Question

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 $(1,\omega ,{\omega }^2)$ and $(1,{\omega }^2,\omega )$. 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

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 $\omega \ne 1$, 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 $P_1=B{B}^{+}(B{B}^{+}{)}^{-1}B$ 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.