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$|\det(A)|=\Pi_{i=1}^n ||\vec{A}_i||\iff A$ has zero columns or $\{\vec{A}_1,\vec{A}_2,...,\vec{A}_n\}$ is an orthogonal set.

Each column of $R$ shows how you write the corresponding column of $A$ as a linear combination of columns in $Q$. That is, $\vec{A}_j = R_{1j} \vec{Q}_1 + \dotsb + R_{nj} \vec{Q}_n$. Since $\vec{Q}_j$ ...
André Caldas's user avatar
1 vote

$|\det(A)|=\Pi_{i=1}^n ||\vec{A}_i||\iff A$ has zero columns or $\{\vec{A}_1,\vec{A}_2,...,\vec{A}_n\}$ is an orthogonal set.

If $0 \ne |\det(R)|=\prod_{i=1}^n \|\vec{R}_i\|$ with an upper triangular matrix $R$ then $R$ is necessarily a diagonal matrix: The left-hand side is the product of the absolute values of the diagonal ...
Martin R's user avatar
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