Matrix A produces orthonomal basis of eigenvectors. Show that $\lambda_1||v||^2\le Av\cdot v ≤ \lambda_n||v||^2 $ for each $v \in\mathbb R^n$

Question

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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 $v \in\mathbb R^n$

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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.

$\Rightarrow \lambda_1 ||v||^2\le Av\cdot v\le\lambda_n||v||^2$

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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 :)

Answer 1

I think this proof is sufficient...

$A$ forms an orthonormal basis of eigenvectors

$\Rightarrow v=\alpha_1 v_1+...+\alpha_n v_n$ where $a_k\in\mathbb R$ and $v_k$ is an eigen vector of $A$.

$\Rightarrow Av=\alpha_1 Av_1+...+\alpha_n Av_n$ using the fact that $v_1,...,v_n$ are eigen vectors.

$\Rightarrow Av=\alpha_1\lambda_1 v_1+...+\alpha_n\lambda_n v_n$ $\Rightarrow Av\cdot v=\alpha_1\lambda_1 v_1\cdot v+...+\alpha_n\lambda_n v_n\cdot v$.

Note: $v\cdot v_k=\alpha_1 v_1 v_k+...+\alpha_k v_k v_k+...+\alpha_n v_n v_k$ since the eigen vectors are orthnormal $\Rightarrow v\cdot v_k=\alpha_k||v_k||^2=\alpha_k$.

So...

$Av\cdot v=\alpha_1^2\lambda_1 +...+\alpha_n^2\lambda_n$

We know that $v\cdot v=||v||^2=\alpha_1 v_1\cdot v+...+\alpha_n v_n \cdot v=\alpha_1^2+...+\alpha_n^2$

$\Rightarrow \lambda_1||v||^2=\lambda_1\alpha_1^2+...+\lambda_1\alpha_n^2\space$ likewise with $\lambda_n||v||^2=\lambda_n\alpha_1^2+...+\lambda_n\alpha_n^2$

$\Rightarrow Av\cdot v-\lambda_n||v||^2=\alpha_1^2(\lambda_1-\lambda_n)+...+\alpha_{n-1}^2(\lambda_{n-1}-\lambda_n)$

We know that: $(\lambda_1-\lambda_n),...,(\lambda_{n-1}-\lambda_n)\le 0$

$\Rightarrow Av\cdot v-\lambda_n||v||^2\le 0 \Leftrightarrow Av\cdot v\le\lambda_n||v||^2$

Similarly...

$\lambda_1||v||^2-Av\cdot v=\alpha_2^2(\lambda_1-\lambda_2)+...+\alpha_n^2(\lambda_1-\lambda_n)$

We know that: $(\lambda_1-\lambda_2),...,(\lambda_{1}-\lambda_n)\le 0$

$\Rightarrow \lambda_1||v||^2-Av\cdot v\le 0$

$\Rightarrow \lambda_1||v||^2\le Av\cdot v$

Hence $\lambda_1||v||^2\le Av\cdot v\le \lambda_n||v||^2$

Answer 2

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 $\lambda_k$.