Orthonormal bases and the spectral theorem?

Question

Let $m, n \in N$. Prove that for any linear transformation $T: R^n \rightarrow R^m$ there exists an orthonormal basis $U = (\vec{u_1},...,\vec{u_n})$ of $R^n$ such that for all $1 \leq i,j \leq n$, if $i \neq j$ then $T(\vec{u_i}) • T(\vec{u_j}) = 0$.

There was a hint that recommended I use the Spectral Theorem, but I'm totally stumped. Can anyone help?

Answer 1

Diagonalize $T^*T$ to obtain an orthonormal basis of eigenvectors $u_1,\ldots, u_n$. Then write $T^*Tu_j = \lambda_j u_j$ for some coefficients $\lambda_j$. Then $$\langle Tu_i,Tu_j\rangle = \langle T^*Tu_i, u_j\rangle = \lambda_i \langle u_i,u_j\rangle = 0$$if $i \neq j$.

Answer 2

Hint: Take $U$ to be an orthonormal basis of eigenvalues of $T^\top T$.