If $A=(a_1, \dots, a_n)^T$ is invertible show that $|\det(A)| = \prod_{i=1}^n ||a_i||$ iff $\{a_1, \dots, a_n\}$ is an orthogonal basis of $\mathbb{R}^n$.

If $\{a_1, \dots, a_n\}$ is an orthogonal basis then $A^TA$ is diagonal, so $|\det(A)| = \sqrt{\det(A)\det(A)} = \sqrt{\det(A^T)\det(A)} = \sqrt{\det(A^TA)} = \sqrt{\prod_{i=1}^n \langle a_i, a_i \rangle} = \prod_{i=1}^n \sqrt{\langle a_i, a_i \rangle} = \prod_{i=1}^n ||a_i||$

For the other direction it is clear, that $\{a_1, \dots, a_n\}$ is a basis, as all columns are linear independent (invertible). How can I prove that they are orthogonal?

  • $\begingroup$ By QR factorisation, you may assume that $A$ is upper triangular and now you may prove the "if and only if" (as well as the inequality $|\det(A)|\le\prod_j\|a_j\|$) in one shot. $\endgroup$ – user1551 Jul 21 '19 at 13:44
  • $\begingroup$ if $A=QR$, $|\det(A)| = \prod_{i=1}^n |r_{ii}|$. But how does this help proving orthogonality? $\endgroup$ – Tim Jul 21 '19 at 13:59
  • $\begingroup$ As $\prod_{i=1}^n |r_{ii}| = \prod_{i=1}^n ||(0, \dots, 0, r_{ii}, 0, \dots, 0)^T|| \leq \prod_{i=1}^n ||(r_{1i}, \dots, r_{ni})^T|| = \prod_{i=1}^n ||r_i|| = \prod_{i=1}^n ||Qr_i|| = \prod_{i=1}^n ||a_i||$ if $|r_{ii}| = ||r_i||$ $\endgroup$ – Tim Jul 21 '19 at 14:31
  • $\begingroup$ that is true iff $r_{ji} = 0$ for $j \neq i$. $\endgroup$ – Tim Jul 21 '19 at 14:48
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    $\begingroup$ … diagonal, so $A=QR=(q_1r_{11}, \dots, q_nr_{nn})^T$, so as $q_i$ are orthogonal, $q_ir_{ii}$ is orthogonal as well. $\endgroup$ – Tim Jul 21 '19 at 14:59

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