In proving Student’s Theorem, a linear transformation by matrix can be used,i.e
\[W=\begin{bmatrix} \mathbf{\bar{X}}\\ \mathbf{Y} \end{bmatrix}=\begin{bmatrix} \mathbf{v'}\\ \mathbf{I-1v'} \end{bmatrix} \] where, \(\mathbf{v'}=(\frac{1}{n},\frac{1}{n},...,\frac{1}{n})=\frac{1}{n}\mathbf{1'}\)
The covariance matrix of the multivariate normal distributed \(\mathbf{W}\) is
\[\Sigma=\begin{bmatrix} \mathbf{v'}\\ \mathbf{I-1v'} \end{bmatrix}\sigma^2\mathbf{I}\begin{bmatrix} \mathbf{v'}\\ \mathbf{I-1v'} \end{bmatrix}' =\sigma^2\begin{bmatrix} \frac{1}{n} &\mathbf{0_n'}\\ \mathbf{0_n}&\mathbf{I-1v'} \end{bmatrix}\tag{1}\]
When we do the matrix transpose,we need to remember that we also need to transpose matrix inside the block matrix, i.e
\[\left[ \begin{array}{c c} A & B \\ C & D \\ \end{array} \right]^\top = \left[ \begin{array}{c c} A^\top & C^\top \\ B^\top & D^\top \\ \end{array} \right]. \]
Therefore, \[\begin{bmatrix} \mathbf{v'}\\ \mathbf{I-1v'} \end{bmatrix}'= \begin{bmatrix} \mathbf{v} & \mathbf{I-v1'} \end{bmatrix} \]
Here, I show how we can get off-diagonal as \(\mathbf{0}s\) in \((1)\)
\(\mathbf{v'}\times (\mathbf{I-v1'})=\mathbf{v'-v'v1'}=(\frac{1}{n},\frac{1}{n},...,\frac{1}{n})-\frac{1}{n}(1,1,...,1)=(0,0,...,0)=\mathbf{0_n}\), similarly, we can show another off-diagonal blocks are zeros.
The paper for the Welch’s t test for unequal varaince t test.
Welch, B. L. (1947). “The generalization of”Student’s” problem when several different population variances are involved”. Biometrika. 34 (1–2): 28–35.