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I understand that a linear transformation from a vector space $V$ to a vector space $W$ is a rank-$2$ tensor. What I would like some help with is how exactly to represent specific linear transformations in tensor notation. I will write below my attempt to link the two; please let me know if it makes sense, and provide whatever suggestions you think might help me.

So, how much, if any, of the following is correct?

Let $(v_1,\ldots,v_n)$ and $(w_1,\ldots,w_m)$ be bases of $V$ and $W$, respectively. Then $v_i \otimes w_j$ is the linear transformation that maps $v_i$ to $w_j$, and $\{v_i \otimes w_j:{i \in \{1,\ldots,n\}, j \in \{1,\ldots,m\}}\}$ is a basis of $V \otimes W$.

If some linear transformation $T$ has the matrix \begin{bmatrix} t_{1,1} & t_{1,2} & \ldots & t_{1,n}\\ t_{2,1} & t_{2,2} & \ldots & t_{2,n}\\ \vdots & \vdots & \ddots & \vdots \\ t_{m,1} & t_{m,2} & \ldots & t_{m,n} \end{bmatrix}

then $$T= \sum_{i \in \{1,\ldots,n\},\\ j \in \{1,\ldots,m\}}(t_{j,i} [v_i \otimes w_j])$$ is the tensor representation of $T$.

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how exactly to represent specific linear transformations in tensor notation.

Indeed every linear map $A:V\rightarrow W$ has a representation as a tensor $T\in V^{*}\otimes W$ such that $\forall x\in V:Ax=T(x)\in W$. In chosen bases,

$$ T=\sum a_{ij}\;v_j^*\otimes w_{i} $$

where $a_{ij}$ is the matrix of the linear map and $\{v_j^*\}$ is the basis in $V^*$ dual to $\{v_i\}$ so that each term is a map reading off the j-th coordinate of the $x$, $x\mapsto a_{ij}x^j\:w_i$.

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Let $E:=\mathbb{R}^n$ and $E^*$ its dual.

Here is a "concrete" way of seeing the equivalence of a linear transform $L:E \rightarrow E$ with a linear application from $L:E \otimes E^* \rightarrow \mathbb{R}$:

It is nothing more that the decomposition of matrix $M=[m_{i,j}]$ of linear mapping $L$ (with respect to canonical bases of $E$ and $E^*$) under the classical form:

$$L=\sum_{i=1}^n\sum_{j=1}^n m_{i,j}E_{i,j}=\sum_{i=1}^n\sum_{j=1}^n m_{i,j}U_i \otimes V_j^* \ \ \ \ (1)$$

where $E_{i,j}$ is the matrix with all its entries equal to $0$, but its $i,j$th entry equal to one, and

$$U_i=\begin{bmatrix}0\\\cdots\\1\\\cdots\\0\end{bmatrix} \in E \ \ and \ \ V_j=\begin{bmatrix}0&\cdots&1&\cdots&0\end{bmatrix} \in E^*$$

(for $U_i$, $i$th coefficient $\neq 0$, for $V_j$, $j$th coefficient $\neq 0$)

Remark: the tensor product could be interpreted either as a Kronecker product, or as an ordinary product by a column vector with a row vector (element of $E^*$)

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