Suppose there is a matrix $A$ that transforms vectors, $$ Y = A x $$ Now express this in some other coordinate system, with $x = B z, \,\, y = B w$, so \begin{align*} & Bw = A B z \\ \Rightarrow & w = B^{-1} A B z \end{align*} So $A$ expressed in the other system is $B^{-1} A B$.

What would be the equivalent in tensor notation, in particular of the $B^{-1}$? Here's what I'm trying \begin{align*} & y^i = A^i_j x_j \\ & \quad\quad x^j = B^j_k z^k \\ & \quad\quad y^i = B^i_m w^m \\ \text{so}\quad & B^i_m w^m = A^i_j B^j_k z_k \end{align*} Now what is the tensor equivalent of premultiplying by $B^{-1}$ on the left, in order to find what $A$ looks like in tensor notation in the new coordinate system?


2 Answers 2


Using a basis, you could use the common method of calculation given by $$A^{-1} = \frac{\text{adj}(A)}{\det(A)} $$ where $\det$ is the determinant, and $\text{adj}$ is the adjugate, i.e. the transpose of the matrix of cofactors $\text{cof}(A)$.

One can show that $$ {(\text{adj}(A))^{i}}_{j} = \frac{\partial \det(A)}{\partial {A^{j}}_{i}}. $$ In order to calculate this, we can use the expression for the determinant in terms of the generalized Kronecker delta $$\det(A) = \frac{1}{n!}\delta^{a_1\dots a_n}_{b_1\dots b_n}{A^{b_1}}_{a_1}\cdots{A^{b_n}}_{a_n}$$ where $n$ is the dimension of the vector space.

Hence the adjugate is

\begin{align*} {(\text{adj}(A))^{a}}_{b} &= \frac{\partial}{\partial {A^{b}}_{a}} \left( \frac{1}{n!}\delta^{c_1\dots c_n}_{d_1\dots d_n}{A^{d_1}}_{c_1}\cdots{A^{d_n}}_{c_n} \right)\\ &= \frac{1}{n!} \delta^{c_1\dots c_n}_{d_1\dots d_n} \frac{\partial}{\partial {A^{b}}_{a}} \left({A^{d_1}}_{c_1}\cdots{A^{d_n}}_{c_n} \right) \end{align*}

If we perform the calculation (for instance, in $n=3$) we get $$\begin{align} {(\text{adj}(A))^{a}}_{b} &= \frac{1}{3!} \delta^{ijk}_{lmn} \frac{\partial}{\partial {A^{b}}_{a}} \left( {A^{l}}_{i}{A^{m}}_{j}{A^{n}}_{k} \right)\\ &= \frac{1}{3!} \delta^{ijk}_{lmn} \left( \delta^{l}_{b}\delta^{a}_{i}{A^{m}}_{j}{A^{n}}_{k} + \delta^{m}_{b}\delta^{a}_{j}{A^{l}}_{i}{A^{n}}_{k} + \delta^{n}_{b}\delta^{a}_{k}{A^{l}}_{i}{A^{m}}_{j} \right)\\ &= \frac{1}{3!} \left( \delta^{ijk}_{lmn} \delta^{l}_{b}\delta^{a}_{i}{A^{m}}_{j}{A^{n}}_{k} + \delta^{ijk}_{lmn} \delta^{m}_{b}\delta^{a}_{j}{A^{l}}_{i}{A^{n}}_{k} + \delta^{ijk}_{lmn} \delta^{n}_{b}\delta^{a}_{k}{A^{l}}_{i}{A^{m}}_{j} \right)\\ &= \frac{1}{3!} \left( \delta^{ajk}_{bmn} {A^{m}}_{j}{A^{n}}_{k} + \delta^{iak}_{lbn} {A^{l}}_{i}{A^{n}}_{k} + \delta^{ija}_{lmb} {A^{l}}_{i}{A^{m}}_{j} \right)\\ &= \frac{1}{2!}\delta^{aij}_{bmn} {A^{m}}_{i}{A^{n}}_{j} \end{align}$$

hence the inverse (in the $n=3$ case) is

$${(A^{-1})^{a}}_{b} = \frac{{(\text{adj}(A))^{a}}_{b}}{\det(A)} = \frac{\frac{1}{2!}\delta^{aij}_{bmn} {A^{m}}_{i}{A^{n}}_{j}} {\frac{1}{3!} \delta^{cde}_{fgh}{A^{f}}_{c}{A^{g}}_{d}{A^{h}}_{e} } = 3\frac{\delta^{aij}_{bmn} {A^{m}}_{i}{A^{n}}_{j}} {\delta^{cde}_{fgh}{A^{f}}_{c}{A^{g}}_{d}{A^{h}}_{e} }$$

You can check that the $n$-dimensional adjugate is $${(\text{adj}(A))^{a}}_{b} = \frac{1}{(n-1)!}\delta^{ac_1\dots c_{n-1}}_{bd_1\dots d_{n-1}} {A^{d_1}}_{c_1}\cdots{A^{d_{n-1}}}_{c_{n-1}}$$

and finally we get the $n$-dimensional inverse $${(A^{-1})^{a}}_{b} = \frac{{(\text{adj}(A))^{a}}_{b}}{\det(A)} = \frac{\frac{1}{(n-1)!}\delta^{ac_1\dots c_{n-1}}_{bd_1\dots d_{n-1}} {A^{d_1}}_{c_1}\cdots{A^{d_{n-1}}}_{c_{n-1}}} {\frac{1}{n!}\delta^{e_1\dots e_n}_{f_1\dots f_n}{A^{f_1}}_{e_1}\cdots{A^{f_n}}_{e_n}} = n\frac{\delta^{ac_1\dots c_{n-1}}_{bd_1\dots d_{n-1}} {A^{d_1}}_{c_1}\cdots{A^{d_{n-1}}}_{c_{n-1}}} {\delta^{e_1\dots e_n}_{f_1\dots f_n}{A^{f_1}}_{e_1}\cdots{A^{f_n}}_{e_n}}$$

Note that all expressions (apart from those with the partial derivative) contain exclusively components of tensors. In these cases we can safely assume that the whole expression is a tensor as well, and we can interpret the indices not as components but as abstract tensor indices.


The tensor equivalent is simply:

$$ (B^{-1})^i_j $$

defined to satisfy

$$ (B^{-1})^i_j B^j_k = \delta^i_k \qquad B^i_j (B^{-1})^j_k = \delta^i_k $$.

So you write

$$ w^m = (B^{-1})^m_i A^i_j B^j_k z^k $$

Though, I would really, really just advice you to use abstract index notation instead of the concrete tensor notation you seem to be using. The advantage is that you treat $z$ now as an element in your vector space $V$ and not as the coordinate representation, and $A$ the linear mapping from $V$ to itself instead of its matrix representation, so you don't have to worry about matrices coming from "changes of variables".

  • 1
    $\begingroup$ I will have to accept tis answer, but it is disappointing: the introductions to tensors seem to say that the notation helps express and keep track of what happens with a change of basis, and here it does not seem to help at all. $\endgroup$
    – beginner
    May 23, 2014 at 8:50
  • 1
    $\begingroup$ @beginner: read the part about abstract index notation. When working with tensors you should almost never explicitly express them relative to a basis. By even starting to think about the matrix $B$ you already set yourself on the wrong path. The geometrically correct thing is that "there is no such thing as a change of basis matrix $B$". $\endgroup$ May 23, 2014 at 8:53

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