(1) Let V be a R−vector space and < , > be an inner product. Prove that if {v1 , · · · , vn } is a set of mutually orthogonal non-zero vectors, then this set is also linearly independent. Proof: We are given (vi , vj ) = 0 if i = j and (vi , vi ) = 1. To prove that the set if linearly independent, we set a1 v1 + · · · + an vn = 0. Now taking inner product with vj on both sides, we get n ai (vi , vj ) = (0, vj ). Since inner products are linear in the first i=1 variable, we get n ai (v1 , vj ) = 0, and this gives aj = 0. Thus we are done. i=1 (2) Let V be a vector space and < , > be an inner product. Then show that (a) < 0, v >= 0 for any v ∈ V . Proof: < 0, v >=< 0 + 0, v >=< 0, v > + < 0, v >. Subtracting < 0, v > from both sides, we get < 0, v >= 0. (b) Show that for a fixed u ∈ V , < u, v >= 0 for any v ∈ V , then u = 0. Proof: Take v = u. Then we have < u, u >= 0, i.e., ||u||2 = 0. This implies that u = 0. (3) Let V be a vector space and < , > be an inner product and B = {v1 , · · · , vn } be an orthonormal basis for V . If c1 , · · · , cn ∈ K are scalars, then show that there is a unique v ∈ V such that < v, vi >= ci . Proof: Let v = n ci vi . Now taking inner product with vj on both sides, we get i=1 (v, vj ) = n ci (vi , vj ). Orthogonality implies that (vi , vj ) = 0 for i = j. Hence we get i=1 (v, vj ) = cj (vj , vj ). Since vj is non-zero (why?), we get ci = (v, vj )/||vj ||2 = (v, vj ). This proves existence. Uniqueness follows from the properties of a basis.
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MATH 4450 - HOME WORK 5
(4) Use the Gram-Schmidt orthogonalisation method to the vectors (1, 0, 1), (1, 0, −1) and (0, 3, 4) to obtain an orthonormal basis for R3 . Proof: Easy. (5) Let V be a vector space equipped with an inner product. Let W be subspace of V . Let W ⊥ be its orthogonal complement. Define a map π : V → V as π(v) = v1 where v = v1 + v2 with v1 ∈ W and v2 ∈ W ⊥ . Show that (a) The map π is well-defined, linear with range W and that π 2 = π. The map π is called the projection map onto the subspace W . (b) Define ν := I − π. Show that ν is the projection map onto W ⊥ i.e. show that ν is a linear map with range W ⊥ and that ν 2 = ν. Proof: (a) Let v ∈ V . Since V = W ⊕ W ⊥ , we see that there exist unique v1 ∈ W, v2 ∈ W ⊥ such that v = v1 + v2 . Thus π(v) = v1 is well-defined. (Note that if the direct sum above was only a sum, then the map is not well-defined in general, since we may have more than one choice of v1 in the above decomposition.) Range(π) = W since for any w ∈ W , we have w = w + 0, and so π(w) = w. Linearity is easily-checked, and so is π 2 = π. Proof of (b) is similar.
