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By Jan Nekovar
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2(1), r = 2 = dim(E), hence N (q) = 0. 2(2), r = 1 and x1 N (q) = 3 x1 − 3x2 = 0 = vect x2 . 2(3), r = 2 and x1 N (q) = x2 x1 + 2x2 − 3x3 = x2 − 7x3 = 0 −11 . 2(4), r = 3 = dim(E), hence N (q) = 0. 5) If the number of non-zero terms r = rk(q) in q(x) = d1 x12 + · · · + dr xr2 is less than n, how do we choose the remaining new coordinates xr+1 , . . , xn in order to make sure that the linear forms x1 , . . , that the matrix P in the formula X = P X is invertible)? For example, if the terms x1 , .
3) Proposition. Let q : E −→ K be a non-degenerate quadratic form on a vector space of dimension n = dim(E) < ∞. If F ⊂ E is a vector subspace, then dim(F ⊥ ) = n − dim(F ), 47 (F ⊥ )⊥ = F. Proof. It is enough to prove the equality dim(F ⊥ ) = n − dim(F ) for all subspaces of E (if we apply it to F and F ⊥ , we obtain dim (F ⊥ )⊥ = n − dim(F ⊥ ) = n − (n − dim(F )) = dim(F ); as F ⊂ (F ⊥ )⊥ by definition, we must have F = (F ⊥ )⊥ ). Fixing a basis of E, we identify E with K n and the polar form f of q with the function f (x, y) = txAy (x, y ∈ K n ).
Deduce that an isometry f : E −→ E is an orthogonal symmetry ⇐⇒ f 2 = Id. 12) Exercise. In this exercise, Rn is equipped with the standard Euclidean scalar product. (1) Show that a matrix S ∈ Mn (R) represents an orthogonal symmetry of Rn ⇐⇒ tS = S and S 2 = I. (2) Show that a matrix P ∈ Mn (R) represents an orthogonal projection on a subspace of Rn ⇐⇒ tP = P and P 2 = P . 5 Isometries of R2 In this section, E = R2 is equipped with the standard Euclidean scalar product (x | y) = x1 y1 + x2 y2 .