Raising the bar

Matrix inversion: doing some housekeeping at elementary level

Any axiomatic treatment of the mathematical material should be preceded by a concrete, I would say naive, exploration.

P.S. Alexandrov (1896-1982)

What candidate would you suggest for the identity matrix?

Think simple, take a 2D case. Some students, even those who have taken matrix algebra, suggest

$latex I=\left(\begin{array}{cc}1&1\\1&1\end{array}\right).$

I suggest analyzing the defining property $latex AI=A$, that is

$latex \left(\begin{array}{cc}a_{11}&a_{12}\\a_{21}&a_{22}\end{array}\right)\left(\begin{array}{cc}
i_{11}&i_{12}\\i_{21}&i_{22}\end{array}\right)=\left(\begin{array}{cc}a_{11}&a_{12}\\a_{21}&a_{22}\end{array}\right).$

Upon multiplication this becomes

$latex \left(\begin{array}{cc}a_{11}i_{11}+a_{12}i_{21}&a_{11}i_{12}+a_{12}i_{22}\\
a_{21}i_{11}+a_{22}i_{21}&a_{21}i_{12}+a_{22}i_{22}\end{array}\right)=\left(\begin{array}{cc}
a_{11}&a_{12}\\a_{21}&a_{22}\end{array}\right).$

This should be true for ANY $latex a_{ij}.$ Selecting $latex a_{11}=0,$ $latex a_{12}=1$ and comparing elements in the upper left corner we see that $latex i_{21}=0.$ Similarly, equating the elements in the lower right corner, we get $latex i_{12}=0.$ The equation simplifies to

$latex \left(\begin{array}{cc}a_{11}i_{11}&a_{12}i_{22}\\a_{21}i_{11}&a_{22}i_{22}\end{array}
\right) =\left(\begin{array}{cc}a_{11}&a_{12}\\a_{21}&a_{22}\end{array}\right).$

This implies $latex i_{11}=i_{22}=1.$ Thus,

$latex I=\left(\begin{array}{cc}1&0\\0&1\end{array}
\right) .$

This easily generalizes to higher dimensions:

Definition. The identity matrix is a square matrix with unities on the main diagonal and zeros outside it. The above exercise does not free us from the necessity of proving $latex AI=A$ for all $latex A$, which I leave to the reader.

Invertibility of matrices

Definition. Let $latex A$ be a square matrix. Its inverse, denoted $latex A^{-1},$ satisfies by definition

(1) $latex AA^{-1}=A^{-1}A=I.$

We say that $latex A$ is invertible if its inverse $latex A^{-1}$ exists.

Remark. (1) includes two equations: $latex AA^{-1}=I$ (right inverse) and $latex A^{-1}A=I$ (left inverse). We need both because of absence of commutativity. In fact, the existence of the right inverse implies the existence of the left inverse, and vice versa. There are many facts equivalent to invertibility.

Surprise #2. The condition $latex A\neq 0$ is not sufficient for $latex A$ to be invertible.

Proof. Take

$latex A=\left(\begin{array}{cc}1&0\\0&0\end{array}\right)\neq 0.$

Suppose it's inverse exists and denote it

$latex B=\left(\begin{array}{cc}b_{11}&b_{12}\\b_{21}&b_{22}\end{array}\right).$

From the equation $latex AB=I$ we have

$latex \left(\begin{array}{cc}1&0\\0&0\end{array}\right)\left(\begin{array}{cc}b_{11}&b_{12}\\
b_{21}&b_{22}\end{array}\right)=\left(\begin{array}{cc}1&0\\0&1\end{array}\right).$

The element in the lower right corner gives $latex 0=1,$ which is impossible.

How to use definition (1)

Rephrasing (1): whenever you have a matrix $latex B$ such that $latex AB=BA=I$, you can conclude that $latex A^{-1}=B.$

Example 1 (analog of $latex (\frac{1}{a})^{-1}=a$, inverse of an inverse) $latex (A^{-1})^{-1}=A$. This is because by (1) $latex B=A$ satisfies $latex BA^{-1}=A^{-1}B=I.$

Example 2 (surprise #3, inverse of a product). If $latex A,B$ are invertible, then $latex (AB)^{-1}=B^{-1}A^{-1}$. To prove this, it suffices to check that the product $latex B^{-1}A^{-1}$ inverts $latex AB$. We do this just for the right inverse:

$latex (AB)(B^{-1}A^{-1})$               (by associativity)

$latex =A(BB^{-1})A^{-1}$             (by (1))

$latex =AIA^{-1}$                         (the identity matrix can be omitted)

$latex =AA^{-1}=I$                    (again by (1)) .

← Previous: Roadmap for studying matrix multiplication | Next: From invertibility to determinants: argument is more important than result →

Get the PDF book “Matrix Algebra” on Lulu