A11_soln - Math 235 Assignment 11 Solutions 1+i 1i 1....

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Math 235 Assignment 11 Solutions 1. Consider C 3 with its standard inner product. Let ~ z = 1 + i 2 - i - 1 + i , ~w = 1 - i - 2 - 3 i - 1 . a) Evaluate h ~ z, ~w i and h ~w, 2 i~ z i . Solution: We have h ~ z, ~w i = (1 + i )(1 + i ) + (2 - i )( - 2 + 3 i ) + ( - 1 + i )( - 1) = 2 i - 1 + 8 i + 1 - i = 9 i h ~w, 2 i~ z i = 2 i h ~ z, ~w i = - 2 i ( - 9 i ) = - 18 b) Find a vector in span { ~ z, ~w } that is orthogonal to ~ z . Solution: A vector orthogonal to ~ z in span { ~ z, ~w } is ~v = perp ~ z ~w = ~w - h ~w,~ z i k ~ z k 2 ~ z = 1 - i - 2 - 3 i - 1 + 9 i 9 1 + i 2 - i - 1 + i = 0 1 - i - 2 - i c) Write the formula for the projection of ~u onto S = span { ~ z, ~w } . Solution: Note that span { ~ z, ~w } = span { ~ z,~v } where ~v is the vector from part b) and { ~ z,~v } is orthogonal over C so we have proj S ~u = < ~u,~ z > k ~ z k 2 ~ z + < ~u,~v > k ~v k 2 ~v 2. Let V be an inner product space, with complex inner product h , i . Prove that if < ~u,~v > = 0, then k ~u + ~v k 2 = k ~u k 2 + k ~v k 2 . Is the converse true? Solution: If < ~u,~v > = 0, then we have k ~u + ~v k 2 = < ~u + ~v,~u + ~v > = < ~u,~u + ~v > + < ~v,~u + ~v > = < ~u,~u > + < ~u,~v > + < ~v,~u > + < ~v,~v > = k ~u k 2 + 0 + 0 + k ~v k 2 = k ~u k 2 + k ~v k 2 The converse is not true. One counter example is: Consider V = C with its standard inner product and let ~u = 1 + i and ~v = 1 - i . Then k ~u + ~v k 2 = k 2 k 2 = 4 and k ~u k 2 + k ~v k 2 = 2 + 2 = 4, but < ~u,~v > = (1 + i )(1 + i ) = 2 i 6 = 0. 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
2 3. Prove that for any n × n matrix A , we have det A = det A . Solution: We prove this by induction. If A is a 1 × 1 matrix, then the result is obvious. Assume the result holds for n - 1 × n - 1 matrices and consider an n × n matrix A . If we expand det A along the first row, we get by definition of the determinant det A = n X i =1 a 1 i C 1 i ( A ) where C 1 i ( A ) represents the cofactors of A . But, each of these cofactors is the determinant of an n - 1 × n - 1 matrix, so we have by our inductive hypothesis that C 1 i ( A ) = C 1 i ( A ). Hence,
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 6

A11_soln - Math 235 Assignment 11 Solutions 1+i 1i 1....

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online