Let $m, n \in \mathbb{N}$. Prove that for any linear transformation $T : \mathbb{R}^n \to \mathbb{R}^m$, there is an orthonormal basis $\mathcal{U} = (\vec{u_1}, \ldots, \vec{u_n})$ of $\mathbb{R}^n$ such that for all $1 \le i, j \le n$, if $i \neq j$ then $T(\vec{u_i}) \cdot T(\vec{u_j}) = 0$. (Hint: use the Spectral Theorem, 8.1.1 in the text).
how to prove for any linear transformation form $\mathbb{R}^n$ to $\mathbb{R}^m$ ,there is an orthonormal basis such that $T(u_i)\cdot T(u_j) = 0$ if $i \neq j$
Recall that $T^* T$ is self-adjoint, and thus by the Spectral Theorem, has an orthogonal basis of eigenvectors $(u_1, \ldots, u_n)$. Let $\lambda_i$ be the eigenvalue corresponding to the eigenvector $u_i$. Then, $$T(u_i) \cdot T(u_j) = (T^* T)(u_i) \cdot u_j = \lambda_i u_i \cdot u_j = 0,$$ whenever $i \neq j$.