Mechanics of Continua and Structures
Definition: Let $\boldsymbol{I}: \mathbb{E}^N\to \mathbb{E}^N$ be such that
\(\begin{align} \boldsymbol{I}\boldsymbol{v}=\boldsymbol{v}, \end{align}\) for all \(\boldsymbol{v}\in \mathbb{E}^N\). Then $\boldsymbol{I}$ is called the identity tensor.
Let $(\hat{\boldsymbol{e}}_i)$ be an orthornormal basis for $\mathbb{E}^N$.
From the definition of $\boldsymbol{I}$ it follows that \(\begin{align} \boldsymbol{I}\hat{\boldsymbol{e}}_1=\hat{\boldsymbol{e}}_1, \boldsymbol{I}\hat{\boldsymbol{e}}_2=\hat{\boldsymbol{e}}_2, \boldsymbol{I}\hat{\boldsymbol{e}}_3=\hat{\boldsymbol{e}}_3. \end{align}\)
Then the components of the identity tensor w.r.t $(\hat{\boldsymbol{e}}_i)$ are:
\[\begin{align} \mathit{I_{ij}}= \hat{\boldsymbol{e}}_i\cdot\boldsymbol{I}\hat{\boldsymbol{e}}_j=\hat{\boldsymbol{e}}_i\cdot\hat{\boldsymbol{e}}_j=\mathit{\delta_{ij}}, \end{align}\]that is,
\[\boldsymbol{I}_{\bullet\bullet} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}=:\usf{I}.\]It is obvious that the identity matrix is the matrix representation of $\boldsymbol{I}$ with respect to all orthornormal bases, and that \(\boldsymbol{T}\boldsymbol{I}=\boldsymbol{I}\boldsymbol{T}=\boldsymbol{T}\) for any tensor \(\boldsymbol{T}\). We also note that if \(\boldsymbol{T}\boldsymbol{a} = \boldsymbol{a}\) for any arbitrary \(\boldsymbol{a}\), then \(\boldsymbol{T}=\boldsymbol{I}\).
Let $(\boldsymbol{e}_i)$ and $(\boldsymbol{f}_i)$, respectively, be bases for $\mathbb{E}^N$ and $\mathbb{F}^N$. Let $\boldsymbol{I}\in \mathcal{L}(\mathbb{E}^N,\mathbb{F}^N;\mathbb{R})$ be defined as $I(\boldsymbol{e}_i)=\boldsymbol{f}_i$. In the context of $\mathcal{L}(\mathbb{E}^N,\mathbb{F}^N;\mathbb{R})$ the operator $\boldsymbol{I}$ defined this way can be though of as Identity tensor.
As can be noted from the definition there does not exist a unique indentity tensor when $\mathbb{E}$ and $\mathbb{F}$ are different.