Notations for vectors and matrices

May 20, 2024

I keep confusing myself when trying to keep track of the upper and lower indexes in matrices when using Einstein Notation. So here’s my reference.

Vectors

We’ll use index notation here for vectors, so summations are implicit.

\[u^i v_i \equiv \sum_i u^i v_i\]

We use a physicist’s notation for vectors, which treats the vector as a geometric object, independent of basis, which happens to have a particular expansion in any particular basis. The basis vectors (written, as physicists do, in boldface) are \(\b{x}_i\), written with lower indexes – lower because subscripts are the ‘default’, and so are used for basis vectors. Lower indexes are used for things which co-vary with the basis vectors (elements of a vector space \(V\)); upper indexes will be used for things which contra-vary (elements of a dual vector space \(V^*\)).

A vector \(\b{a}\) expands as a linear combination of basis elements, written:

\[\b{a} = a^i \b{x}_i = \begin{pmatrix} a^1 \\ a^2 \\ \vdots \\ a^{\| V \|} \end{pmatrix}\]

Confusingly, this means that the components of a vector are contravariant and written with upper indexes. Upper indexes in general correspond to varying rows when we write vectors out as matrices.

And a dual vector \(\b{b}^T \in V^*\) can be written as a sum over the dual basis \(\{ \b{x}^i \}\), and its components get lower indexes and identify columns:

\[\b{b}^T = b_i \b{x}^i = ( b_1 , b_2, \ldots, b_{\| V \|})\]

A dot product requires one of each, and when we write it with two vectors we are implicitly taking the transpose of its left argument, resulting in a matrix multiplication.

\[\b{b} \cdot \b{a} \equiv \b{b}^T \b{a} = (b_1, \ldots, b_n) \begin{pmatrix} a^1 \\ \vdots \\ a^n \end{pmatrix} = b_i a^i\]

It’s convenient to define that, if one of the arguments in the dot product is already a dual vector, we’ll skip the transposes, so \(\b{b}^T \cdot \b{a} = \b{b} \cdot \b{a}^T = b^i a_i\). At least, it’s not usually not that confusing?

The dot product may be seen as letting the basis components of each vector cancel out, with the effect of settings \(i=j\):

\[\b{b} \cdot \b{a} = (b^i \b{x}_i)^T (a^j \b{x}_j) = b_i a^j (\b{x}^i \cdot \b{x}_j) = b_i a^i\]

Matrices

So: a subscript index labels a vector component and a column, and a superscript index labels a covector component and a row. A matrix has one of each.

A matrix \(A: U \ra V\) can be written as a vector in the tensor product space \(U^* \otimes V\), and gets a lower index for its \(U^*\) component and an upper index for its \(V\) component:

\[A = A^j_i \b{u}^i \o \b{v}_j = \begin{pmatrix} A^1_1 & \cdots & A^1_i & \cdots & A^1_{\|U\|} \\ \vdots & & \vdots & & \vdots \\ A^j_1 & \cdots & A^j_i & \cdots & A^j_{\|U\|} \\ \vdots && \vdots && \vdots \\ A^{\|V\|}_1 & \cdots & A^{\|V\|}_i & \cdots & A^{\|V\|}_{\|U\|} \end{pmatrix}\]

\(A^j_i\) is \(i\)‘th column and the \(j\)‘th row, and it’s the scalar value given by \(A_i^j = \b{v}^j A \b{u}_i\).

Since we tend to write matrix multiplication with the \(U\) on the right side of \(A\), and the \(V\) on the left, it’s tempting to write this as \(A = \b{v}_j A_i^j \b{u}^i\).

It’s useful to refer to particular rows and columns of a matrix in isolation, by omitting one of the indexes. When we do this I like to write a vector symbol, \(\vec{A}\), to remind ourselves that we’re dealing with a vector. (Unfortunately it’s not common to write a vertical vector arrow, but that seems like it would be appropriate to distinguish column- vs row- vectors.) A lower index only, \(A_i\), means the \(i\)‘th column vector; an upper index only, \(A^j\), means the \(j\)‘th row vector. Therefore:

\[A = ( \vec{A}_1 \; \cdots \; \vec{A}_i \; \cdots \; \vec{A}_{\|U\|} ) = \begin{pmatrix} \vec{A}^1 \\ \vdots \\ \vec{A}^j \\ \vdots \\ \vec{A}^{\|V \|} \end{pmatrix}\]

We can think of a particular column of a matrix like \(\vec{A}^1\) as its action on a particular basis vector from \(U\): \(\vec{A}_1 = A(\b{u}_1) = A_1^j \b{v}_j\). Similarly a particular row is given by the action of \(A\) on a particular basis covector from \(V^*\): \(\vec{A}^1 = (\b{v}^1)^T A = A_i^1 \b{u}^i\).


A shorthand

I find it helpful to use \(a^x\) as the scalar value of a vector \(\b{a}\)’s \(\b{x}\) component, and to use \(\b{a}^x\) as a vector value, as

\[\b{a}^x = a^x \b{x} = (a \cdot x) \b{x}\]

and, in \(\bb{R}_3\):

\[\b{a} = \b{a}^x + \b{a}^y + \b{a}^z\]

Although I usually am not too concerned with upper and lower indexes and would just write \(\b{a}_x\).