Matrix inverse ($A^{-1}$)

A matrix inverse satisfies $A{A}^{-1}=A^{-1}A=I$.

The inverse of a block matrix can be expressed through the inverse of one diagonal block and the inverse of its Schur complement. This reduces many deletion and conditioning identities to smaller matrix inversions.

If $A$ is invertible and $1+v^T{A}^{-1}u\ne 0$, then

The Moore-Penrose inverse is the unique matrix $A^{+}$ satisfying $A{A}^{+}A=A$, $A^{+}A{A}^{+}=A^{+}$, and symmetry of $A{A}^{+}$ and $A^{+}A$. If $A=UD{V}^T$ is a singular value decomposition, then $A^{+}=V{D}^{+}{U}^T$.

For a bounded operator $A:X\to Y$ between Hilbert spaces, $A^{†}$ has domain $\mathcal{R}(A)\oplus \mathcal{R}(A{)}^{\perp }$, maps into $\mathcal{N}(A{)}^{\perp }$, and inverts the restriction of $A$ to $\mathcal{N}(A{)}^{\perp }$. It is bounded exactly when $\mathcal{R}(A)$ is closed.

The Moore--Penrose inverse is characterized by

The identity $(AB{)}^{†}=B^{†}{A}^{†}$ does not hold for arbitrary matrices. A sufficient condition is that $A$ have full column rank and $B$ have full row rank.

For a rectangular matrix $X$ and $\lambda >0$,Multiplying either side by $X{X}^T+\lambda I$ verifies the identity.