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Matrix inversion.

Proposition 7.2.1   Let $\phi : U \longrightarrow V$ be a linear mapping. If this mapping is invertible, then $\phi^{-1}: V \longrightarrow U$ is a linear mapping.

Example 7.2.2       

As a consequence of  1.6, we have:

If $\phi : \; U \longrightarrow V$ is invertible, then $\dim U = \dim V$ (eventually infinite).
Thus, the matrix (w.r.t. any basis) of an invertible linear application is a square matrix. Denote by A the matrix of $\phi$ w.r.t. some bases of U and V; the matrix of $\phi^{-1}$ is A-1.

How to compute A-1?

The sequence of elementary operations which transforms the matrix A into the matrix I transforms also the matrix I into the matrix A-1.

\fbox{
$\begin{matrix}A\\ I \end{matrix} \;
\overrightarrow{\underrightarrow{\text{elementary operations}}}
\;\begin{matrix}I\\ A^{-1} \end{matrix}$ }

Example 7.2.3   $A=\begin{pmatrix}1 & 1 & 1 \\ 1 & 1 & -1 \\ 1 & -1 & 1 \end{pmatrix}$.

$\underbrace{\begin{matrix}1 & 1 & 1 \\ 1 & 1 & -1 \\ 1 & -1 & 1 \end{matrix}}_{A}$ $\underbrace{\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}}_{I} $
$\downarrow \quad \begin{matrix}R_2 \leftarrow R_2-R_1 \\ R_3 \leftarrow R_3 -R_1 \end{matrix}$
$\begin{matrix}1 & 1 & 1 \\ 0 & 0 & -2 \\ 0 & -2 & 0 \end{matrix}$ $\begin{matrix}1 & 0 & 0 \\ -1 & 1 & 0 \\ -1 & 0 & 1 \end{matrix}$

$\downarrow \quad R_2 \leftrightarrow R_3$
$\begin{matrix}1 & 1 & 1 \\ 0 & -2 & 0 \\ 0 & 0 & -2\end{matrix}$ $\begin{matrix}1 & 0 & 0 \\ -1 & 0 & 1 \\ -1 & 1 & 0\end{matrix} $
$\downarrow \quad \begin{matrix}R_2 \leftarrow - \frac 12 R_2 \\ R_3 \leftarrow - \frac 12 R_3 \end{matrix}$
$
\begin{matrix}1 & 1 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}$ $\begin{matrix}1 & 0 & 0 \\ \frac 12 & 0 & - \frac 12 \\ \frac 12 & - \frac 12 & 0\end{matrix}$

$\downarrow \quad R_1 \leftarrow R_1 -R_3$
$
\begin{matrix}1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{matrix}$ $\begin{matrix}\frac 12 & \frac 12 & 0 \\ \frac 12 & 0 & - \frac 12 \\ \frac 12 & - \frac 12 & 0\end{matrix}$
$\downarrow \quad R_1 \leftarrow R_1 -R_2$
$\underbrace{\begin{matrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}}_{I} $ $\underbrace{\begin{matrix}0 & \frac 12 & \frac 12 \\ \frac 12 & 0 & - \frac 12 \\ \frac 12 & - \frac 12 & 0 \end{matrix}}_{A^{-1}} $

We have: $A^{-1}= \begin{pmatrix}0 & \frac 12 & \frac 12 \\ \frac 12 & 0 & - \frac 12 \\ \frac 12 & - \frac 12 & 0 \end{pmatrix} $.

Let us summarize:

\fbox{ $ ( I \vert A ) \longrightarrow ( A^{-1} \vert I )$\space }


next up previous contents
Next: More on systems of Up: The algebra of matrices. Previous: How to compute the
Noah Dana-Picard
2001-02-26