[proofplan] We first verify that $X^\top X$ is invertible, so the displayed formula for $H$ is well-defined. Symmetry follows from the transpose rule and the symmetry of $X^\top X$, while idempotence follows by multiplying $H$ by itself and cancelling $X^\top X$. The diagonal bounds are obtained by testing $H$ and $I_n-H$ on standard basis vectors, using symmetry and idempotence to rewrite the resulting quadratic forms as squared Euclidean norms. Finally, the trace is computed directly using cyclicity of trace for finite matrices. [/proofplan]

[step:Show that $X^\top X$ is invertible] Let \begin{align*} A := X^\top X \in \mathbb{R}^{p \times p}. \end{align*} The matrix $A$ is symmetric because \begin{align*} A^\top = (X^\top X)^\top = X^\top (X^\top)^\top = X^\top X = A. \end{align*} To prove that $A$ is invertible, let $v \in \mathbb{R}^p$ satisfy $Av=0$. Multiplying on the left by $v^\top$ gives \begin{align*} 0 = v^\top A v = v^\top X^\top Xv = (Xv)^\top(Xv)=|Xv|^2. \end{align*} Hence $Xv=0$. Since $X$ has full column rank, its columns are linearly independent, so $Xv=0$ implies $v=0$. Thus $\ker A=\{0\}$. Since $A$ is a square $p \times p$ matrix, it is invertible. [/step]

[step:Compute the transpose of $H$] Since $A=X^\top X$ is symmetric and invertible, its inverse $A^{-1}$ is also symmetric: \begin{align*} (A^{-1})^\top = (A^\top)^{-1}=A^{-1}. \end{align*} Using the transpose rule $(BCD)^\top=D^\top C^\top B^\top$, we obtain \begin{align*} H^\top &= \bigl(XA^{-1}X^\top\bigr)^\top \\ &= (X^\top)^\top (A^{-1})^\top X^\top \\ &= XA^{-1}X^\top \\ &= H. \end{align*} Therefore $H$ is symmetric. [/step]

[step:Multiply $H$ by itself to prove idempotence] Using associativity of matrix multiplication and the identity $A=X^\top X$, we compute \begin{align*} H^2 &= \bigl(XA^{-1}X^\top\bigr)\bigl(XA^{-1}X^\top\bigr) \\ &= XA^{-1}(X^\top X)A^{-1}X^\top \\ &= XA^{-1}AA^{-1}X^\top \\ &= XA^{-1}X^\top \\ &= H. \end{align*} Thus $H$ is idempotent. [/step]

[step:Bound each diagonal entry using squared Euclidean norms] Let $I_n \in \mathbb{R}^{n \times n}$ be the identity matrix. For each $i \in \{1,\dots,n\}$, let $e_i \in \mathbb{R}^n$ denote the $i$th standard basis vector. By definition of diagonal entries, \begin{align*} h_{ii}=e_i^\top H e_i. \end{align*} Since $H^\top=H$ and $H^2=H$, we have \begin{align*} h_{ii} = e_i^\top H e_i = e_i^\top H^2 e_i = (He_i)^\top(He_i) = |He_i|^2 \ge 0. \end{align*} Next define \begin{align*} K := I_n-H \in \mathbb{R}^{n \times n}. \end{align*} Then $K$ is symmetric and idempotent, because \begin{align*} K^\top = I_n-H^\top=I_n-H=K \end{align*} and \begin{align*} K^2 &= (I_n-H)^2 \\ &= I_n-2H+H^2 \\ &= I_n-H \\ &= K. \end{align*} Therefore \begin{align*} 1-h_{ii} &= e_i^\top I_n e_i-e_i^\top H e_i \\ &= e_i^\top K e_i \\ &= e_i^\top K^2 e_i \\ &= (Ke_i)^\top(Ke_i) \\ &= |Ke_i|^2 \\ &\ge 0. \end{align*} Hence $h_{ii}\le 1$. Combining the two inequalities gives \begin{align*} 0 \le h_{ii}\le 1. \end{align*} [/step]

[step:Compute the sum of the diagonal entries by cyclicity of trace] For a square matrix $M \in \mathbb{R}^{m \times m}$, write \begin{align*} \operatorname{tr}(M):=\sum_{j=1}^m M_{jj}. \end{align*} Thus \begin{align*} \sum_{i=1}^n h_{ii}=\operatorname{tr}(H). \end{align*} For matrices $B \in \mathbb{R}^{r \times s}$ and $C \in \mathbb{R}^{s \times r}$, direct expansion gives \begin{align*} \operatorname{tr}(BC) &= \sum_{a=1}^r (BC)_{aa} \\ &= \sum_{a=1}^r \sum_{b=1}^s B_{ab}C_{ba} \\ &= \sum_{b=1}^s \sum_{a=1}^r C_{ba}B_{ab} \\ &= \operatorname{tr}(CB). \end{align*} Applying this cyclic trace identity first to $X$ and $A^{-1}X^\top$, and then using $A=X^\top X$, we get \begin{align*} \operatorname{tr}(H) &= \operatorname{tr}(XA^{-1}X^\top) \\ &= \operatorname{tr}(A^{-1}X^\top X) \\ &= \operatorname{tr}(A^{-1}A) \\ &= \operatorname{tr}(I_p) \\ &= p. \end{align*} Therefore \begin{align*} \sum_{i=1}^n h_{ii}=p. \end{align*} This proves all asserted identities. [/step]