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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 $i \neq j$

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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 $i \neq j$.

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  • $\begingroup$ Is there any other way to prove that? Self-adjoint seems new to me. $\endgroup$ – uther Apr 10 '18 at 15:25
  • $\begingroup$ What does the Spectral Theorem say? I suspect that it'll say more or less the same thing I said, but possibly with different language. $\endgroup$ – Theo Bendit Apr 10 '18 at 15:43
  • $\begingroup$ it says if there is a symmetric matrix, then it could be orthogonally diagonalizable. $\endgroup$ – uther Apr 10 '18 at 15:47
  • $\begingroup$ can we construct a ATA as it is symmetric? $\endgroup$ – uther Apr 10 '18 at 15:48
  • $\begingroup$ Yeah, my argument translates directly. I'm using operators, rather than matrices. Adjoints of operators turn into transposes of matrices (under certain conditions), so a self-adjoint operator is one with a matrix that is its own transpose, i.e. a symmetric matrix. The same argument will work: you just need to use $(Av) \cdot w = v \cdot (A^T w)$. $\endgroup$ – Theo Bendit Apr 10 '18 at 17:35

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