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awarded  Yearling
Jun
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revised Orbit , trajectory, dynamical system
Latex input and grammar/vocabulary corrections
Jun
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suggested approved edit on Orbit , trajectory, dynamical system
Jun
1
comment Orbit , trajectory, dynamical system
@Evgeny I would suggest that your turn your comment into an answer.
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awarded  Notable Question
Jan
11
revised Prove that there is a symmetric matrix B, such that BX=Y
Latex coding for improved reading
Jan
11
suggested approved edit on Prove that there is a symmetric matrix B, such that BX=Y
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Mar
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accepted From a vector to a skew symmetric matrix
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Oct
17
comment Component-free formula for the determinant of a tensor
Instructive answer. I was hoping for something simpler, in a sense that it would require a lighter background.
Oct
17
accepted Component-free formula for the determinant of a tensor
Oct
17
revised Component-free formula for the determinant of a tensor
added 231 characters in body
Oct
17
comment Component-free formula for the determinant of a tensor
@MarianoSuárez-Alvarez $\mathbf{m},\mathbf{n},\mathbf{p}$ being an orthonormal basis of $\mathbb{R}^3$, consider the determinant of $\mathbb{A}=\mathbf{m}\otimes\mathbf{n}+\mathbf{n}\otimes\mathbf{m}+\mathbf{p} \otimes \mathbf{p}$.
Oct
17
comment Component-free formula for the determinant of a tensor
@MarianoSuárez-Alvarez Yes, this is one way to find the determinant for this example. How about a non-singular tensor?
Oct
17
comment Component-free formula for the determinant of a tensor
@Muphrid good point. I want to write $\mathbf{a}$ but I wrote $\mathbf{m}$. The post is corrected.
Oct
17
revised Component-free formula for the determinant of a tensor
edited body