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For a linear operator on a finite-dimensional vector space over a field $K$, all polynomial kernels stabilize at their first power exactly when its minimal polynomial $m$ has no repeated irreducible polynomial factor in $K[X]$. This formulation does not require those factors to be linear.

To prove it, the Bézout's identity for polynomials gives $\ker q(T)=\ker \gcd (q,m)(T)$: a greatest common divisor is a combination of $q$ and $m$, while $q$ is its multiple. If $m$ is square-free, $\gcd (p^n,m)=\gcd (p,m)$, proving the displayed equality. Conversely, if an irreducible factor $f$ occurs to exponent $e\ge 2$, its primary decomposition theorem summand has minimal polynomial $f^e$. On this summand $f(T)$ is nonzero, while $f(T{)}^e=0$. Thus the kernels of those two powers differ. An eigenbasis follows from this criterion precisely when the distinct factors of $m$ are also linear over $K$; a real rotation through a right angle has square-free $m=X^2+1$ but no real eigenvalues.

The sudden approximation treats a rapid, finite change of a Hamiltonian operator as leaving the quantum state approximately unchanged during the switch. The energy measurement can nevertheless change immediately: the same quantum state has different probability amplitudes in the new eigenbasis. Compute their inner products with the new energy eigenstates, then apply the Born rule. The two-level sudden quench below makes this distinction quantitative without assuming the change is small.

An ideal finite step has exact quantum state continuity at a finite Hamiltonian step; a finite-duration implementation has a small evolution error under the bounded-model conditions below. An impulsive quantum kick is different: its Hamiltonian operator grows as its duration shrinks, retaining a finite integrated action and a finite change of quantum state. A sudden approximation therefore concerns short evolution during switching, not conservation of energy during externally driven change. The subsequent unitary time evolution can be substantial.^[ref]

Over a field of characteristic different from two, three linear projections summing to the identity annihilate one another in both orders. Each pair sum is the complementary projection to the third. The sum of two linear projections criterion therefore proves zero products. Every vector is the sum of its three images; applying each projection to a zero sum proves that these images form a direct sum. Choosing a basis in each image gives a simultaneous eigenbasis, with triples of eigenvalues $(1,0,0)$, $(0,1,0)$ or $(0,0,1)$. This is algebraic orthogonality, and does not assume that the projections are orthogonal projections for a specified inner product.

Take an isolated two-level system, or a spin one-half qubit, with Pauli matrices ${\sigma }_x,{\sigma }_z$. Let ${\sigma }_z|0⟩=|0⟩$ and ${\sigma }_z|1⟩=-|1⟩$. The initial and final Hamiltonian operators areThe energy gaps are ${\Delta }_i,{\Delta }_f$. Prepare the initial ground state $|0⟩$, with energy eigenvalue $-{\Delta }_i/2$. A final orthonormal eigenbasis isDirect matrix multiplication gives $H_f|g_f⟩=-{\Delta }_f|g_f⟩/2$ and $H_f|e_f⟩=+{\Delta }_f|e_f⟩/2$. The unchanged initial quantum state is $|0⟩=c|g_f⟩-s|e_f⟩$. Its new probability amplitudes are the inner products $c,-s$; the Born rule gives $p_g=c^2,p_e=s^2$. The angle between the real amplitude vectors is $\theta /2$, whereas the Hamiltonian axis on the Bloch sphere rotates through $\theta$.

At $\theta =\pi /3$, $c=\sqrt{3}/2$ and $s=1/2$, so the final energy measurement gives $-{\Delta }_f/2$ with probability $3/4$ and $+{\Delta }_f/2$ with probability $1/4$. At $\theta =0$ the initial ground state remains the final ground state; at $\theta =\pi$ it is the final excited energy eigenstate. These statements are about the new energy labels, not an instantaneous physical rotation of the quantum state.