1.2.51 Type One One Tensor Definition

A Type One One Tensor maps vectors to scalars, forming a key element in tensor algebra.

Type One One Tensor Definition is the characterization of a tensor with exactly one contravariant factor and exactly one covariant factor, the case in which the type $(r, s)$ equals $(1, 1)$ . A type $(1, 1)$ tensor is, by definition, canonically identified with a linear map from a vector space to itself, so that ordinary linear operators and their matrix representations are themselves instances of tensors, forming the first example of a genuinely mixed tensor within the type-graded hierarchy.

Formal Definition

Let $V$ be a vector space over a field $F$ , with dual space $V^{*}$ . Setting $r = 1$ and $s = 1$ gives the tensor product space

T_{1}^{1} (V) = V \otimes V^{*}

An element of this space is a type $(1, 1)$ tensor, carrying one contravariant index and one covariant index, written in components as $T_{j}^{i}$ .

Canonical Identification with Linear Maps

The Isomorphism

For finite-dimensional $V$ , there is a canonical isomorphism

V \otimes V^{*} ≅ Hom (V, V)

sending an elementary tensor $v \otimes ω$ , for $v \in V$ and $ω \in V^{*}$ , to the rank-one linear map

u \mapsto ω (u) v

and extended by linearity to all of $V \otimes V^{*}$ . Under this identification, an arbitrary linear map on $V$ , decomposed as a sum of such rank-one pieces, corresponds to a general, not-necessarily-decomposable type $(1, 1)$ tensor.

Component Correspondence

Relative to a basis, this isomorphism identifies the components $T_{j}^{i}$ of a type $(1, 1)$ tensor exactly with the matrix entries of the corresponding linear map, with the upper index playing the role of the row index and the lower index the column index, so that the action of the map on a vector $u$ with components $u^{j}$ is given by ordinary matrix-vector multiplication,

{(T u)}^{i} = \sum_{j} T_{j}^{i} u^{j}

Transformation Law and the Trace

The components of a type $(1, 1)$ tensor transform under a change of basis $A$ by

{\tilde{T}}_{j}^{i} = \sum_{k, l} {(A^{-1})}_{k}^{i} A_{j}^{l} T_{l}^{k}

which is precisely the similarity transformation used in linear algebra to express how the matrix of a linear operator changes between bases. Contracting the single upper index with the single lower index,

tr (T) = \sum_{i} T_{i}^{i}

produces a type $(0, 0)$ tensor, a scalar, which is exactly the trace of the corresponding linear map. This confirms directly that the trace, ordinarily introduced as a basis-independent scalar associated to a matrix, is basis-independent precisely because it arises as a full contraction of a type $(1, 1)$ tensor.

Comparison with Type (2, 0) and Type (0, 2)

Type $(1, 1)$ tensors, along with type $(2, 0)$ and type $(0, 2)$ tensors, are all order-two tensors, but each is a distinct kind of order-two object with a different natural interpretation: type $(0, 2)$ tensors are bilinear forms, transforming by congruence, while type $(1, 1)$ tensors are linear operators, transforming by similarity. These two transformation behaviors are genuinely different, and only the availability of a metric tensor allows one to move between a bilinear form and a linear operator by raising or lowering the appropriate index.

Role Within Tensor Algebra

Type $(1, 1)$ tensors provide the tensor-algebraic home for the entire theory of linear operators on $V$ , unifying matrices, endomorphisms, and their invariants — trace, eigenvalues, and characteristic polynomial among them — within the same framework used for tensors of every other type. Because contraction between a single upper and single lower index is always available for a type $(1, 1)$ tensor, this type occupies a distinguished position as the simplest mixed type on which the fundamental tensor operation of contraction can be applied to produce a genuinely new, basis-independent scalar invariant.