representing tensors with rank $\le$ 2 as vectors and matrices

Question

Why is this tensor expression:

$$U_{rs} \bar{J}^r_i \bar{J}^s_j$$

Equivalent to this matrix expression:

$$\bar{J}^T U \bar{J}$$

Answer 1

How to represent tensors (rank $\le$ 2) as vectors and matrices:

Matrix-Matrix product

$$\begin{matrix}AB \equiv a^i_r b^r_j & AB \equiv a_{ir}b_{rj} & AB \equiv a^{ir}b^{rj}\end{matrix}$$

$$\begin{matrix}BA \equiv a^r_i b^j_r & BA \equiv a_{ri}b_{jr} & BA \equiv a^{ri}b^{jr}\end{matrix}$$

Vector-Matrix product

$$\begin{matrix}u^T A \equiv u_r A^r_i & u^T A \equiv u_r A_{ri} & u^T A \equiv u^r A^{ri}\end{matrix}$$

$$\begin{matrix} Au \equiv u_r A^i_r & Au \equiv u_r A_{ir} & Au \equiv u^r A^{ir}\end{matrix}$$

Vector-Vector products

$$\begin{matrix}x^T y \equiv x_i y_i & \text{(scalar)}\end{matrix}$$

$$\begin{matrix}x^T A~ x \equiv a_{ij} x_i x_j & \text{(scalar)}\end{matrix}$$

Matrix Transpose

$$\begin{matrix}A \equiv a^i_j \rightarrow A^T \equiv \dot{a}^j_i \\ A \equiv a_{ij} \rightarrow A^T \equiv \dot{a}_{ji}\\ A \equiv a^{ij} \rightarrow A^T \equiv \dot{a}^{ji}\end{matrix}$$

The dot over the "a" is for book keeping to show that A is transposed if you convert the tensor back to matrix form, however, the dot is not needed if the tensor remains a tensor. For instance, $\dot{a}_{ji} = a_{ji}$ from a tensor point of view.

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$$U_{rs} \bar{J}^r_i \bar{J}^s_j = \bar{J}^r_i U_{rs} \bar{J}^s_j $$

$$U_{rs} \bar{J}^r_i \bar{J}^s_j = \dot{\bar{J}}^i_r U_{rs} \bar{J}^s_j $$

$$U_{rs} \bar{J}^r_i \bar{J}^s_j \equiv \bar{J}^T U \bar{J}$$