[proofplan] We compute directly from the definition $H=X(X^\top X)^{-1}X^\top$. Full column rank ensures that $X^\top X$ is invertible, so the formula is well-defined. Symmetry and idempotence give the projection algebra needed to control the quadratic forms $e_i^\top H e_i$ on standard basis vectors $e_i \in \mathbb{R}^n$. The trace is then computed by cyclically rearranging the finite-dimensional matrix trace. [/proofplan] [step:Verify that the hat matrix is well-defined] Since $X$ has full column rank $p$, its columns are linearly independent. Hence, for every nonzero vector $v \in \mathbb{R}^p$, \begin{align*} v^\top X^\top X v = |Xv|^2 > 0. \end{align*} Therefore $X^\top X \in \mathbb{R}^{p \times p}$ is symmetric positive definite, so it is invertible. Thus \begin{align*} H = X(X^\top X)^{-1}X^\top \end{align*} is a well-defined element of $\mathbb{R}^{n \times n}$. [/step] [step:Compute symmetry and idempotence from the defining formula] Let \begin{align*} A := (X^\top X)^{-1} \in \mathbb{R}^{p \times p}. \end{align*} Let $I_p \in \mathbb{R}^{p \times p}$ denote the identity matrix. Since $X^\top X$ is symmetric, its inverse $A$ is also symmetric. Therefore \begin{align*} H^\top &= \bigl(XAX^\top\bigr)^\top \\ &= XA^\top X^\top \\ &= XAX^\top \\ &= H. \end{align*} For idempotence, use associativity of matrix multiplication and the identity $(X^\top X)A=I_p$: \begin{align*} H^2 &= \bigl(XAX^\top\bigr)\bigl(XAX^\top\bigr) \\ &= XA(X^\top X)AX^\top \\ &= XAI_pX^\top \\ &= XAX^\top \\ &= H. \end{align*} Thus $H$ is symmetric and idempotent. [/step] [step:Bound each diagonal entry by projecting a standard basis vector] For each $i \in \{1,\dots,n\}$, let $e_i \in \mathbb{R}^n$ denote the $i$-th standard basis vector. The $(i,i)$-entry of $H$ is \begin{align*} h_{ii} = e_i^\top H e_i. \end{align*} Using symmetry and idempotence, \begin{align*} h_{ii} &= e_i^\top H e_i \\ &= e_i^\top H^\top H e_i \\ &= (He_i)^\top(He_i) \\ &= |He_i|^2. \end{align*} Hence $h_{ii} \geq 0$. To prove the upper bound, decompose \begin{align*} e_i = He_i + (I_n-H)e_i, \end{align*} where $I_n \in \mathbb{R}^{n \times n}$ is the identity matrix. The two summands are orthogonal because \begin{align*} (He_i)^\top((I_n-H)e_i) &= e_i^\top H^\top(I_n-H)e_i \\ &= e_i^\top H(I_n-H)e_i \\ &= e_i^\top(H-H^2)e_i \\ &= 0. \end{align*} By the Pythagorean identity in $\mathbb{R}^n$, \begin{align*} |e_i|^2 = |He_i|^2 + |(I_n-H)e_i|^2. \end{align*} Since $|e_i|^2=1$ and $|(I_n-H)e_i|^2 \geq 0$, we get \begin{align*} h_{ii}=|He_i|^2 \leq 1. \end{align*} Therefore $0 \leq h_{ii} \leq 1$ for every $i \in \{1,\dots,n\}$. [/step] [step:Compute the trace by moving conformable finite matrices cyclically] For finite matrices $B \in \mathbb{R}^{n \times p}$ and $C \in \mathbb{R}^{p \times n}$, the trace identity \begin{align*} \operatorname{tr}(BC)=\operatorname{tr}(CB) \end{align*} follows by expanding entries: \begin{align*} \operatorname{tr}(BC) &= \sum_{i=1}^{n} (BC)_{ii} \\ &= \sum_{i=1}^{n}\sum_{j=1}^{p} B_{ij}C_{ji} \\ &= \sum_{j=1}^{p}\sum_{i=1}^{n} C_{ji}B_{ij} \\ &= \sum_{j=1}^{p} (CB)_{jj} \\ &= \operatorname{tr}(CB). \end{align*} Apply this identity with $B=X$ and $C=(X^\top X)^{-1}X^\top$. Then \begin{align*} \operatorname{tr}(H) &= \operatorname{tr}\bigl(X(X^\top X)^{-1}X^\top\bigr) \\ &= \operatorname{tr}\bigl((X^\top X)^{-1}X^\top X\bigr) \\ &= \operatorname{tr}(I_p) \\ &= p. \end{align*} This proves all asserted properties of the hat matrix. [/step]