use the spectral decomposition to show that $x^T A x \le \lambda _1 ||x||^2$

Question

use the spectral decomposition to show that $x^T A x \le \lambda _1 ||x||^2$

The spectral decomposition says that $A = \sum \lambda _i v_i v_i ^T$ Multiplying both sides by $x^T$ and $x$, we get $x^T A x = x^T \sum \lambda _i v_i v_i ^T x$ Where do I proceed from here?

Answer 1

The family $(v_i)$ is an orthonormal basis so $x=\sum_i x_i v_i$ and so

$$x^TAx=\sum_j\sum_i x_j\lambda_i v_j^Tv_iv_i^Tx=\sum_j\sum_i x_j\lambda_i \delta_{i,j}v_i^Tx=\sum_{i}x_i\lambda_iv_i^Tx\\=\sum_i\sum_jx_i\lambda_iv_i^Tx_jv_j=\sum_i x_i^2\lambda_i\le \lambda_1\sum_i x_i^2=\lambda_1\Vert x\Vert ^2$$ where $$\lambda_1=\max_i(\lambda_i)$$