Let A be a 3 x 3 orthogonal matrix with column vectors C, C2, C3. Use properties of orthogonal matrices and the inner product to prove that the vector v = c +2c2 + 2c3 has length ||v|| = 3. Provide references to theorems from the course notes for any such properties that you use in your proof. 15 Inner product and orthogonality If u = (u,..., un) and v = (v,..., Un) are vectors in R" then we define their inner product by Definition 15.1. u v = Uv + ... + Un Vn Notice that if we write u, v as 1 x n row matrices then u v = uv and if we write them as n x 1 column matrices then u. v = utv. Some important properties of the inner-product are Theorem 15.1. Let u, v, w ER" and a ER, then (a) u. v = v. u, (b) (u + v) wu.w+v.w, (c) (au) v a(u. v), (d) v v0, and v v=0 if and only if v = 0. Proof. The first three are straightforward to prove using the definition or the fact that u v = utv and known results from matrix multiplication. The third is also not hard but important so let us check it . V. V = VV + VV + + Un vn = (v) + (v) + ... + (vn) Being a sum of non-negative numbers this is equal to zero if and only if v = V = = Vn = 0. That is if and only if v = 0. In R and R the inner product is the dot product and the quantity v.v is the length of the vector v. So we make that definition in any dimension. Definition 15.2. If ve Rn we define the length of v to be ||v|| v, u Rn we define the distance between v and u to be ||vu||. 0 = arccos = Notice that a vector has length zero if and only if it is the zero vector. Recall that in R2 if we have two vectors v and u then v u = ||v||||u|| cos(0) where is the angle between them as in Figure 15.1. In Rn we don't really know what the angle between two vectors is but we do know v u, |v|| and ||u||. So we can try and use these to define the angle. Obviously we would like to define v.u ||v||||u|| v. v. If