# Show Tensor Determinants Equality

• hotvette
In summary: I don't know the name for it, but you're basically supposed to make it look like the definition for det(A)*det(B).

Homework Helper

## Homework Statement

show that $\det(\underline{\bf{A}})\det(\underline{\bf{B}}) = \det(\underline{\bf{AB}})$

## Homework Equations

\begin{align*}
&\underline{\bf{A}} = A_{ij} \underline{e}_i \otimes \underline{e}_j \\
&\underline{\bf{B}} = B_{mn} \underline{e}_m \otimes \underline{e}_n \\
&\underline{\bf{A}}\underline{\bf{B}} = A_{ij}B_{mn} \underline{e}_i \otimes \underline{e}_n \\
& \det({\underline{\bf{A}}})=\epsilon_{pqr}A_{p1}A_{q2}A_{r3} \\
& \det({\underline{\bf{B}}})=\epsilon_{xyz}B_{x1}B_{y2}B_{z3} \\
& \det({\underline{\bf{A}}})(\det{\underline{\bf{B}}})=\epsilon_{pqr}A_{p1}A_{q2}A_{r3}\epsilon_{xyz}B_{x1}B_{y2}B_{z3}
\end{align*}

## The Attempt at a Solution

\begin{align*}
&\underline{\bf{C}}={\underline{\bf{A}}}{\underline{\bf{B}}} \\
& \underline{\bf{C}} = A_{im}B_{mj} \underline{e}_i \otimes \underline{e}_j \\
& C_{ij} = A_{im}B_{mj} \\
& \det(\underline{\bf{AB}}) = \det(\underline{\bf{C}}) = \epsilon_{stu}C_{s1}C_{t2}C_{u3} \\
& C_{s1} = A_{sm}B_{m1} \\
& C_{t2} = A_{tn}B_{n2} \\
& C_{u3} = A_{uk}B_{k3} \\
& \det(\underline{\bf{AB}}) =\det(\underline{\bf{C}}) = \epsilon_{stu} A_{sm}A_{tn}A_{uk}B_{m1}B_{n2}B_{k3} \\
& \det(\underline{\bf{AB}}) =\det(\underline{\bf{C}}) = \epsilon_{pqr} A_{px}A_{qy}A_{rz}B_{x1}B_{y2}B_{z3}
\end{align*}

Which means I need to show that $A_{px}A_{qy}A_{rz}=\epsilon_{xyz} A_{p1}A_{q2}A_{r3}$. That's where I'm stuck. Any hints?

Last edited:
What you stated as what you need to show would be sufficient if it were true but is not always true.
(for example $A_{11}A_{11}A_{11}$ is not always equal to $\varepsilon_{111}A_{11}A_{12}A_{13} ( = 0)$
The necessary condition is weaker and provable. Remember the common indices in your equations are summed over. The sums are equal not every term in them.

Lets see...
$$det(AB) = \varepsilon_{pqr} A_{px}B_{x1}A_{qy}B_{y2}A_{rz}B_{z3}$$
which must be shown to be equal to:
$$det(A)det(B) = \varepsilon_{p'q'r'}A_{p'1}A_{q'2}A_{r'3} \varepsilon_{uvw}B_{u1}B_{v2}B_{w3}$$

I think you're going to need to expand the sums here, or at least some of them. It isn't so bad as there are only six non-zero terms for the Levi-Civita tensors 3 indices.
Hmmm...
$$det(A)det(B) = \left[(A_{11}A_{22}A_{33} + A_{21}A_{32}A_{13} + A_{31}A_{12}A_{13})-(A_{21}A_{12}A_{33} + A_{11}A_{32}A_{23} A_{31}A_{22}A_{13})\right]det(B) = \cdots$$

Its that or do some heavy hitting with the symmetric equivalents of your equation, or manipulate explicitly stated sums over very specific cases. You might also find it useful to count terms and see how much combination/cancelation must occur.
$det(AB)$ has 6 iterations of the Levi-Citiva tensor indices times the 3x3x3 iterations of the three product contractions between A and B. That's 162 terms.
$det(A)det(B)$ on the other hand has just 6x6 = 36 terms. One concludes there's quite a bit of cancelation going on here.

Hmmm 162 terms...
$$det(AB) = A_{1x}B_{x1}A_{2y}B_{y2}A_{3z}B_{z3} + A_{2x}B_{x1}A_{3y}B_{y2}A_{1z}B_{z3} + A_{3x}B_{x1}A_{1y}B_{y2}A_{2z}B_{z3} - A_{2x}B_{x1}A_{1y}B_{y2}A_{3z}B_{z3} - A_{1x}B_{x1}A_{3y}B_{y2}A_{2z}B_{z3} - A_{3x}B_{x1}A_{2y}B_{y2}A_{1z}B_{z3}$$

 Try factoring some here:
$$det(AB) = A_{1x}B_{x1}\left[A_{2y}B_{y2}A_{3z}B_{z3} - A_{3y}B_{y2}A_{2z}B_{z3}\right] + A_{2x}B_{x1}\left[A_{3y}B_{y2}A_{1z}B_{z3} - A_{1y}B_{y2}A_{3z}B_{z3} \right]+ A_{3x}B_{x1}\left[A_{1y}B_{y2}A_{2z}B_{z3}- A_{2y}B_{y2}A_{1z}B_{z3} \right]$$
and continue from there?

hotvette said:

## Homework Statement

show that $\det(\underline{\bf{A}})\det(\underline{\bf{B}}) = \det(\underline{\bf{AB}})$

## Homework Equations

\begin{align*}
&\underline{\bf{A}} = A_{ij} \underline{e}_i \otimes \underline{e}_j \\
&\underline{\bf{B}} = B_{mn} \underline{e}_m \otimes \underline{e}_n \\
&\underline{\bf{A}}\underline{\bf{B}} = A_{ij}B_{mn} \underline{e}_i \otimes \underline{e}_n \\
& \det({\underline{\bf{A}}})=\epsilon_{pqr}A_{p1}A_{q2}A_{r3} \\
& \det({\underline{\bf{B}}})=\epsilon_{xyz}B_{x1}B_{y2}B_{z3} \\
& \det({\underline{\bf{A}}})(\det{\underline{\bf{B}}})=\epsilon_{pqr}A_{p1}A_{q2}A_{r3}\epsilon_{xyz}B_{x1}B_{y2}B_{z3}
\end{align*}

hotvette said:

## Homework Statement

show that $\det(\underline{\bf{A}})\det(\underline{\bf{B}}) = \det(\underline{\bf{AB}})$

## Homework Equations

\begin{align*}
&\underline{\bf{A}} = A_{ij} \underline{e}_i \otimes \underline{e}_j \\
&\underline{\bf{B}} = B_{mn} \underline{e}_m \otimes \underline{e}_n \\
&\underline{\bf{A}}\underline{\bf{B}} = A_{ij}B_{mn} \underline{e}_i \otimes \underline{e}_n \\
& \det({\underline{\bf{A}}})=\epsilon_{pqr}A_{p1}A_{q2}A_{r3} \\
& \det({\underline{\bf{B}}})=\epsilon_{xyz}B_{x1}B_{y2}B_{z3} \\
& \det({\underline{\bf{A}}})(\det{\underline{\bf{B}}})=\epsilon_{pqr}A_{p1}A_{q2}A_{r3}\epsilon_{xyz}B_{x1}B_{y2}B_{z3}
\end{align*}

## The Attempt at a Solution

\begin{align*}
&\underline{\bf{C}}={\underline{\bf{A}}}{\underline{\bf{B}}} \\
& \underline{\bf{C}} = A_{im}B_{mj} \underline{e}_i \otimes \underline{e}_j \\
& C_{ij} = A_{im}B_{mj} \\
& \det(\underline{\bf{AB}}) = \det(\underline{\bf{C}}) = \epsilon_{stu}C_{s1}C_{t2}C_{u3} \\
& C_{s1} = A_{sm}B_{m1} \\
& C_{t2} = A_{tn}B_{n2} \\
& C_{u3} = A_{uk}B_{k3} \\
& \det(\underline{\bf{AB}}) =\det(\underline{\bf{C}}) = \epsilon_{stu} A_{sm}A_{tn}A_{uk}B_{m1}B_{n2}B_{k3} \\
& \det(\underline{\bf{AB}}) =\det(\underline{\bf{C}}) = \epsilon_{pqr} A_{px}A_{qy}A_{rz}B_{x1}B_{y2}B_{z3}
\end{align*}

Which means I need to show that $A_{px}A_{qy}A_{rz}=\epsilon_{xyz} A_{p1}A_{q2}A_{r3}$. That's where I'm stuck. Any hints?

The definition you give for AB is different in sections 2 and 3, and I think that 3 gives it correctly. In that case, AB is basically the same as the matrix product of the two 3x3 matrices A and B, while the determinant is the same as the standard determinant of a matrix. So, you can look at the proof of det(AB) = det(A) * det(B) for matrices, and use the same type of argument.

Oops, yep I made a goof in (2). I've been able to work it out based on the above replies. Thanks!

## What is the concept of tensor determinants equality?

The concept of tensor determinants equality refers to the property of two tensors having equal determinants. This means that the determinant of one tensor can be obtained by multiplying the determinants of its components, and the same applies to the other tensor. This property is important in tensor algebra and is used in various mathematical and scientific applications.

## What is the significance of tensor determinants equality in physics?

In physics, tensor determinants equality is important because it helps in simplifying complex equations involving tensors. It also allows for the calculation of determinants of higher-dimensional tensors by breaking them down into lower-dimensional tensors. This concept is used in various fields of physics, such as electromagnetism, quantum mechanics, and general relativity.

## How is tensor determinants equality related to symmetry?

Tensor determinants equality is closely related to symmetry. A tensor is considered symmetric if its components are invariant under certain transformations. In such a case, the tensor determinants equality property holds, and the determinant of the tensor remains the same regardless of the transformation. This is because the determinants of symmetric tensors are equal to the products of the determinants of their invariant components.

## Can tensor determinants equality be applied to tensors of any dimension?

Yes, tensor determinants equality can be applied to tensors of any dimension. For example, the determinant of a 3-dimensional tensor can be obtained by multiplying the determinants of its 3x3 components. Similarly, the determinant of a 4-dimensional tensor can be calculated by multiplying the determinants of its 4x4 components, and so on.

## What are some real-life applications of tensor determinants equality?

Tensor determinants equality has various real-life applications, such as in computer graphics, image processing, and data compression. It is also used in engineering, particularly in the analysis of stress and strain in materials. In addition, this concept is also important in the study of fluid mechanics and other branches of physics.