Prove that $\lambda _1 ||v||^2 \leq Av \cdot v \leq \lambda _n ||v||^2$. Given that $A \in \mathbb{M}^{n \times n}(\mathbb{R})$ admits an orthonormal basis of eigenvectors with eigenvalues $\lambda _1 \leq \lambda _2 \leq \dots \leq \lambda _n$. Show that $\lambda _1 ||v||^2 \leq Av \cdot v \leq \lambda _n ||v||^2$ for each $v \in$ $\mathbb{R}^n$.
My suggested (wrong) solution:
We know that:
\begin{eqnarray*}
Av_i= λ_i e_i \quad \text{ for } \quad  i=1,2,…,n
\end{eqnarray*}
and
\begin{eqnarray*}
v= \sum^n_{i=1} α_i e_i,
\end{eqnarray*}
where $\alpha \in \mathbb{R}$ are scalars.
Thus from $Av⋅v$, we get
\begin{eqnarray*}
Av⋅v &=& 
A \sum^n_{i=1} α_i e_i ⋅\sum^n_{i=1} α_i e_i\\
&=& \sum^n_{i=1} \lambda_i α_i e_i ⋅\sum^n_{i=1} α_i e_i\\
&=& \sum^n_{i=1} \lambda_i \cdot \left(\sum^n_{i=1} α_i e_i\right)^2\\
&=& \sum^n_{i=1} \lambda_i \cdot \left(\sum^n_{i=1} α_i\right)^2 \cdot \left(\sum^n_{i=1} e_i\right)^2.
\end{eqnarray*}
Since the eigenvectors $e_i$ form an orthonormal basis, then $e_i⋅e_i=||e_i||^2=1$, which makes
\begin{eqnarray*}
\left(\sum^n_{i=1} e_i\right)^2 &=& 1 \\
\Rightarrow Av⋅v &=& \sum^n_{i=1} \lambda_i \cdot \left(\sum^n_{i=1} α_i\right)^2.
\end{eqnarray*}
And, 
\begin{eqnarray*}
||v||^2&=&v⋅v\\
&=& \sum^n_{i=1} \alpha_i e_i \cdot \sum^n_{i=1} \alpha_i e_i\\
&=& \left(\sum^n_{i=1} \alpha_i\right)^2 \cdot \left(\sum^n_{i=1} e_i\right)^2\\
&=&  \left(\sum^n_{i=1} \alpha_i\right)^2 .
\end{eqnarray*}
So I end up with from $λ_1 ||v||^2\leq Av⋅v\leq  λ_n ||v||^2$ to
\begin{eqnarray*}
\lambda_1 \left(\sum^n_{i=1} \alpha_i\right)^2 \leq \sum^n_{i=1} \lambda_i \cdot \left(\sum^n_{i=1} α_i\right)^2
\leq \lambda_n \left(\sum^n_{i=1} \alpha_i\right)^2 
\end{eqnarray*}
Factoring out the $\left(\Sigma_{i=1}^n \alpha\right)^2,$
\begin{eqnarray*}
\lambda_1 \leq \sum^n_{i=1} \lambda_i \leq \lambda_n,
\end{eqnarray*}
which is obviously wrong. Can someone please point out where my mistake is?
 A: Your error comes from the fact that in general :
$$\left(\sum^n_{i=1} α_i e_i\right) \ne 
\left(\sum^n_{i=1} α_i\right) \cdot \left(\sum^n_{i=1} e_i\right)$$
Besides, the inequalities you want to prove are a property of "Rayleigh quotient" 
$$\lambda_{min} \leq R(v):=\dfrac{Av.v}{\|v\|^2} \leq \lambda_{max}$$ 
(https://en.wikipedia.org/wiki/Rayleigh_quotient) which can be proven by an elementary and elegant barycentric argumentation.
A: With
$Ae_i = \lambda_i e_i, \tag 1$
where the
$e_1, e_2, \ldots, e_n \in \Bbb R^n \tag 2$
form an orthonormal basis, we may write any
$0 \ne v \in \Bbb R^n \tag 3$
in the form
$v = \displaystyle \sum_1^n \alpha_i e_i, \tag 4$
whence
$Av = A \left  (\displaystyle \sum_1^n \alpha_i e_i \right ) = \displaystyle \sum_1^n \alpha_i A e_i = \sum_1^n \alpha_i \lambda_i e_i,  \tag 5$
and
$Av \cdot v = \left ( \displaystyle \sum_1^n \alpha_i \lambda_i e_i \right ) \cdot \left ( \displaystyle \sum_1^n \alpha_i e_i \right )$ $= \displaystyle \sum_{i,j = 1}^n \lambda_i \alpha_i \alpha_j e_i \cdot e_j = \sum_{i,j = 1}^n \lambda_i \alpha_i \alpha_j \delta_{ij}, \tag 6$
since the orthonormality of the $e_i$ may be written
$e_i \cdot e_j = \delta_{ij}; \tag 7$
then (6) becomes
$Av \cdot v = \displaystyle \sum_1^n \lambda_i \alpha_i^2 \ge \sum_1^n  \lambda_1 \alpha_i^2  = \lambda_1 \sum_1^n \alpha_i^2 = \lambda_1 v \cdot v =\lambda_1  \Vert v \Vert^2,  \tag 8$
since (4) yields
$\Vert v \Vert^2 = \displaystyle \sum_1^n \alpha_i^2, \tag 9$
as the reader may easily verify.  In a manner similar to (8) we also have
$Av \cdot v = \displaystyle \sum_1^n \lambda_i \alpha_i^2 \le \sum_1^n  \lambda_n \alpha_i^2  = \lambda_n \sum_1^n \alpha_i^2 = \lambda_n v \cdot v =\lambda_n  \Vert v \Vert^2; \tag{10}$
combining (8) and (10) we obtain
$\lambda_1 \Vert v \Vert^2 \le Av \cdot v \le \lambda_ n \Vert v \Vert^2, \tag{11}$
the sought-for result.
A: You made a mistake here
$$ \sum^n_{i=1} \lambda_i α_i e_i ⋅\sum^n_{i=1} α_i e_i= \sum^n_{i=1} \lambda_i \cdot \left(\sum^n_{i=1} α_i e_i\right)^2.$$
To fix it, one has to change $\sum^n_{i=1} α_i e_i=\sum^n_{j=1} α_j e_j$ and $e_i\cdot e_j=\delta_{ij}$. In fact
\begin{eqnarray*}
Av⋅v &=& 
A \sum^n_{i=1} α_i e_i ⋅\sum^n_{i=1} α_i e_i\\
&=& \sum^n_{i=1} \lambda_i α_i e_i ⋅\sum^n_{j=1} α_j e_j\\
&=& \sum^n_{i,j=1} \lambda_iα_iα_j( e_i\cdot e_j)\\
&=& \sum^n_{i,j=1} \lambda_i α_iα_j\delta_{ij}\\
&=&\sum^n_{i,j=1} \lambda_i(α_i)^2.
\end{eqnarray*}
Since $\lambda_1\le\lambda_i\le\lambda_n$, one has
$$ \sum^n_{i,j=1} \lambda_1(α_i)^2\le Av\cdot v\le \sum^n_{i,j=1} \lambda_n(α_i)^2 $$
or
$$ \lambda_1\|v\|^2\le Av\cdot v\le \lambda_n\|v\|^2. $$
