[proofplan] The matrix $X$ defines a [linear map](/page/Linear%20Map) from $\mathbb{R}^p$ to $\mathbb{R}^n$. Since the domain has larger dimension than the codomain, this map has a nonzero kernel vector. Evaluating the quadratic form of $G=\frac{1}{n}X^\top X$ on that vector gives zero, while the same quadratic form is nonnegative in every direction. Thus $G$ is positive semidefinite and has a nonzero null vector, forcing its smallest eigenvalue to be exactly $0$. [/proofplan]

[step:Find a nonzero vector annihilated by $X$]Let \begin{align*} L_X:\mathbb{R}^p \to \mathbb{R}^n, \qquad L_X(v)=Xv \text{ for every } v \in \mathbb{R}^p \end{align*} be the linear map represented by $X$. Write the columns of $X$ as $x_1,\dots,x_p \in \mathbb{R}^n$. Since $p>n$, the $p$ vectors $x_1,\dots,x_p$ in the $n$-dimensional [vector space](/page/Vector%20Space) $\mathbb{R}^n$ are linearly dependent. Hence there exists a vector \begin{align*} v=(v_1,\dots,v_p) \in \mathbb{R}^p \end{align*} with $v \neq 0$ and \begin{align*} \sum_{j=1}^{p} v_j x_j = 0. \end{align*} By the definition of matrix multiplication, this is exactly $Xv=0$.[/step]

[guided]The key point is that $X$ maps a higher-dimensional space into a lower-dimensional one. To make this precise, define \begin{align*} L_X:\mathbb{R}^p \to \mathbb{R}^n, \qquad L_X(v)=Xv \text{ for every } v \in \mathbb{R}^p. \end{align*} This is a linear map because matrix multiplication is linear in the vector argument. Let $x_1,\dots,x_p \in \mathbb{R}^n$ denote the columns of $X$. For a vector \begin{align*} v=(v_1,\dots,v_p) \in \mathbb{R}^p, \end{align*} matrix multiplication gives \begin{align*} Xv=\sum_{j=1}^{p} v_j x_j. \end{align*} Since $p>n$, there are more columns than the dimension of the ambient space $\mathbb{R}^n$. Therefore the family $x_1,\dots,x_p$ is linearly dependent. Consequently there are scalars $v_1,\dots,v_p \in \mathbb{R}$, not all zero, such that \begin{align*} \sum_{j=1}^{p} v_j x_j = 0. \end{align*} Equivalently, the vector $v=(v_1,\dots,v_p)$ is nonzero and satisfies \begin{align*} Xv=0. \end{align*} Thus the kernel of $L_X$ contains a nonzero vector.[/guided]

[step:Show the Gram matrix is positive semidefinite and singular] Define the Gram matrix $G \in \mathbb{R}^{p \times p}$ by $G=\frac{1}{n}X^\top X$. Throughout the rest of the proof, $|\cdot|$ denotes the Euclidean norm on the Euclidean space containing its argument; explicitly, $|y|^2=y^\top y$ for every $y \in \mathbb{R}^n$ and $|z|^2=z^\top z$ for every $z \in \mathbb{R}^p$. For every $w \in \mathbb{R}^p$, associativity of matrix multiplication and the definition of the Euclidean norm give \begin{align*} w^\top G w = w^\top \left(\frac{1}{n}X^\top X\right)w = \frac{1}{n}(Xw)^\top(Xw) = \frac{1}{n}|Xw|^2 \ge 0. \end{align*} Hence $G$ is positive semidefinite. Applying the definition of $G$ to the nonzero vector $v \in \mathbb{R}^p$ constructed above gives \begin{align*} Gv = \frac{1}{n}X^\top Xv = \frac{1}{n}X^\top 0 = 0. \end{align*} Thus $0$ is an eigenvalue of $G$. [/step]

[step:Identify the smallest eigenvalue] The matrix $G$ is symmetric. Indeed, transposition preserves scalar multiplication and reverses products, so \begin{align*} G^\top = \left(\frac{1}{n}X^\top X\right)^\top. \end{align*} Since $(X^\top X)^\top = X^\top X$, this gives \begin{align*} G^\top = \frac{1}{n}X^\top X = G. \end{align*} Because $G \in \mathbb{R}^{p \times p}$ is real and symmetric, it satisfies the hypothesis of the [Spectral Theorem for Real Symmetric Matrices](/page/Spectral%20Theorem), so the eigenvalues of $G$ are real. Since $G$ is positive semidefinite, every eigenvalue $\lambda$ of $G$ satisfies $\lambda \ge 0$: if $u \in \mathbb{R}^p$ is a nonzero eigenvector with $Gu=\lambda u$, then \begin{align*} 0 \le u^\top G u = u^\top(\lambda u)=\lambda |u|^2, \end{align*} and $|u|^2>0$, so $\lambda \ge 0$. From the previous step, $0$ is an eigenvalue of $G$. Therefore all eigenvalues of $G$ are at least $0$, and one eigenvalue is equal to $0$. Hence \begin{align*} \lambda_{\min}(G)=0. \end{align*} This proves the theorem. [/step]