Read an operator through preserved subspaces

Try

The matrix A=[[4,2],[0,1]] preserves the horizontal axis because A(x,0)=(4x,0). A vector off that axis can acquire an additional horizontal component. Draw e₁ and e₂, then their images (4,0) and (2,1). The preserved axis supports a smaller operator, while the vertical class modulo that axis supports another. These are the two diagonal entries, even though the whole matrix is not diagonal.

Predict

Predict whether every complement to an invariant subspace is also invariant. The vertical axis complements the horizontal axis in this example, but A(0,1)=(2,1) leaves it. A complement supplies coordinates; invariance imposes a dynamical condition. Here span((−2,3)) is an invariant complement because A(−2,3)=(−2,3). The existence of this particular complement should not be generalized to every operator.

Experiment

Compare the matrix in the standard basis with its matrix in the ordered basis ((1,0),(−2,3)). Compute both basis images explicitly. In the second basis the matrix is diag(4,1), so applying the operator repeatedly scales the two components independently. Now compare N=[[0,1],[0,0]]. Its horizontal axis is invariant, but a complementary vector (a,b) with b≠0 maps to (b,0); no complementary line is invariant. Record this failure as a counterexample to the tempting rule that every invariant subspace splits into independent dynamics.

Understand

A subspace U is T-invariant when T(U)⊆U. It suffices to test images of a basis because T is linear. In a basis beginning with U, the matrix has block form [[B,C],[0,D]]. B represents the restricted operator on U, and D represents the induced operator on V/U. The latter is well-defined precisely because U is invariant. A complement W is invariant exactly when its off-diagonal block C vanishes in that decomposition. Thus block diagonalization requires both subspaces to be invariant, whereas block triangularization requires only the first. The quotient map forgets the component entering U; it does not imply that this component is zero in V.

Apply

One useful source of invariant spaces is commutation. If S commutes with T and Tv=λv, then T(Sv)=S(Tv)=λSv, so each eigenspace of T is S-invariant. More generally ker p(T) is invariant under T, since T commutes with every polynomial in itself. Check this directly by applying p(T)T to a vector annihilated by p(T). These observations will organize generalized eigenspaces and simultaneous spectral arguments. As a numerical action, use N above to compare ker N and ker N²: the first is a line and the second is the whole plane.

A useful determinant consequence is det(tI−T)=det(tI−B)det(tI−D) in the block basis. This follows from the block-triangular determinant formula, not from a claim that C is absent. Hence restriction and quotient spectra combine to give the full spectrum even when dynamics remain coupled. The next lesson will use exactly this distinction to compare algebraic multiplicity with the dimension of an eigenspace.

A differential operator with an invariant degree ladder

A polynomial space supplies a concrete invariant chain: constants, affine polynomials, quadratics, and so on. Let Dp=(ax²+bx+c)p″+(dx+e)p′+fp, with real constants a,b,c,d,e,f. Differentiation lowers degree enough that D preserves every space Pₘ of polynomials of degree at most m. Applied to xʲ, the coefficients of xʲ, xʲ⁻¹ and xʲ⁻² are respectively aj(j−1)+dj+f, bj(j−1)+ej and cj(j−1); terms with negative exponents are absent. Thus the matrix in ascending monomials is upper triangular, and its diagonal entries are λⱼ=aj(j−1)+dj+f.

If a≥0 and d>0, then λⱼ₊₁−λⱼ=2aj+d>0, so the diagonal entries are distinct. For each j, D−λⱼI is invertible on Pⱼ₋₁: its triangular diagonal entries are nonzero there. Solve uniquely for a lower-degree correction q so that (D−λⱼI)(xʲ+q)=0. These monic eigenpolynomials have different degrees, hence form a basis. This is an actual eigenvector construction, not merely a list of eigenvalues. The sufficient inequalities are not necessary. Repeated diagonal entries require more investigation: p↦p″ on P₂ is nonzero nilpotent and not diagonalizable, while a scalar multiple of identity is diagonalizable despite complete repetition.

Why these eigenpolynomials are also orthogonal

Take Lp=(x²−1)p″+2xp′, equivalently the derivative of (x²−1)p′, on [−1,1]. Its eigenvalues are j(j+1). The first modes, normalized to value one at x=1, are 1, x, (3x²−1)/2, (5x³−3x)/2 and (35x⁴−30x²+3)/8. Substitution verifies the displayed modes; the degree-ladder construction proves existence and uniqueness up to scaling at every degree.

For real polynomials p and q, integration by parts gives ∫(Lp)q=−∫(x²−1)p′q′=∫p(Lq). Both boundary terms vanish because x²−1 is zero at the endpoints and polynomial derivatives are finite there. Distinct-eigenvalue polynomials are therefore orthogonal for the integral inner product: subtract the two identities to get (λⱼ−λₖ)∫pⱼpₖ=0. Each degree-j eigenpolynomial is orthogonal to all lower degrees. Gram–Schmidt applied to monomials produces the same direction, because there is only one monic degree-j polynomial orthogonal to Pⱼ₋₁: the difference of two would be lower-degree and orthogonal to itself.

The vanishing boundary coefficient is essential. For the superficially similar Mp=(x²p′)′ on [−1,1], take p=x² and q=1. Then ∫(Mp)q=4 but ∫p(Mq)=0; the missing boundary term is four. Merely writing a differential expression in divergence form does not establish self-adjointness for the chosen functions. Here we prove symmetry on finite polynomial spaces. We do not assert a complete unbounded-operator domain theorem or invoke regular Sturm–Liouville theory at the degenerate endpoints.

One basis, two constructions

Predict which degree changes improve the integral fit of an even function. Compare polynomial differential eigenvectors with orthogonal projection coefficients. Coefficient lists run from constant through highest degree.

Target |x|x⁴ Projection degree 01234 Inspect eigenmode 01234

Selected mode coefficients: -0.5, 0, 1.5. Eigenvalue 6; differential image coefficients: -3, 0, 9. Integral squared norm 0.40000000.

Projection coefficients: 0.18750000, 0.00000000, 0.93750000, 0.00000000, 0.00000000. Integrated squared error: 0.01041667. This is an L² optimum, not a pointwise or uniform error guarantee; the fit can overshoot the target.

Mode n	Eigenvalue	Norm squared	Target inner product	Projection weight	Included?
0	0	2.00000000	1.00000000	0.50000000	Yes
1	2	0.66666667	0.00000000	0.00000000	Yes
2	6	0.40000000	0.25000000	0.62500000	Yes
3	12	0.28571429	0.00000000	0.00000000	No
4	20	0.22222222	-0.04166667	-0.18750000	No

Residual inner products with included modes: 0.00000000, 0.00000000, 0.00000000. The integral formulas use polynomial moments and parity, not plot samples. Floating-point roundoff below 10⁻¹³ in the squared error is displayed as zero; the raw norm difference is 1.042e-2.

Reset

Five modes and 161 plot samples bound the work. The all-degree proof is separate. State is temporary and awards no mastery.

Exact integral fit, not pointwise interpolation

For a continuous target g, projection onto Pₘ is the sum of the mode pⱼ multiplied by ∫g pⱼ / ∫pⱼ². Orthogonality makes the residual perpendicular to every polynomial of degree at most m. The Pythagorean identity then proves this is the unique minimizer of the integrated squared error. This uses the existing finite-dimensional projection theorem inside the inner-product space of continuous functions; that space need not be complete for this finite-dimensional argument.

For g(x)=|x| and degree two, the odd-mode coefficient vanishes by parity. The constant mode has inner product1 and squared norm2, giving weight1/2. The quadratic mode has inner product1/4 and squared norm2/5, giving weight5/8. Thus the best quadratic is 3/16+15x²/16. Its squared norm is 1/2+5/32=21/32, while ∫g²=2/3, so its integrated squared error is 1/96. At x=0 the error is 3/16 and at x=1 the fit is 9/8, overshooting the target1. An L² optimum need not interpolate or minimize maximum error.

Predict why adding degree three leaves this even target’s fit unchanged. Then compare x⁴ at degrees two and four, checking the residual inner products rather than only the graph. Write a proof that the all-degree eigenpolynomials and Gram–Schmidt directions coincide, and identify the exact boundary step that would fail for M.

For preparation, return to orthogonal projection and its best-approximation proof. For a different domain and boundary contract, compare Sturm–Liouville eigenproblems.

Continue

The block viewpoint separates a proof strategy from a lucky choice of eigenvectors. It prepares triangularization, characteristic polynomials and primary decompositions. An invariant subspace need not have an invariant complement; additional hypotheses, such as a normal operator together with an inner product, will restore stronger decompositions. Avoid interpreting an arbitrary upper-triangular matrix as independent coordinates: its upper-right block records coupling that can accumulate under powers.

Review Linear Algebra I when a prerequisite needs practice. Continue to Find spectral information that survives coordinates.