Showing a matrix is invertible.

Question

Let $x = \begin{bmatrix} 1 \\ x_2 \\ x_3 \end{bmatrix}$ where $x_3 = \alpha_1 + \alpha_2x_2$ is a linear function of $x_2$. Show that $Q_{xx} = E[xx']$ is not invertible.

I multiplied out the $xx'$ to get the matrix $xx' = \begin{bmatrix} 1 & x_2 & x_3 \\ x_2 & x_2^2 & x_2x_3 \\ x_3 & x_3x_2 & x_3^2 \end{bmatrix}$. Now I've played around and substituted in the expression for $x_3$ and tried taking expectations. But I am still unsure of how to show it is not invertible.

Answer 1

\begin{align} \begin{bmatrix} 1 & \mathbb{E}[X_2] & \mathbb{E}[X_3] \\\mathbb{E}[X_2] & \mathbb{E}[X_2^2] & \mathbb{E}[X_2X_3]\\ \mathbb{E}[X_3] & \mathbb{E}[X_2X_3] &\mathbb{E}[X_3^2]\end{bmatrix} &= \begin{bmatrix} 1 & \mathbb{E}[X_2] & \alpha_1+\alpha_2\mathbb{E}[X_2] \\\mathbb{E}[X_2] & \mathbb{E}[X_2^2] & \alpha_1\mathbb{E}[X_2]+\alpha_2\mathbb{E}[X_2^2]\\ \alpha_1+\alpha_2\mathbb{E}[X_2] & \alpha_1\mathbb{E}[X_2]+\alpha_2\mathbb{E}[X_2^2] &\mathbb{E}[(\alpha_1+\alpha_2X_2)^2]\end{bmatrix} \end{align}

Check that $\alpha_1$ multiplied to the first column plus $\alpha_2$ multiplied to the second column gives you the third column.