Polynomial Signals
Exercise PLAN-505
primary source nl only. Additive answer successor with distinct content-bound review; historical rejected answer retained only for audit. No new Lean verification is claimed for this NL view.
Assumptions and conventions
- A is non-normal; eigenvalues and singular values differ
- Degree-two singular value transformation uses right singular vectors on both sides (even parity)
- T_2(x) = 2 x^2 - 1
Question
Consider the two-by-two operator A that sends the first basis vector to zero and the second basis vector to the first (the strictly upper-triangular nilpotent). A tempting inference says: since the only eigenvalue of A is zero, any Chebyshev transformation of A should equal the Chebyshev value at zero times a projector pattern, so the degree-two singular value transformation of A should be the degree-two Chebyshev value at zero, namely minus one, times the identity. Show by explicit computation that this inference is FALSE: compute the singular value decomposition of A, compute the degree-two singular value transformation, and exhibit the discrepancy. State the correct principle that resolves the paradox.
Hint: mathematical route
For A=[[0,1],[0,0]], the even singular-value transform T_2^SV(A)=2A†A-I=diag(-1,1), whereas the ordinary polynomial T_2(A)=2A^2-I=-I. Hermitian definite-parity agreement is sufficient; normality alone fails, as B=(i/2)I shows.
- The nilpotent A=[[0,1],[0,0]] has SVD W=I, Sigma=diag(1,0), V=X, since Sigma X=A.
- The scalar polynomial T_2(x)=2x^2-1 gives T_2(Sigma)=diag(1,-1).
- The stipulated right-even transform is X diag(1,-1)X†=diag(-1,1)=2A†A-I.
- The ordinary polynomial equals T_2(A)=2A^2-I=-I because A^2=0; this is not a general rule determining defective-matrix functions from their eigenvalue list.
- The actual singular-value output has a+1 second diagonal entry, so it differs from the proposed -I output.
- Definite-parity SVT agrees with spectral evaluation for Hermitian matrices, with unit-modulus sign completion at zero; normality alone does not ensure agreement, since B=(i/2)I has T_2^SV(B)=-I/2 but T_2(B)=-3I/2.
Reference solution
Use the ordered standard basis e_1,e_2. The operator is A=[[0,1],[0,0]], so A^2=0 and both eigenvalues are zero. A singular value decomposition is A=W Sigma V† with W=I_2, Sigma=diag(1,0), and V=X=[[0,1],[1,0]]. Indeed Sigma X=A. Equivalently A†A=diag(0,1), so the right singular vector for singular value1 is e_2 and that for singular value0 is e_1.
The stipulated even singular-value transformation uses the right singular vectors on both sides. Since T_2(x)=2x^2-1, T_2(Sigma)=diag(1,-1). Therefore T_2^SV(A)=V T_2(Sigma)V†=X diag(1,-1)X=diag(-1,1). Its second diagonal entry is+1, so it is not -I_2. The same result follows directly from the even-transform identity T_2^SV(A)=2A†A-I_2.
For comparison, the ordinary polynomial evaluated on A is T_2(A)=2A^2-I_2=-I_2. Thus the alleged output -I_2 is the ordinary polynomial in this particular example, not the output of the specified singular-value transform. One must not generally determine a function of a defective matrix merely from its eigenvalue list: this A is not diagonalizable, and for example T_1(A)=A is nonzero even though T_1(0)=0. Here the ordinary degree-two result happens to reduce to -I because A^2=0. The resolution is that an even singular-value transformation is V f(Sigma)V†, whereas ordinary polynomial evaluation forms f(A); these are different operations for general A.
A sufficient agreement statement is: for a Hermitian H and a function of definite parity, the parity-defined singular-value transformation agrees with the spectral function f(H). To see the sign boundary correctly, write H=Q D Q†, Sigma=|D| and choose a diagonal unitary J with J_jj=sign(D_jj) at nonzero eigenvalues and J_jj=+1 at zero. Then H=(QJ)Sigma Q† is an SVD with unitary left factor. For even f, f(|D|)=f(D), so the right transformation agrees. For odd f, f(0)=0 and J f(|D|)=f(D), so the balanced transformation agrees too. No zero-eigenvalue division or nonunitary sign(0)=0 factor is needed.
Normality alone is not such a sufficient condition. For the normal contraction B=(i/2)I_2, the singular values are1/2 and T_2^SV(B)=(-1/2)I_2, while the ordinary polynomial T_2(B)=2B^2-I_2=(-3/2)I_2. These differ in magnitude, not just a sign or global phase. General normal matrices may have complex eigenphases, whereas Hermitian eigenvalues have only real signs; definite parity handles those real signs, not arbitrary complex phases. Conversely normality is not necessary for every individual coincidence, since the odd degree-one transform equals A for any square A. The relevant principle for the question is therefore the precise right-even singular-value formula, which produces diag(-1,1), not a blanket rule equating eigenvalue and singular-value transformations for normal matrices.
These are exact matrix identities and the projected-block meaning of the stated transformation. No deterministic normalized-state preparation, gate compilation, query bound, or approximation-error guarantee is inferred from the matrix calculation.
Assessment rubric
- Exhibits and verifies A=I_2 diag(1,0)X†, with singular values1 and0 and X as the right-singular-vector matrix. (1 weight)
- Computes T_2(Sigma)=diag(1,-1) from T_2(1)=1 and T_2(0)=-1, using the stipulated even singular-value convention. (1 weight)
- Pulls back by the right singular vectors to obtain diag(-1,1), explicitly differing from -I_2; distinguishes this from the ordinary polynomial T_2(A)=2A^2-I_2=-I_2. (1 weight)
- Resolves the paradox by distinguishing singular-value/right-vector transformation from ordinary polynomial evaluation, without inferring general defective-matrix functions from eigenvalues alone or asserting that normality alone ensures agreement. (1 weight)
Graph-grounded prerequisites
These are support/source anchors, not an automatically extracted theorem-dependency proof.
- Matrix analysis (frozen curriculum anchor; see graph registry)
- Operator and semigroup theory (frozen curriculum anchor; see graph registry)
- Approximation theory (frozen curriculum anchor; see graph registry)
Lean view and evidence boundary
No aligned Lean view has been admitted for this native core in this textbook checkpoint. This is not a claim that the underlying mathematics lacks a formal proof. Contributions must preserve this exact target contract and split.
Sources and provenance
- website/research/sources.json
lin-wiebe-2026-qasc
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Item digest: a973d61b0a43ba5ce483b75042576d8cffdfac75f5b85302df219bb85102afdf
Graph snapshot: qcb-snapshot-ffa644f0da3031a1ce754b9a
Authored support steps: 6. This is not measured graph distance or model difficulty.
Answer revision lineage
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