Inverse of a diagonal matrix
Let’s say we have a shape (2, 2) diagonal matrix:
\[\begin{split}\newcommand{A}{\boldsymbol A}
\newcommand{AI}{\boldsymbol A^{-1}}
\newcommand{L}{\boldsymbol L}
\newcommand{I}{\boldsymbol I}
\A =
\begin{bmatrix}
p & 0 \\
0 & q
\end{bmatrix}\end{split}\]
We want to find \(\AI\), the left inverse of \(\A\), such that:
\[\I = \AI \A\]
where \(\I\) is a shape (2, 2) identity matrix:
\[\begin{split}\I =
\begin{bmatrix}
1 & 0 \\
0 & 1
\end{bmatrix}\end{split}\]
As always when trying to find the inverse, we are solving a system of
simultaneous equations. In the case of a diagonal matrix, the equations are
easier to solve because of the zeros off the diagonal.
Write \(\AI\) as:
\[\begin{split}\AI =
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}\end{split}\]
From the definition of matrix multiplication, we now have:
\[\begin{split}\I = \AI \A =
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
\begin{bmatrix}
p & 0 \\
0 & q
\end{bmatrix}
=
\begin{bmatrix}
a p + b 0 & a 0 + b q \\
c p + d 0 & c 0 + d q
\end{bmatrix}\end{split}\]
Comparing \(\I\) with this result, we have:
\[\begin{split}1 = a p + b 0 \\
0 = a 0 + b q \\
0 = c p + d 0 \\
1 = c 0 + d q\end{split}\]
From the central two equations, \(b = c = 0\). Substituting, we have:
\[\begin{split}1 = a p \implies a = \frac{1}{p} \\
1 = d q \implies d = \frac{1}{q} \\\end{split}\]
\[\begin{split}\AI =
\begin{bmatrix}
\frac{1}{p} & 0 \\
0 & \frac{1}{q}
\end{bmatrix}\end{split}\]
This result generalizes for any shape (\(n, n\)) diagonal matrix. To see this,
consider a shape (3, 3) or larger diagonal matrix, its inverse and the
identity. Each zero in the identity matrix requires a zero in the
corresponding position in the inverse, otherwise the corresponding row /
column dot product will pick up a non-zero value from the diagonal.
\[\begin{split}\boldsymbol D =
\begin{bmatrix}
d_1 & 0 & 0 & ... & 0 \\
0 & d_2 & 0 & ... & 0 \\
0 & 0 & d_3 & ... & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & ... & d_n
\end{bmatrix}\end{split}\]
\[\begin{split}\boldsymbol D^{-1} =
\begin{bmatrix}
\frac{1}{d_1} & 0 & 0 & ... & 0 \\
0 & \frac{1}{d_2} & 0 & ... & 0 \\
0 & 0 & \frac{1}{d_3} & ... & 0 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & ... & \frac{1}{d_n}
\end{bmatrix}\end{split}\]