Question

10. (3 marks) Let A= | 1 /1 41 ( x y ): *). For what values of x and y does (u, v) = ut Av define an inner product on R2?

Answer 1

DEFINITION: Let V be a real vector space. Suppose to each pair of vectors u, v ∈ V there is assigned a real number, denoted by ⟨u, v⟩. This function is called a (real) inner product on V if it
satisfies the following axioms:
(1)(Linear Property): ⟨au + bv, w⟩ = a⟨u, w⟩ + b⟨u, w⟩.
(2) (Symmetric Property): ⟨u, v⟩ = ⟨v, u⟩.
(3)(Positive Definite Property): ⟨u, u⟩ ≥ 0.; and ⟨u, u⟩ = 0 if and only if u = 0.
The vector space V with an inner product is called a (real) inner product space.

Let V be a real inner product space with basis S = {u​​​​​​1, u​​​​​​2, ... , u​​​​​​n}. The matrix
A = [a​​​​​​ij], where a​​​​​​ij = ⟨ui, uj⟩
is called the matrix representation of the inner product on V relative to the basis S. Observe that A is symmetric, because the inner product is symmetric; that is, ⟨ui, uj⟩ = ⟨uj, ui⟩.

efinite matrix wi, wi) for any for so Theorem. Let A be the matrix representation of any inner product on V. Then A is a positive definite matrix. Because (wi, w;) = (W;, W;) for any basis vectors W; and w;, the matrix A is symmetric. Let X be any nonzero vector in R". Then (u) = X for some nonzero vector u EV. Theorem 7.16 tells us that x"AX = [u] A[u] = (u, u) > 0. Thus, A is positive definite.

A Theorem A = is positive definite if and only if a and d are positive and = ad – b2 is positive. 6 cl Let u = (x,y)". Then f(u) = u Au = [x, y) = ax + 2bxy + dy? Ily] Suppose f(u) >0 for every u +0. Then f(1,0) = a > 0) and f(0,1) = d > 0. Also, we have f(b, -a) = a(ad – 62) > 0. Because a > 0, we get ad – b2 > 0. Conversely, suppose a > 0, b = 0, ad – b2 > 0. Completing the square gives us 2bb2 2 ad - 52 f(u) = a (x2 + 20 xy + y2) + dy? – + a a / ap Accordingly, f(u) > 0 for every u 70.

Hence (u, v) = uTAv is an inner product if A is symmetric and positive definite.

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