Let $U_1$ and $U_2$ be two independent uniform random variables on $[0,1]$. Suppose that $X= (X_1,X_2,X_3,X_4)′$where $X_1=U_1$, $X_2=U_2$, $X_3=U_1+U_2$ and $X_4=U_1−U_2$.
(a) Compute the variance covariance matrix $\Sigma$ of $X$.
(b) Verify that $v_1=\frac{1}{\sqrt{3}}(0,1,1,−1)$, $v_2=\frac{1}{\sqrt{3}}(1,0,1,1)$, $v_3=\frac{1}{\sqrt{6}}(0,2,−1,1)$, and $v_4=\frac{1}{\sqrt{6}}(2,0,−1,−1)$ are orthonormal eigenvectors of $\Sigma$. Find out the corresponding eigenvalues.
We have that $$\Sigma=\begin{bmatrix}\frac{1}{12}&0&\frac{1}{12}&\frac{1}{12}\\0&\frac{1}{12}&\frac{1}{12}&-\frac{1}{12}\\ \frac{1}{12}&\frac{1}{12}&\frac{1}{6}&0\\ \frac{1}{12}&-\frac{1}{12}&0&\frac{1}{6}\end{bmatrix}$$
since $\mathsf{Var}(X_1)=\mathsf{Var}(X_2)=\mathsf{Var}(U_1)=\frac{1}{12}$, $\mathsf{Var}(X_3)=\mathsf{Var}(U_1+U_2)=\frac{1}{6}$, $\mathsf{Cov}(X_1,X_2)=0$, $\mathsf{Cov}(X_1,X_3)=\mathsf{Cov}(U_1,U_1+U_2)=\mathsf{Cov}(U_1,U_1)+\mathsf{Cov}(U_1,U_2)=\mathsf{Var}(U_1)=\frac{1}{12}$, $\mathsf{Cov}(X_1,X_4)=\mathsf{Cov}(U_1,U_1-U_2)=\mathsf{Cov}(U_1,U_1)-\mathsf{Cov}(U_1,U_2)=\mathsf{Var}(U_1)=\frac{1}{12}$, $\mathsf{Cov}(X_2,X_3)=\mathsf{Cov}(U_2,U_1+U_2)=\mathsf{Cov}(U_1,U_2)+\mathsf{Cov}(U_2,U_2)=\mathsf{Var}(U_2)=\frac{1}{12}$, $\mathsf{Cov}(X_2,X_4)=\mathsf{Cov}(U_2,U_1-U_2)=\mathsf{Cov}(U_1,U_2)-\mathsf{Cov}(U_2,U_2)=-\mathsf{Var}(U_2)=-\frac{1}{12}$, and $\mathsf{Cov}(X_3,X_4)=\mathsf{Cov}(U_1+U_2,U_1-U_2)=\mathsf{Cov}(U_1,U_1)-\mathsf{Cov}(U_1,U_2)+\mathsf{Cov}(U_1,U_2)-\mathsf{Cov}(U_2,U_2)=\frac{1}{12}-\frac{1}{12}=0$
Software then gives that
$$\begin{align*} det(\Sigma-\lambda I) &=det\left(\begin{bmatrix}\frac{1}{12}-\lambda&0&\frac{1}{12}&\frac{1}{12}\\0&\frac{1}{12}-\lambda&\frac{1}{12}&-\frac{1}{12}\\ \frac{1}{12}&\frac{1}{12}&\frac{1}{6}-\lambda&0\\ \frac{1}{12}&-\frac{1}{12}&0&\frac{1}{6}-\lambda\end{bmatrix}\right)\\\\ &=det\left(\begin{pmatrix}\frac{1}{12}-\lambda&0&\frac{1}{12}&\frac{1}{12}\\ 0&\frac{1}{12}-\lambda&\frac{1}{12}&-\frac{1}{12}\\ 0&0&\frac{3\lambda\left(4\lambda-1\right)}{1-12\lambda}&0\\ 0&0&0&\frac{3\lambda\left(4\lambda-1\right)}{1-12\lambda}\end{pmatrix}\right)\\\\ &=\frac{\lambda^2\left(1-4\lambda\right)^2}{16} \end{align*}$$
Hence $\lambda_1=\frac{1}{4}$ and $\lambda_2=0$
We have that for $\lambda_2=0$,
$$\begin{bmatrix}\frac{1}{12}&0&\frac{1}{12}&\frac{1}{12}\\0&\frac{1}{12}&\frac{1}{12}&-\frac{1}{12}\\ \frac{1}{12}&\frac{1}{12}&\frac{1}{6}&0\\ \frac{1}{12}&-\frac{1}{12}&0&\frac{1}{6}\end{bmatrix}\begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix}=\begin{bmatrix}0\\0\\0\\0\end{bmatrix}$$
means $x_2=-x_3+x_4$ and $x_1=-x_3-x_4$. Letting $x_3=x_4=1$ we have that $x_2=0$ and $x_1=-2$
Hence the associated eigenvector is $\frac{1}{\sqrt{6}}(-2,0,1,1)$ which is equivalent to $v_4$ which was provided.
Similarly, we find that for $\lambda_1=\frac{1}{4}$ we get an associated eigenvector of $\frac{1}{\sqrt{3}}(1,0,1,1)$ which is $v_2$.
Where do $v_1$ and $v_3$ come from?