Introduction to dimensioned linear algebra

This document provides a brief introduction to working with dimensioned scalars, vectors, and matrices. It is based on and somewhat summarizes sections 2.4–5 of Multidimensional Analysis¹.

It should be clear that sums of dimensioned quantities are not always defined. Letting $\text m$ denote meters and $\text s$ denote seconds, we see that $$ 1 \text{ m} + 2 \text{ m} = 3 \text{ m}, $$ but $$ 1 \text{ m} + 2 \text{ s} = \;? $$

In what follows we'll be concerned with the physical dimensions of scalars, vectors, and matrices, not their numeric components, so while the definitions are stated while working with real numbers $(\mathbb R)$, we could as easily work with complex numbers $(\mathbb C)$.

Terminology and notation

Denote the physical dimension of a scalar with $\sim$; for example, if $a$ represents a quantity of meters, write $a \sim \text{meters}$. Thus, if $b$ is some other quantity, $a + b$ is defined exactly when $a \sim b$. In this case we say $a$ and $b$ have the same dimensional form. We write the dimensionless quantity as simply $1$.

Additionally, we say $a$ and $b$ are dimensionally parallel if there is some dimensioned scalar $c$ such that $a \sim c b$. We write this as $a \approx b$.

It'll also be useful to define the dimensional inverse of a scalar, denoted $a^\sim$ and defined by $a^\sim a \sim a a^\sim \sim 1$. In words, the product of a scalar and its dimensional inverse is dimensionless. For example, if $a \sim \text{ m} \cdot \text{s}^{-1}$, then $a^\sim \sim \text{ m}^{-1} \cdot \text{s}$.

Vectors

When is a dot product between (column) vectors defined? Assuming $$ \mathbf a = \begin{bmatrix} \mathbf a_1 \\ \mathbf a_2 \\ \vdots \\ \mathbf a_n \end{bmatrix} \in \mathbb R^n $$ (and similar for $\mathbf b$), the dot product $$ \mathbf a \cdot \mathbf b = \mathbf a^\top \mathbf b = \sum_{i=1}^n \mathbf a_i \mathbf b_i $$ is defined exactly when $\mathbf a_i \mathbf b_i \sim \mathbf a_j \mathbf b_j$ for all $i, j$, or equivalently, there is some dimensioned scalar $c$ such that $\mathbf a_i \mathbf b_i \sim c$ for all $i$ (and in which case $\mathbf a^\top \mathbf b \sim c$). This is true exactly when $\mathbf a_i \sim c \mathbf b_i^\sim$, (i.e., $\mathbf a_i \approx \mathbf b_i^\sim$) for all $i$. (Note it is also true that $\mathbf a_i \approx \mathbf b_i$.)

We may extend the definitions of having the same dimensional form and being dimensionally parallel to vectors by requiring that, for two vectors (of the same shape), their corresponding components have the same dimensional form or are dimensionally parallel, respectively.
(Later we will extend this component-wise definition to matrices as well.)

We also extend the definition of dimensional inverse: $ \mathbf a^\sim \coloneqq \begin{bmatrix} \mathbf a_1^\sim & \cdots & \mathbf a_n^\sim \end{bmatrix} $. Note that the shape is transposed, so the dimensional inverse of a column vector is a row vector. It follows that $\mathbf a^\sim \mathbf a \sim 1$.

Then we can restate the condition for two vectors to have a dot product: $\mathbf a^\top \mathbf b$ is defined exactly when $\mathbf a \approx \mathbf b^{\sim \top}$.

Examples of vector dot products

In the first example, observe that $\mathbf a \sim \text{m s } \cdot \mathbf b^{\sim \top}$ (so $\mathbf a \approx \mathbf b^{\sim \top}$), and in the second example $\mathbf a \sim 1 \cdot \mathbf b^{\sim \top}$, but in the third example there is no dimensioned scalar $c$ such that $\mathbf a \sim c \mathbf b^{\sim \top}$.

\[ \begin{align*} \mathbf a &= \begin{bmatrix} 1 \text{ m} \\ 2 \text{ m} \\ 3 \text{ m} \end{bmatrix} \\ \mathbf b &= \begin{bmatrix} 1 \text{ s} \\ 2 \text{ s} \\ 3 \text{ s} \end{bmatrix} \\ \mathbf a^\top \mathbf b &= 1 \text{ m} \cdot \text{s} + 4 \text{ m} \cdot \text{s} + 9 \text{ m} \cdot \text{s} \\ &= 14 \text{ m} \cdot \text{s} \\ \\ \mathbf a &= \begin{bmatrix} 1 \text{ m}^1 \\ 2 \text{ m}^2 \\ 3 \text{ m}^3 \end{bmatrix} \\ \mathbf b &= \begin{bmatrix} 1 \text{ m}^{-1} \\ 2 \text{ m}^{-2} \\ 3 \text{ m}^{-3} \end{bmatrix} \\ \mathbf a^\top \mathbf b &= 1 \text{ m}^1 \cdot \text{m}^{-1} + 4 \text{ m}^2 \cdot \text{m}^{-2} + 9 \text{ m}^3 \cdot \text{m}^{-3} \\ &= 14 \\ \\ \mathbf a &= \begin{bmatrix} 1 \text{ m}^1 \\ 2 \text{ m}^2 \\ 3 \text{ m}^3 \end{bmatrix} \\ \mathbf b &= \begin{bmatrix} 1 \text{ m}^{-1} \\ 2 \text{ m}^{-1} \\ 3 \text{ m}^{-1} \end{bmatrix} \\ \mathbf a^\top \mathbf b &= 1 \text{ m}^1 \cdot \text{m}^{-1} + 4 \text{ m}^2 \cdot \text{m}^{-1} + 9 \text{ m}^3 \cdot \text{m}^{-1} \\ &= 1 + 4 \text{ m}^1 + 9 \text{ m}^2 \\ &= \; ? \end{align*} \]

Matrices

We're finally ready to discuss dimensioned matrices. First recall from Vectors how we defined two vectors having the same dimensional form or being dimensionally parallel in terms of corresponding components. We use these same component-wise definitions for matrices, and note now that vectors are a special case of matrices. Likewise the dimensional inverse of a matrix is obtained by inverting the dimensions of components and then taking the transpose of the matrix.

When is the product of two matrices defined? Again assuming $\mathbf a_{(i)}, \mathbf b_{(j)} \in \mathbb R^n$ for all $i, j$, we'll define two matrices $\mathbf A$ and $\mathbf B$ in terms of the $\{\mathbf a_{(i)}\}$ and $\{\mathbf b_{(j)}\}$:

\[ \begin{align*} \mathbf A &= \begin{bmatrix} \mathbf a_{(1)}^\top \\ \vdots \\ \mathbf a_{(m)}^\top \end{bmatrix} \in M_{m \times n} \\ \\ \mathbf B &= \begin{bmatrix} \mathbf b_{(1)} & \cdots & \mathbf b_{(k)} \end{bmatrix} \in M_{n \times k}. \end{align*} \]

Then the $(i, j)$ entry of the product $\mathbf A \mathbf B$ is

\[ [\mathbf A \mathbf B]_{i, j} = \mathbf a_{(i)}^\top \mathbf b_{(j)} \]

which we know is defined exactly when $\mathbf a_{(i)} \approx \mathbf b_{(j)}^{\sim \top}$. For a fixed $j$, this must be true for all $i$, so $\mathbf b_{(1)}^{\sim \top} \approx \mathbf a_{(1)} \approx \cdots \approx \mathbf a_{(m)}$. So all the rows of $A$ are dimensionally parallel, and we can write $\mathbf A$ as an outer product

\[ \mathbf A \sim \begin{bmatrix} 1 \\ c_2 \\ \vdots \\ c_m \end{bmatrix} \begin{bmatrix} \text{the first row of } \mathbf A \end{bmatrix} = \mathbf c \mathbf a_{(1)}^\top \]

where $ \mathbf c = \begin{bmatrix} 1 & c_2 & \cdots & c_m \end{bmatrix}^\top $ is some dimensioned column vector.

The important point is that for $\mathbf A$ to be multiplied on the right by a matrix, its dimensions must take the form of an outer product. It is useful to write this as $\mathbf A \sim \mathbf u \mathbf v^\sim$ for some $\mathbf u, \mathbf v$. We may similarly find that for $\mathbf B$ to be multiplied on the left by a matrix, we must have $\mathbf B \sim \mathbf x \mathbf y^\sim$ for some $\mathbf x, \mathbf y$.

So the condition for a matrix to be multipliable is that its dimensions take the form of an outer product, i.e., $\mathbf A \sim \mathbf u \mathbf v^\sim$, and in this case it is multipliable on both the left and the right. We refer to these two vectors $\mathbf u, \mathbf v$ as dimension vectors for $\mathbf A$, but note that they are not unique. For example, if $c$ is any non-zero dimensioned scalar,

\[ \begin{align*} \mathbf A &\sim \mathbf u \mathbf v^\sim \\ &\sim (\mathbf u / c) (c \mathbf v^\sim) \\ &\sim (\mathbf u / c) (\mathbf v / c)^\sim. \end{align*} \]

Finally, for two matrices to be multiplied together, we find

\[ \begin{align*} \mathbf A \mathbf B &\sim (\mathbf u \mathbf v^\sim)(\mathbf x \mathbf y^\sim) \\ &\sim \mathbf u (\mathbf v^\sim \mathbf x) \mathbf y^\sim \end{align*} \]

which is defined exactly when $\mathbf x \approx \mathbf v$ (since $\mathbf (\mathbf v^\sim \mathbf x)^\top = \mathbf x^\top \mathbf v^{\sim \top}$).

Now we have condition for two matrices to be multiplied: to compute $\mathbf A \mathbf B$ with $\mathbf A \sim \mathbf u \mathbf v^\sim$ and $\mathbf B \sim \mathbf x \mathbf y^\sim$, we must have $\mathbf x \approx \mathbf v$.

Hart, G. W. (1995). Multidimensional Analysis: Algebras and Systems for Science and Engineering. Springer-Verlag. ↩