Solving Det(\vec{V} \otimes\vec{V})=0 in 10 Seconds: Tips and Tricks

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SUMMARY

The discussion focuses on demonstrating that Det(\vec{V} \otimes \vec{V})=0 efficiently. The key insight is utilizing the Levi-Civita symbol to express the determinant in three dimensions, specifically using the formula \det A = \sum_{ijk} \epsilon_{ijk} A_{1i} A_{2j} A_{3k}. By applying this method to the dyadic product of vector \vec{V}(x,y,z), it becomes clear that the determinant vanishes, confirming the non-invertibility of the matrix.

PREREQUISITES
  • Understanding of tensor products, specifically \vec{V} \otimes \vec{V}.
  • Familiarity with the Levi-Civita symbol and its properties.
  • Basic knowledge of determinants in linear algebra.
  • Experience with three-dimensional vector spaces.
NEXT STEPS
  • Study the properties of the Levi-Civita symbol in detail.
  • Learn about tensor products and their applications in linear algebra.
  • Explore advanced determinant techniques, including cofactor expansion.
  • Investigate the implications of non-invertible matrices in various mathematical contexts.
USEFUL FOR

Students and professionals in mathematics, particularly those studying linear algebra, tensor analysis, and vector calculus, will benefit from this discussion.

A-prime
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Homework Statement


The probem is to show within 10 seconds that Det([tex]\vec{V}[/tex] [tex]\otimes[/tex][tex]\vec{V}[/tex])=0

with [tex]\vec{V}[/tex](x,y,z)=(Vx([tex]\vec{r}[/tex]),Vy([tex]\vec{r}[/tex]),Vz([tex]\vec{r}[/tex]))

The Attempt at a Solution


So I thought if I show the matrix is not invertible I would be done. But even with that I can't easily solve it in a few seconds...

Tips?
 
Last edited:
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You can use the Levi-Civita symbol to write down the determinant. In 3d, this is

[tex]\det A = \sum_{ijk} \epsilon_{ijk} A_{1i} A_{2j} A_{3k}.[/tex]

If you apply this to your dyadic, you should be able to see that it vanishes quickly.
 

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