Linear

# Algebra 2 [Lecture notes] by Jan Nekovar

By Jan Nekovar

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4) Conversely, 2u1 = x1 − x2 , 2u2 = x2 − x1 . 3) Cubic equations. , by replacing x + a/3 by x. 2), which gives 0 = (u + v)3 + p(u + v) + q = u3 + v 3 + (3uv + p)(u + v) + q. 2) simplifies as u3 + v 3 + q = 0, hence u6 + qu3 − p 3 3 v 6 + qv 3 − = 0, p 3 3 = 0. 3) are equal to u3 et v 3 (of course, T1 T2 = (−p/3)3 = u3 v 3 ). 1 are the cubic roots of −q. 4) hence 3u1 = x1 + ρ2 x2 + ρx3 , 3v1 = x1 + ρx2 + ρ2 x3 . 4) Quartic equations. Let us consider a general quartic equation x4 + ax3 + bx2 + cx + d = 0.

In this section we recall the classical approach to solving equations of degree n ≤ 4. After that we reformulate it using Lagrange’s idea of resolvents and show that this method does not permit to solve equations of degree n ≥ 5. The history of the subject can be found in [Ti 2]. 2) Quadratic equations. 1) is solved by completing the square: p 2 Equivalently, one can write x = u + v, which yields 0= x+ 2 +q− p 2 2 . 0 = (u + v)2 + p(u + v) + q = u2 + u(2v + p) + v 2 + pv + q. 2) simplifies as p 2 p p 2 +p − + q = u2 + q − .

The polynomials sk = xk1 + · · · + xkn satisfy recursive relations sk − σ1 sk−1 + · · · + (−1)k−1 σk−1 s1 + (−1)k kσk = 0 (k ≥ 1) (of course, σk = 0 for k > n). 11) Discriminant. Let n ≥ 2. The polynomial (xi − xj ) ∈ Z[x1 , . . , xn ] ∆ := i