Diagonal elements of the projection matrix I am having some problem trying to prove that the diagonal elements of the hat matrix $h_{ii}$ are between $1/n$ and $1$.
Suppose that $\mathrm{Range}(X_{n,k})=K $ the number of columns of our matrix of data with a constant.$\implies H_{k,k}$
$H=X(X' X)^{-1}X'   \implies H'=H  ;H^{2}=H $
If $y = \beta x + \epsilon  \implies HY  =\hat Y; (I-H)=\epsilon $
$\mathbf{H}=\begin{bmatrix}h_{11} &...  &h_{1n}\\⋮ & ⋱ &⋮\\
h_{n1} & ... & h_{nn}\end{bmatrix}$
If $\boldsymbol 1=(1, \ldots, 1) \in  X \implies 1H^2=1H=1 \implies \sum h_{i1}^2=\sum h_{i1}=1 , h_{ii}\leq 1$
So $∑h_{i1}^2\sum h_{i2}^2...\sum h_{in}^2=1 \implies h_{11}^2h_{22}^2...h_{nn}^2\leq 1 $
How to prove that  $h_{ii} \geq(1/n)$?
(exercise 3.4 from Meyer "Classical and modern regression with applications") Let  $h_{ii}$ be the ith diagonal of the Hat matrix H. (a) prove that for a multiple regression model with a constant term, $h_{ii} \geq(1/n)$ ; (b) show that $h_{ii}\leq 1$ (Hint: make the use of the fact that H is idempotent)
 A: This is several years later, but I found the notation very difficult in the asker's question and self-answer, so here's a cleaner solution. 
We have $\mathbf{H} = \mathbf{X}(\mathbf{X}^T\mathbf{X})^{-1}\mathbf{X}^T$ where $(1,...,1)^T$ is a column of $\mathbf{X}$. We want to show that the diagonals $h_{ii}$ of $\mathbf{H}$ have $h_{ii} \geq 1/n$. 
Define $\mathbf{P} = \mathbf{H} - \mathbf{C}$, where 
$$\mathbf{C} = \frac{1}{n}\begin{pmatrix}1 &  \dots & 1 \\ \vdots & \ddots & \vdots \\ 1 & \dots & 1 \end{pmatrix}$$
the matrix consisting of only $1/n$. This is the projection matrix onto the space spanned by $(1, ..., 1)$. Then 
$$\mathbf{P}^2 = \mathbf{H}^2 - \mathbf{H}\mathbf{C} - \mathbf{C}\mathbf{H} + \mathbf{C}^2 = \mathbf{H} - \mathbf{H}\mathbf{C} - \mathbf{C}\mathbf{H} + \mathbf{C}$$
However, $\mathbf{H}$ orthogonally projects onto $\text{Col}(\mathbf{X})$, and $\mathbf{C}$ orthogonally projects onto $\text{span}\{(1,...,1)\} \subset \text{Col}(\mathbf{X})$, so obviously $\mathbf{H}\mathbf{C} = \mathbf{C}$. Still intuitively, but less obviously, $\mathbf{C}\mathbf{H} = \mathbf{C}$. To see this, we can compute $\mathbf{C} = \mathbf{C}\big(\mathbf{H} + (\mathbf{I} - \mathbf{H})\big)$, and note that $\mathbf{C}(\mathbf{I} - \mathbf{H}) = 0$ because $\mathbf{I} - \mathbf{H}$ projects onto $\text{Col}(\mathbf{X})^\perp$. 
Therefore we have $\mathbf{P}^2 = \mathbf{H} - \mathbf{C} = \mathbf{P}$. So $\mathbf{P}$ is also a projection matrix.
So $h_{ii} = p_{ii} + c_{ii} = p_{ii} + 1/n$. Since projection matrices are always positive semidefinite, the diagonals of $\mathbf{P}$ satisfy $p_{ii} \geq 0$. (In fact, you can show that since $\mathbf{P}$ is symmetric and idempotent, it satisfies $0 \leq p_{ii} \leq 1$.)
Then $h_{ii} \geq 1/n$ as needed.
A: For prove that  $h_{ii} \geq (1/n)$, we can center $H_c=X(X_c' X_c)^{-1}X_c'$ ,
$\mathbf{H_c}=\begin{bmatrix}x_{11}-\bar x_1 &...  &x_{1n}-\bar x_1
\\⋮ & ⋱ &⋮\\
x_{n1}-\bar x_n & ... & x_{nn}-\bar x_n\end{bmatrix}$
$y=\alpha1+ X_c'\beta +\epsilon⇒ \hat y=\hat \alpha1+ X_c'\hat\beta  ⇒
\hat y=\bar y+ X_c'\hat\beta= \bar y+ X_c'(X_c' X_c)^{-1}X_c'y⇒ 
\hat y=[(1/n) 1'y]1+H_cy$
$=[1/n\begin{bmatrix}1&...  &1\\⋮ & ⋱ &⋮\\1 & ... & 1\end{bmatrix}+H_c ] y=Hy $
Then  $ H=1/n\begin{bmatrix}1&...  &1\\⋮ & ⋱ &⋮\\1 & ... & 1\end{bmatrix}+H_c  $⇒
$h_{ii} \geq (1/n)$ because $H_c$ is a positive definite matrix.
A: Here is another answer that that only uses the fact that all the eigenvalues of a symmetric idempotent matrix are at most 1, see one of the previous answers or prove it yourself, it's quite easy.
Let $H$ denote the hat matrix. The $i$th diagonal element of the hat matrix is given by 
$$h_{ii} = \mathbf{e}_i^{t} \mathbf{H} \mathbf{e}_i,$$ 
where $\mathbf{e}_i^{t}$ is the vector whose $i$th element is 1 and the rest are 0s. Consider the quadratic form on the unit sphere given by
$$ f(\mathbf{x}) = \frac{\mathbf{x}^{t} \mathbf{H} \mathbf{x}}{\mathbf{x}^{t} \mathbf{x}}. $$
It is well known that the maximum of this expression is $\lambda_n$, the largest eigenvalue of the matrix $\mathbf{H}$. Returning to the diagonal elements of the hat matrix, one therefore has
$$h_{ii} = \mathbf{e}_i^{t} \mathbf{H} \mathbf{e}_i = \frac{\mathbf{e}_i^{t} \mathbf{H} \mathbf{e}_i}{\mathbf{e}_i^{t} \mathbf{e}_i} \underbrace{\mathbf{e}_i^{t} \mathbf{e}_i}_{ = 1} \leq \lambda_n \leq1 $$
and this gives us what we need.
A: Unlike the clever "centering" proof given by Drew N, the proof below is kind of brutal-force (but it has the advantage of giving even a sharper lower bound than $n^{-1}$, see $(1)$).
Partition the design matrix $X \in \mathbb{R}^{n \times p}$ to be $X = \begin{bmatrix} e & Z\end{bmatrix}$, where $e$ is the intercept term consisting of $n$ ones, $Z$ is the matrix consisting of $X$'s remaining $p - 1$ columns. By block matrix inversion formula,
\begin{align}
 & (X'X)^{-1} = \begin{bmatrix} n & e'Z \\ Z'e & Z'Z \end{bmatrix}^{-1} \\
=& \begin{bmatrix}
 n^{-1} + n^{-2}e'Z(Z'PZ)^{-1}Z'e & -n^{-1}e'Z(Z'PZ)^{-1} \\
-n^{-1}(Z'PZ)^{-1}Z'e & (Z'PZ)^{-1}
\end{bmatrix},
\end{align}
where $P = I_{(n)} - n^{-1}ee'$ is idempotent, hence $A := Z'PZ$ is invertible (hence positive definite. The invertibility of $A$ is proved in detail at the end). It then follows by $h_{ii} = x_i'(X'X)^{-1}x_i$ that
\begin{align}
& h_{ii} = \begin{bmatrix} 1 & z_i' \end{bmatrix}
\begin{bmatrix}
 n^{-1} + n^{-2}e'ZA^{-1}Z'e & -n^{-1}e'ZA^{-1} \\
-n^{-1}A^{-1}Z'e & A^{-1}
\end{bmatrix}
\begin{bmatrix}
1 \\
z_i
\end{bmatrix} \\
=& \begin{bmatrix}
n^{-1} + n^{-2}e'ZA^{-1}Z'e - n^{-1}z_i'A^{-1}Z'e & 
- n^{-1}e'ZA^{-1} + z_i'A^{-1}
\end{bmatrix}
\begin{bmatrix}
1 \\
z_i
\end{bmatrix} \\
=& n^{-1} + n^{-2}e'ZA^{-1}Z'e - n^{-1}z_i'A^{-1}Z'e 
- n^{-1}e'ZA^{-1}z_i + z_i'A^{-1}z_i \\
\geq & n^{-1} + \left(n^{-1}\sqrt{e'ZA^{-1}Z'e} - \sqrt{z_iA^{-1}z_i}\right)^2 \tag{1} \\ 
\geq & n^{-1}.
\end{align}
In $(1)$, we used Cauchy-Schwarz inequality
\begin{align}
|z_i'A^{-1}Z'e|^2 \leq z_i'A^{-1}z_i \times e'ZA^{-1}Z'e.  
\end{align}

Proof of $A$ is invertible:  Since $P$ is idempotent, to show $A$ is invertible, it suffices to show $\operatorname{rank}(PZ) = p - 1$, which (by rank-nullity theorem) is implied by $PZx = 0$ only has $0$ solution.  If $PZx = 0$, then $Zx \in \operatorname{Ker}(I_{(n)} - n^{-1}ee') = \operatorname{span}(e)$. But since $X$ has full column rank, $e$ and $Zx \in \operatorname{span}(Z)$ are linearly independent, whence $Zx = 0$, which implies $x = 0$ due to $\operatorname{rank}(Z) = p - 1$.  This completes the proof.
A: Here is another simpler (and perhaps more illuminating) proof that is based on QR decomposition of the design matrix $X$.
Suppose the QR decomposition of $X$ is $X = QR$, where $Q \in \mathbb{R}^{n \times p}$ is a matrix whose columns are orthogonal (so that $Q'Q = I_{(p)}$), $R \in \mathbb{R}^{p \times p}$ is an upper-triangular matrix.  Since $\operatorname{rank}(X) = p$, $\operatorname{rank}(R)$ must be at least $p$, thus $R$ is invertible.  Also note that since the first column of $X$ is $e$, the Gram-Schmidt procedure implies that the first column of $Q$ is $\frac{1}{\sqrt{n}}e$.  Denote by $e_i$ the length-$n$ column vector of all zeros but the $i$-th position $1$, it then follows that
\begin{align}
 & h_{ii} = e_i'X(X'X)^{-1}X'e_i = e_i'QR(R'Q'QR)^{-1}R'Q'e_i \\
=& e_i'QR(R'R)^{-1}R'Q'e_i = e_i'QRR^{-1}(R')^{-1}R'Q'e_i \\
=& e_i'QQ'e_i = \tilde{q}_i'\tilde{q}_i \\
=& n^{-1} + Q_{i2}^2 + \cdots + Q_{ip}^2 \geq n^{-1},
\end{align}
where $\tilde{q}_i' = \begin{bmatrix}\frac{1}{\sqrt{n}} & Q_{i2} & \cdots & Q_{ip}\end{bmatrix}$ denotes the $i$-th row of $Q$. This completes the proof.

More details on QR decomposition:  Suppose $X = \begin{bmatrix} e & x_2 & 
\cdots & x_p \end{bmatrix}$.  By assumption, $\{e, x_2, \ldots, x_p\}$ are linearly independent, which allows us to apply the Gram-Schmidt procedure to  obtain an orthonormal group $\{q_1, q_2, \ldots, q_p\}$ based on $\{e, x_2, \ldots, x_p\}$ as follows:
\begin{align}
& z_1 = e, \; q_1 = \frac{z_1}{\|z_1\|}, \tag{1} \\
& z_2 = x_2 - \frac{x_2'z_1}{z_1'z_1}z_1, \; q_2 = \frac{z_2}{\|z_2\|}, \\
& \cdots \cdots \cdots \\
& z_p = x_p - \frac{x_p'z_{n - 1}}{z_{n - 1}'z_{n - 1}}z_{n - 1} - \cdots 
- \frac{x_p'z_1}{z_1'z_1}z_1, \; q_p = \frac{z_p}{\|z_p\|}.  
\end{align}
In matrical form, the above transformation can be recorded as
\begin{align}
X = \begin{bmatrix} e & x_2 & 
\cdots & x_p \end{bmatrix} &=
\begin{bmatrix} q_1 & q_2 & 
\cdots & q_p \end{bmatrix}
\begin{pmatrix}
\|z_1\| & \frac{x_2'z_1}{\|z_1\|} & \cdots & \frac{x_p'z_1}{\|z_1\|} \\
        & \|z_2\|                 & \cdots & \frac{x_p'z_2}{\|z_2\|} \\
        &                         & \ddots & \vdots                  \\
        &                         &        & \|z_p\| 
\end{pmatrix}\\
&=: QR. \tag{2}
\end{align}
It is thus evident from $(1)$ and $(2)$ that $q_1 = \frac{1}{\sqrt{n}}e$.
A: $H = H^\top H$ implies $h_{ii} = h_{ii}^2 + \sum_{j \neq i} h_{ij}^2$, so $h_{ii} \geq h_{ii}^2$, hence $h_{ii} \leq 1$.
Assuming presence of an intercept, we have $h_{ii} = 1/n + D_i^2/\left(n-1\right) \geq 1/n$, where $D_i^2$ is the squared Mahalanobis distance between the $i$-th row of a design matrix $X$ with 0-centered regressors and the origin.
