[proofplan] The proof is a direct matrix computation. Orthogonality of $T$ implies that multiplication by $T$ preserves the Gram matrix of the loading matrix. Since the uniqueness matrix $\Psi$ is left unchanged, the covariance matrix computed from $(\Lambda T,\Psi)$ is identical to the covariance matrix computed from $(\Lambda,\Psi)$. [/proofplan] [step:Use orthogonality to simplify the rotated loading covariance] Because $T \in \mathbb{R}^{k \times k}$ is orthogonal, we have $T^\top T = I_k$. This identity says that $T^\top$ is a left inverse of the square matrix $T$. Hence $T$ is invertible and its inverse is $T^{-1} = T^\top$. Multiplying the identity $T^{-1} = T^\top$ on the left by $T$ gives \begin{align*} T T^\top = T T^{-1} = I_k. \end{align*} Now compute the loading covariance associated to the rotated loading matrix $\Lambda T$: \begin{align*} (\Lambda T)(\Lambda T)^\top &= (\Lambda T)(T^\top \Lambda^\top) \\ &= \Lambda (T T^\top) \Lambda^\top \\ &= \Lambda I_k \Lambda^\top \\ &= \Lambda \Lambda^\top. \end{align*} The first equality uses the transpose rule $(AB)^\top = B^\top A^\top$, and the second uses associativity of matrix multiplication. [/step] [step:Add the unchanged uniqueness matrix] Define the covariance matrix associated to the rotated pair $(\Lambda T,\Psi)$ by \begin{align*} \Sigma_T := (\Lambda T)(\Lambda T)^\top + \Psi. \end{align*} Using the identity proved in the previous step, \begin{align*} \Sigma_T &= (\Lambda T)(\Lambda T)^\top + \Psi \\ &= \Lambda \Lambda^\top + \Psi \\ &= \Sigma. \end{align*} Thus $(\Lambda T,\Psi)$ determines the same covariance matrix as $(\Lambda,\Psi)$. [/step]