Diagonalize the quadratic form $Q(x_1,\dots,x_n)=\sum_{i=1}^n a_ix_i^2+(\sum_{i=1}^n b_ix_i)^2$

Question

In my linear algebra class, I happened to meet an interesting question which asks me to diagonalize the quadratic form $$Q(x_1,\dots,x_n)=\sum_{i=1}^n a_ix_i^2+(\sum_{i=1}^n b_ix_i)^2,$$ where $a_i,b_i$ are real numbers. So I need to find an $n\times n$ invertiable matrix $C$ such that the change of variables $(x_1,\dots,x_n)^T=C(y_1,\dots,y_n)^T$ transforms the quadratic form $Q(x_1,\dots,x_n)$ into a diagonal form $Q(y_1,\dots,y_n)=\sum_{i=1}^n \lambda_iy_i^2$. But I have no idea how to find such an invertiable matrix?

Answer 1

The insight is to see that $Q$ can be written as a quadratic form, i.e., $$Q(x_1,\dots,x_n) = \sum_{i=1}^n a_ix_i^2+(\sum_{i=1}^n b_ix_i)^2 = x^T \text{diag}(a) x + x^T (bb^T) x = x^T \left(\text{diag}(a)+bb^T\right) x.$$

If you let $A = \text{diag}(a)+bb^T$, then since $A$ is symmetric, $A$ can be eigen-decomposed as $A = U \Lambda U^T$, where $\Lambda$ is a diagonal matrix. Hence

$$Q(x) = x^T A x = (Ux)^T \Lambda (Ux)$$

Let $y = Ux$, and you get the required form.