As the title suggests. For example if I have a matrix $A = \begin{pmatrix} a & b\\ c& d \end{pmatrix}$ and all elements consist of variables with units $kg$ and then I take the inverse of the matrix is the resulting units simply $kg^{-1}$? How can this be the case if not all matrices have inverses?

Somewhat related to my other question about unit quantities in other matrix equations.


2 Answers 2


$ \def\m#1{\left[\begin{array}{c}#1\end{array}\right]} $The wonderful book $\,$Multidimensional Analysis (by George W Hart)$\,$ addresses this question in detail. It is surprisingly difficult to get it right.

For example, here is the dimensional sketch of a rectangular matrix and that of its pseudoinverse $$\eqalign{ A &= \m{ (m\cdot C^{-1}) & (m\cdot s\cdot K^{-1}) \\ (kg\cdot C^{-1}) & (kg\cdot s\cdot K^{-1}) \\ (m\cdot s^{-1}\cdot C^{-1})&(m\cdot K^{-1})} \\\\ A^{+} &= \m{ (C\cdot m^{-1}) & (C\cdot kg^{-1}) & (C\cdot s\cdot m^{-1}) \\ (K\cdot s^{-1}\cdot m^{-1}) & (K\cdot s^{-1}\cdot kg^{-1}) & (K\cdot m^{-1}) } \\\\ }$$ Matrices which are squareable must have a special dimensional structure, e.g. $$\eqalign{ B &= \m{ ({\tt1}) & (m\cdot s^{-1}) \\ (s\cdot m^{-1}) & ({\tt1}) } \quad\implies\quad B^2 &\overset{\Delta}{\;=\;} B \\ }$$ In this case, the diagonal elements carry no units, while the units of the other elements are the reciprocal of those in the transposed position. All powers of $B$ carry the same units.

Likewise, functions of $B$ such as the square root or exponential, carry the same units as $B$.

Note that for the rectangular matrix above $$\eqalign{ AA^+ &= \m{ ({\tt1}) & (m\cdot kg^{-1}) & (s) \\ (kg\cdot m^{-1}) & ({\tt1}) & (kg\cdot s\cdot m^{-1}) \\ (s^{-1}) & (m\cdot s^{-1}\cdot kg^{-1}) & ({\tt1}) } \\\\ A^+A &= \m{ ({\tt1}) & (C\cdot s\cdot kg^{-1}) \\ (kg\cdot s^{-1}\cdot C^{-1}) & ({\tt1}) \\ } \\ }$$ So these projection matrices are squareable.

Also, an identity matrix has no dimensions on its diagonal elements, but does carry dimensions in its off-diagonal elements. An equation like $(I+B)$ only makes sense if $I$ carries the same units as $B$.

So there is not one, but an infinite number of $2\times 2$ identity matrices when units are included.


As you know from the cofactor formula, the entries of the inverse matrix are homogeneous rational functions of degree $-1$ in the entries of the original matrix, so their unit will be $\mathrm{kg}^{-1}$.

As for your objection, I don't see your point. Are you fazed by the fact that $\sqrt{\bullet}$ is only defined for positive real numbers when you solve for time in, say, a uniformly accelerated motion?


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .