Inverse of Diagonal Matrix

Theorem

Let:

$\mathbf D = \begin{bmatrix} a_{11} & 0 & \cdots & 0 \\ 0 & a_{22} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & a_{nn} \\ \end{bmatrix}$

be an $n \times n$ diagonal matrix.

Then its inverse is given by:

$\mathbf D^{-1} = \begin{bmatrix} \dfrac 1 {a_{11}} & 0 & \cdots & 0 \\ 0 & \dfrac 1 {a_{22}} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \dfrac 1 {a_{nn}} \\ \end{bmatrix}$

provided that none of the diagonal elements are zero.

If any of the diagonal elements are zero, $\mathbf D$ is singular.

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Proof

Without loss of generality, consider the right inverse of $\mathbf D$.

Suppose none of the diagonal elements are zero.

Then by the definition of inverse, our assertion is that the matrix product of the two matrices in question is the unit matrix of order $n$.

Now, observe that:

\(\ds \begin {bmatrix} a_{11} & 0 & \cdots & 0 \\ 0 & a_{22} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & a_{nn} \\ \end {bmatrix} \begin {bmatrix} \frac 1 {a_{11} } & 0 & \cdots & 0 \\ 0 & \frac 1 {a_{22} } & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \frac 1 {a_{nn} } \\ \end {bmatrix}\)

\(=\)

\(\ds \begin{bmatrix} \frac{a_{11} }{a_{11} } & 0 & \cdots & 0 \\ 0 & \frac{a_{22} }{a_{22} } & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \frac{a_{nn} }{a_{nn} } \\ \end {bmatrix}\)

\(\ds \)

\(=\)

\(\ds \begin {bmatrix} 1 & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \\ \end {bmatrix}\)

\(\ds \)

\(=\)

\(\ds \mathbf I\)

$\Box$

Now suppose one of the diagonal elements is zero.

Then $\map \det {\mathbf D} = 0$, where $\det$ indicates the determinant of $\mathbf D$.

From Determinant of Inverse Matrix it would follow that:

$\map \det {\mathbf D^{-1} } = \dfrac 1 {\map \det {\mathbf D} }$

But this equation has no solution, and so $\mathbf D$ cannot admit an inverse.

$\blacksquare$