,2t;0,1]. Differentiate this expression and check its initial value; a nonexistent second eigenvector is unnecessary. The named exact repeated-root controls use this branch explicitly instead of classifying a tiny floating-point discriminant by its sign.

For constant A, uniqueness also proves E(t+s) = E(t)E(s), because both sides as functions of t solve the same matrix IVP with value E(s) at t = 0. Thus E(−t) is the inverse of E(t). Invertibility of a finite-time flow does not imply long-time stability; an unstable flow is still reversible mathematically.

Apply

Construct E for a diagonalizable example, a defective example and a changed coordinate basis. In each case check the same three things: derivative, initial identity and the requested initial state. A plausible portrait or correct eigenvalues alone do not establish all three.

Continue

Use parameter families and boundary portraits to classify the inspector’s entire real parameter family. Return to linear systems for eigenvector decomposition and defective portraits for the meaning of the polynomial factor.

Source scope: matrix-exponential construction and initialized vector solutions appear in MIT 18.03SC’s final materials, Fall 2011, consulted 28 October 2026. The matrices and verification tasks here are original.