Matrix Inversion¶
Recall (doc/postdoc). Not a first course.
Job: \(A^{-1}\) in \(\mathrm{GL}_n(K)\); computational vs structural criteria.
Core¶
\(A\) invertible \(\Leftrightarrow\) \(\det A\neq0\) \(\Leftrightarrow\) full rank \(\Leftrightarrow\) \(\ker A=0\) \(\Leftrightarrow\) columns basis \(\Leftrightarrow\) \(0\) not eigenvalue.
\[
A^{-1}=\frac{1}{\det A}\operatorname{adj}A,\quad
(\operatorname{adj}A)_{ij}=(-1)^{i+j}M_{ji}
\]
| Method | Notes |
|---|---|
| Gauss–Jordan | $[A |
| \(2\times2\) | \(\frac1{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}\) |
| Block | Schur complements when blocks invertible |
| Neumann | \((I-N)^{-1}=\sum N^k\) if \(\|N\|<1\) (Banach) |
Left/right inverse: for square over fields, equivalent. Pseudoinverse: Moore–Penrose for least squares when singular.
Used later for¶
- Systems: unique solution \(x=A^{-1}b\)
- Eigenvalues: diagonalization \(P^{-1}AP\)
- ODE: fundamental matrices
- Numerical analysis: condition \(\kappa(A)=\|A\|\|A^{-1}\|\)
Check¶
If \(A\) is invertible, is \(A+E\) invertible for all small \(E\)?
Yes in norm topology: \(\mathrm{GL}_n\) open. Explicitly if \(\|A^{-1}E\|<1\) then \(A+E=A(I+A^{-1}E)\) invertible.