[proofplan] Represent the Wishart matrix as $W = X^\top X$ using its defining Gaussian sample matrix. The $j$th diagonal entry of $W$ is the sum of squares of the $j$th coordinates of the independent Gaussian rows. After normalizing those coordinates by $\sqrt{\sigma_{jj}}$, we obtain $n$ independent standard normal random variables, so the normalized diagonal entry is their sum of squares, which is precisely a chi-squared random variable with $n$ degrees of freedom. [/proofplan] [step:Extract the $j$th coordinate variables from the Gaussian sample matrix] Fix $j \in \{1,\dots,p\}$ with $\sigma_{jj} > 0$. For each $i \in \{1,\dots,n\}$, define the real-valued random variable \begin{align*} Z_i: \Omega &\to \mathbb{R} \\ \omega &\mapsto \frac{X_{ij}(\omega)}{\sqrt{\sigma_{jj}}}. \end{align*} Since $X_i \sim \mathcal{N}_p(0,\Sigma)$, its $j$th coordinate $X_{ij}$ has one-dimensional Gaussian distribution $\mathcal{N}(0,\sigma_{jj})$. Because the rows $X_1,\dots,X_n$ are independent, the coordinate random variables $X_{1j},\dots,X_{nj}$ are independent. Therefore $Z_1,\dots,Z_n$ are independent random variables with distribution $\mathcal{N}(0,1)$. [guided] Fix an index $j \in \{1,\dots,p\}$ satisfying $\sigma_{jj} > 0$. The positivity assumption is needed so that division by $\sqrt{\sigma_{jj}}$ is defined. For each row index $i \in \{1,\dots,n\}$, define \begin{align*} Z_i: \Omega &\to \mathbb{R} \\ \omega &\mapsto \frac{X_{ij}(\omega)}{\sqrt{\sigma_{jj}}}. \end{align*} The row $X_i$ has distribution $\mathcal{N}_p(0,\Sigma)$. Hence its $j$th coordinate $X_{ij}$ is a real-valued Gaussian random variable with mean $0$ and variance equal to the $j$th diagonal covariance entry $\sigma_{jj}$. Thus \begin{align*} X_{ij} \sim \mathcal{N}(0,\sigma_{jj}). \end{align*} Scaling a centered Gaussian random variable by the reciprocal of its standard deviation gives a standard normal random variable, so \begin{align*} Z_i = \frac{X_{ij}}{\sqrt{\sigma_{jj}}} \sim \mathcal{N}(0,1). \end{align*} It remains to check independence. The random vectors $X_1,\dots,X_n$ are independent by the defining construction of the Wishart distribution. Each $X_{ij}$ is obtained from $X_i$ by applying the $j$th coordinate map, and independence is preserved under measurable transformations applied separately to independent random vectors. Therefore $X_{1j},\dots,X_{nj}$ are independent, and hence $Z_1,\dots,Z_n$ are independent standard normal random variables. [/guided] [/step] [step:Identify the diagonal entry as a normalized sum of Gaussian squares] By matrix multiplication, \begin{align*} w_{jj} = (X^\top X)_{jj} = \sum_{i=1}^n X_{ij}X_{ij} = \sum_{i=1}^n X_{ij}^2. \end{align*} Dividing by $\sigma_{jj}$ and using the definition of $Z_i$ gives \begin{align*} \frac{w_{jj}}{\sigma_{jj}} = \sum_{i=1}^n \frac{X_{ij}^2}{\sigma_{jj}} = \sum_{i=1}^n Z_i^2. \end{align*} Since $Z_1,\dots,Z_n$ are independent standard normal random variables, the defining characterization of the chi-squared distribution with $n$ degrees of freedom gives \begin{align*} \sum_{i=1}^n Z_i^2 \sim \chi_n^2. \end{align*} Therefore \begin{align*} \frac{w_{jj}}{\sigma_{jj}} \sim \chi_n^2, \end{align*} as required. [/step]