Recipe

Classify a tensor's type and transform its components

Use when: you're given a tensor, or its transformation rule under a basis change $P$ , and need its $(p,q)$ type or its correctly transformed components

Goal: the tensor's $(p,q)$ type, and its components transformed correctly using $P^{-1}$ for each contravariant factor and $P^{\mathsf T}$ for each covariant factor

Taught in Covariant, contravariant, and mixed tensors

Quick recipe

Identify how the tensor is built or how its rule transforms
Report the tensor's type as (p, q)
Transform each contravariant factor using P^{-1}
Transform each covariant factor using P^T

1
Look at how the tensor is described: either as an object built from copies of $V$ and $V^{*}$ , or as a transformation rule already written in terms of $P$ and $P^{-1}$ .

Example

$T'=P^{-1}TP$ has exactly one $P^{-1}$ factor and one $P$ factor.
- given as a named object — recall its known type directly: vector $(1,0)$ , functional $(0,1)$ , bilinear form $(0,2)$ , operator $(1,1)$ .
- given only as a transformation rule, such as $T'=P^{-1}TP$ — count how many factors of $P^{-1}$ (contravariant) and how many factors of $P$ or $P^{\mathsf T}$ (covariant) appear.
Why: the $(p,q)$ type is defined entirely by how many contravariant and covariant factors build the tensor, and both routes recover the same count.
2
Write $p$ (the count of contravariant factors) first, then $q$ (the count of covariant factors), as the ordered pair $(p,q)$ .

Example

One $P^{-1}$ factor and one $P$ factor gives type $(1,1)$ , a mixed tensor, exactly a linear operator's own transformation rule.

Why: naming the type is what tells you, in the next steps, exactly which transformation rule each factor of the tensor's components must follow.
3
For every index built from a copy of $V$ , transform that factor's components by left-multiplying with $P^{-1}$ , the same rule an ordinary vector follows.

Why: a contravariant index is, by definition, one that transforms exactly the way a vector's own components do under a basis change.
4
For every index built from a copy of $V^{*}$ , transform that factor's components by multiplying with $P^{\mathsf T}$ , the same rule a covector follows.

Example

For covector $\varphi=(2,3)$ and $P=\begin{bmatrix}1&1\\0&1\end{bmatrix}$ : $P^{\mathsf T}=\begin{bmatrix}1&0\\1&1\end{bmatrix}$ , so $P^{\mathsf T}\varphi=(2,5)$ .

Why: a covariant index transforms "with" the basis change, opposite a contravariant one, exactly the rule chapter 03's dual space forced onto a linear functional's own components.

Common mistakes

Applying $P$ instead of $P^{-1}$ to a contravariant factor, or vice versa, swapping the two rules.
Forgetting that a $(0,2)$ tensor needs $P^{\mathsf T}$ applied on BOTH sides, not just one.
Reading off $(p,q)$ backwards, reporting covariant factors before contravariant ones.
Assuming a rule with two $P$ -type factors is symmetric in how they're computed, when multiplication order still matters.