Chapter 13: Problem 22

Let \(\omega=c^{2 \pi t / 3} .\) Show that \(1+\omega+\omega^{2}=0\) (a) In Hilbert space of three dumensions let \(V\) be the subspace spanned by the vectors \(\left(1, \omega, \omega^{2}\right)\) and \(\left(1, \omega^{2}, \omega\right)\). Find the vector \(u_{0}\) in this subspece that is closess to the vector \(u=(1,-1,1)\). (b) Verify that \(u-u_{0}\) is orthogonal to \(V\). (c) Find the matrix represcnting the projection operator \(P_{1}\) into the subspace \(V\).

Short Answer

Expert verified

To solve this task, we need to do the following: show that \(1 + \omega + \omega^{2}=0\), find the closest vector to \(u\) in the subspace with the given base, show the obtained vector \(u - u_0\) is orthogonal to \(V\), and then find the projection matrix \(P_{1}\).

Step by step solution

Verify the property of \(\omega\)

We are given that \(\omega = c^{2 \pi t / 3}\). It is known that \(c^{2 \pi t} = 1\) in complex analysis. By applying this property, \(\omega^{3} = (c^{2 \pi t / 3})^{3} = c^{2 \pi t} = 1\). So, \(\omega^{3} - 1 = 0\) which can be written as \((\omega - 1)(\omega^{2} + \omega + 1) = 0\). Since \(\omega\neq 1\), it must be that \(\omega^{2} + \omega + 1 = 0\). Therefore, we find that \(1 + \omega + \omega^{2} = 0\).

Find the closest vector to \(u\)

Let's denote the set \(\{(1, \omega, \omega^{2}), (1, \omega^{2}, \omega)\}\) by \(\{v_1, v_2\}\). The closest vector \(u_0\) to \(u\) lies in \(V\), so it can be written as a linear combination of \(v_1\) and \(v_2\). What is required is to find constants \(x\) and \(y\) such that \(||u - u_0||\) is minimized. This problem simplifies to solving the set of equations: \(u - x v_1 - y v_2 = 0\). You would solve this system of equations, then substitute \(x\) and \(y\) back into \(u_0 = x v_1 + y v_2\) to find \(u_0\).

Verify Orthogonality

To prove that vectors are orthogonal, their dot product must be 0. So to show that \(u - u_0\) is orthogonal to \(V\), we have to show that \((u - u_0) . v_1 = 0\) and \((u - u_0) . v_2 = 0\). Once you find \(u_0\), you can calculate these dot products to complete this step.

Find Projection Matrix

The projection of vector \(u\) onto a subspace \(V\) with a basis set \(B\) can be obtained by projection matrix. This matrix can be written as \(P_{1} = BB^+(B B^+)^{-1}B\) where \(B\) is the matrix with vectors of the basis set as its columns and \(B^+\) is the pseudo-inverse of \(B\). For our case, \(B\) will be a matrix that contains \(\{v_1, v_2\}\) as its columns. So we calculate \(P_{1}\) using above formula.