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24 Nov 2023: If A is the matrix of the first of these (say), find a non-singular matrix P such that PT AP is diagonalwith entries 1. ... an be real numbers such that a1 an = 0 and a21 a2n = 1.

8 Nov 2023: 12. (a) Show that if x ̸= y are vectors in the finite dimensional vector space V , then there is a linear func-tional θ V such that θ(x) ̸= ... Let A, B V. Prove that A B if and only if Ao Bo.Show that A = V if and only if Ao = {0}.

23 Oct 2023: Show that:. (a) There is a nondeterministic automaton N with n 1 states such that L(N) = Ln and(b) if D is a deterministic automaton with fewer than 2n states, ... Prove that there is a proper excerpt e of v such that xey L.

28 Nov 2023: Operator-valued measures. A resolution of the identity of H over K is amap P : B B(H) such that. ... over K. Then there is an isometric, unital -isomorphism Φ of L(P) onto acommutative, unital C-subalgebra of B(H) such that.

11 Mar 2024: Proof 2: more words. Let ϵ > 0. Choose N N such that N > 1/ϵ2. ... f1 When f : A B is a function, and C is a subset of B, thesymbol f1(C) denotes the preimage of the set C: the set ofelements a A such

23 May 2024: Tho15, Lemma 2.8]):. Lemma 2.11. There exists a non-zero N0 OF such that for all primes p and for all b B(Op) wehave N0 κb V (Op). ... There exists b′ B(Op) such that ordp (b tb′) = 1,where t is a uniformiser of Op.

29 Sep 2023: The discriminant = (a, b)=16(4a327b2)is non-zero. Conversely, for any pair (a, b) O2K such that (a, b) , 0, the equation(2.1) defines an elliptic curve. ... ap bp = cp. to the Fermat equation in exponent p; thus a, b, c Z are coprime non-zero

10 Oct 2023: Suppose that α has minimal polynomial f over Q, and let β = α4.Find a, b, c Q such that β = a bα cα2. ... 9. Let K and L be subfields of a field M such that M/K is finite.

8 Mar 2024: αd OK such that. OK = α1Z. αdZ.Such a collection of elements α1,. , ... 14 PÉTER P. VARJÚ. of this form. So no such π1 may exist.

14 Sep 2023: If χ: A B is an algebra homomorphismthen there is b B such that χ(x) = bxb1 for all x A. ... Evidently such a pair is also an element of AutA(M/X/G). (b) Given a group homomorphism σ : G AutA(M/X/G) such that pr1 σ =1G, the restriction

28 Nov 2023: ii) If A is compact and B is closed, then there exists f X such that supA f <infB f. ... An algebra norm on A is a norm ‖‖ on A such that ‖ab‖ 6 ‖a‖‖b‖ for alla,b A.

29 Nov 2023: 10. )〉. So. R2 =〈(. 10. )〉〈(. 01. )〉is the only such decomposition in this case. ... Ws such that W = sj=1Wj. Thus. V =ri=1. Ui sj=1.

31 Jan 2024: G = H1 BH2 BH3 BH4 B BHn = {e}. such that each quotient Hi1/Hi is simple. ... Since they have the same cycle type, there is a σ Sn such that (a b c) = σ(1 2 3)σ1.

25 Oct 2023: Then there is a point y / E such. that B(y,r) E 6= whenever r > 0. ... Since Uj1 isopen we can find a δj1 > 0 with δj1 δj/4 such that B(xj1,δj1) Uj1.

20 Dec 2023: Proposition 2.1.1. ([25, Exercise 2F]) For every ring homomorphism φ: A Band element b B such that bφ(a)φ(a)b = φ(δ(a)) for all a A, ... there is a uniquering homomorphism ψ : A[T; δ] B such that ψ|A = φ and ψ(T) = b.

25 Oct 2023: Theorem 4.2. If f : [a,b] [a,b] is a continuous function, then we canfind a w [a,b] such that f(w) = w. ... Supposethat we can find a a0 < a1 < < an b such that, writing σ = ‖fP‖we have either.

24 Oct 2023: Vn such that V ' V1 Vn. Theorem 4.12. Suppose that G is finite and |G| 6= 0 in F. ... central character, such that z v = λ(z) v forall z Z(G) and v V.

31 Jan 2024: Theorem. There is no group homomorphism F : {0} Z such that the composition. ... such that. f g ' IdY f ′ g′ ' IdZ g f ' IdX g′ f ′ ' IdY.

29 Jan 2024: 4Perhaps the Revelations are not flagged as such? Perhaps checking that something isa revelation involves establishing its independence from earlier revelation? ... Ramsey is such a theorem; Van der Waerdenisn’t! vdW does not prove itself in that way;

24 Nov 2023: Let a,b K such that the polynomials f = Xm a, g = Xm b are irreducible. ... Use asuitable Lagrange resolvent to find a Q(ζ3) such that K = Q(ζ3, 3.