# Prove the determinant of a tensor is invariant

Given is a second-order tensor $T$, and three arbitrary vectors, $u$, $v$ and $w$, defined in Euclidean point space $\mathcal{E}$.

Prove that the determinant of the tensor $T$

$\det T=\frac{Tu.(Tv \times Tw)}{u.(v \times w)}$

is independent of $u$, $v$ and $w$, as long as they form a base in the 3-dimensional vector space $\mathcal{V}$ defined on $\mathcal{E}$ (that is, vectors $u$, $v$ and $w$ are linearly independent).

This is basically proving that the determinant of a tensor is invariant of the particular basis it happens to be resolved in.

I've tried to work through this using index notation, but it got pretty cumbersome quickly, so I thought there might be another way to do it. I'm not looking for the full treatment here, but rather as a general advice how to tackle this. I think I can manage the details myself.

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Since you use the cross and dot products in your question, you are working over an inner product space, therefore $\mathcal{V}$ is canonically identified with its dual. In particular this means that we can treat $T$, a rank-two tensor, as a linear transformation on $\mathcal{V}$. Therefore $T$ extends to a linear transformation on the exterior algebra of $\mathcal{V}$ and in particular you have a map $\tilde{T}:\Lambda^3V \to\Lambda^3V$. Using the inner-product, you can take the Hodge dual identifying $*: \Lambda^3V \to \mathbb{R}$ satisfying $** = 1$. So pre- and post-composing with $*$, you can extend $T$ to a linear map $\hat{T}:\mathbb{R}\to\mathbb{R}$ (which is just multiplication by a real number).

You can check that your expression is in fact

$$\det T = \frac{*[\tilde{T}(u\wedge v \wedge w)]}{*(u\wedge v \wedge w)} = \frac{\hat{T}s}{s}$$

which is clearly independent of the chosen basis of $\mathcal{V}$.

The statement in the end reflects the fact that the determinant of a linear transformation on a vector space represents the numerical factor by which the volume element in the vector space changes under the linear transformation.

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Well, one should observe that $u \cdot (v \times w)$ is equal to the determinant, in an oriented orthonormal basis, of the matrix U with $u, v, w$ as its columns. So, if we assume the determinant is multiplicative, we have $Tu \cdot (Tv \times Tw) = \det TU = \det T \det U = (\det T) u \cdot (v \times w)$, which is what we wanted to prove.