Inverse of a matrix
Dependencies:
Let $F$ be a field and $A \in \mathbb{M}_{n, n}(F)$.
If a matrix $L$ exists such that $LA = I_n$, then $L$ is a left inverse of $A$. If a matrix $R$ exists such that $AR = I_n$, then $R$ is a right inverse of $A$. If $A$ has both a left inverse and a right inverse, it can be proven that all left inverses and right inverses are identical. Therefore, such a matrix is simply called the inverse of $A$ and is denoted as $A^{-1}$.
A matrix whose inverse exists is called an invertible matrix.
$AB = BA = I \Rightarrow$ $A$ and $B$ are inverses of each other. Therefore, $A = (A^{-1})^{-1}$.
Proof
Suppose $A$ has a left inverse $L$ and a right inverse $R$.
\[ L = LI_n = L(AR) = (LA)R = I_nR = R \]
Hence, every left inverse is equal to every right inverse. Therefore, they are all equal.
(It can actually be proven that a left inverse is also a right inverse).
Dependency for:
- A is diagonalizable iff there are n linearly independent eigenvectors
- RREF([A|I]) = [I|inv(A)] iff A is invertible
- Full-rank square matrix is invertible
- Elementary row operation is matrix pre-multiplication
- Inverse of product
Info:
- Depth: 4
- Number of transitive dependencies: 7