Question
Assume that $A \in \mathbb M^{n×n}(\mathbb R)$ admits an orthonormal basis of
eigenvectors with eigenvalues $\lambda_1 \le \lambda_2 \le … \le \lambda_n$. Show that $\lambda_1||v||^2\le Av\cdot v ≤ \lambda_n||v||^2 $ for each $v \in\mathbb R^n$
My attempt
Since $\lambda_1 \le \lambda_2 \le … \le \lambda \le …\le \lambda_n$ Where $\lambda$ is the eigenvalue for $v$ $\Rightarrow \lambda_1\le\lambda\le\lambda_n$ $\Leftrightarrow \lambda_1 ||v||^2\le\lambda||v||^2\le\lambda_n||v||^2$.
$\lambda||v||^2=\lambda(v\cdot v)=Av\cdot v $ since $\lambda$ is an eigen value.
$\Rightarrow \lambda_1 ||v||^2\le Av\cdot v\le\lambda_n||v||^2$
Thoughts
I believe this was too easy; this question is the last question on my linear algebra worksheet and I have not used the fact that A admits an orthnormal basis of eigen vectors. Any ideas or straight up proofs would be really appreciated 🙂
Best Answer
It is wrong because the satement says “for each $v\in\mathbb R^n$” and you assumed that $v$ is an eigenvector.
Simply use that fact that any $v\in\mathbb R^n$ can be written as $\alpha_1v_1+\alpha_2v_2+\cdots+\alpha_nv_n$ where $v_k$ is an eigenvector corresponding to the eigenvalue $\lambda_k$.