Show Tensor Determinants Equality

[hotvette]

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?

 

[jambaugh]

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

[Edit] 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?

 

[Ray Vickson]

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

Are you sure about your given definition of $\underline{\bf{A}} \underline{\bf{B}}$?

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.

 

[hotvette]

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