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Proving the det(AB) = det(A)det(B) relation using the Levi-Cevita tensor

I recently had to prove that $$\det(\mathbf{A}\mathbf{B}) = \det(\mathbf{A})\det(\mathbf{B})$$

I thought that it would be an interesting exercise to show this using the standard definition of the determinant of a matrix that physicists usually give (using Einstein summation):

$$\det(\mathbf{A}) \triangleq \varepsilon^{i_{1}\cdots i_{n}} a_{1 i_{1}} \cdots a_{n i_{n}}$$

We note that if we denote $\mathbf{C} = \mathbf{A}\mathbf{B}$, then we have $c_{ij} = a_{ik}b_{kj}$, and thus:

$$\begin{align*}\det(\mathbf{C}) &= \varepsilon^{i_{1}\cdots i_{n}}c_{1 i_{1}}\cdots c_{n i_{n}} \\
&= \varepsilon^{i_{1}\cdots i_{n}} a_{1 k_{1}}b_{k_{1} i_{1}} \cdots a_{n k_{n}}b_{k_{n} i_{n}} \\
&= a_{1 k_{1}}\cdots a_{n k_{n}} \varepsilon^{i_{1} \cdots i_{n}}b_{k_{1} i_{1}}\cdots b_{k_{n} i_{n}}\end{align*}$$

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