Generalized Inverse¶

Definition¶

Let $A$ be a real $m\times n$ matrix. A generalized inverse of $A$ is any $n\times m$ matrix $A^{I}$ satisfying

The matrix products are defined because $A^{I}$ has the shape of $A^{\top}$. No condition on $m$, $n$ or the rank of $A$ is imposed.

Interpretation¶

An inverse matrix exists only for square nonsingular matrices. The condition $AA^{I}A=A$ keeps only what is needed to solve equations: whenever $\boldsymbol{b}=A\boldsymbol{x}$ is reachable, $A^{I}\boldsymbol{b}$ is a vector that $A$ maps back to $\boldsymbol{b}$, because $A(A^{I}A\boldsymbol{x})=A\boldsymbol{x}$$A\mathbf{x}$.

Because one equation is a weak requirement, a generalized inverse is in general far from unique, and the notation $A^{I}$ means "some generalized inverse". Adding the three further Penrose conditions singles out one of them, the Moore-Penrose inverse $A^+$. A common mistake is to assume properties of $A^+$, such as symmetry of $AA^{I}$, for an arbitrary $A^{I}$.

One-sided inverses are special cases: if $LA=I_n$ (left inverse, full column rank) then $ALA=A$$A$; likewise for a right inverse.

Key Results¶

• Existence: every matrix has a generalized inverse. If $A=P\begin{pmatrix}I_r&O\\O&O\end{pmatrix}Q$ with $P,Q$ invertible, then the generalized inverses are exactly $Q^{-1}\begin{pmatrix}I_r&X\\Y&Z\end{pmatrix}P^{-1}$ with $X,Y,Z$ arbitrary.
• Uniqueness: $A^{I}$ is unique if and only if $A$ is square and invertible, and then $A^{I}=A^{-1}$$A^{-1}$.
• $AA^{I}$ and $A^{I}A$ are idempotent, $\operatorname{rank}(AA^{I})=\operatorname{rank}(A^{I}A)=\operatorname{rank}A$$\mathrm{rank}A$, and $\operatorname{rank}A^{I}\ge\operatorname{rank}A$$\mathrm{rank}A$.
• Consistency: $A\boldsymbol{x}=\boldsymbol{b}$ has a solution if and only if $AA^{I}\boldsymbol{b}=\boldsymbol{b}$$\mathbf{b}$.
• General solution: if the system is consistent, its solutions are exactly $\boldsymbol{x}=A^{I}\boldsymbol{b}+(I_n-A^{I}A)\boldsymbol{z}$$A^{I}A)\mathbf{z}$, $\boldsymbol{z}\in\mathbb{R}^n$${\mathbb{R}}^n$, an affine subspace.

Examples¶

Let $A=\begin{pmatrix}1&1\\1&1\end{pmatrix}$ (rank $1$, singular).

Two different generalized inverses. For $G=\begin{pmatrix}1&0\\0&0\end{pmatrix}$$\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)$: $AG=\begin{pmatrix}1&0\\1&0\end{pmatrix}$ and $AGA=\begin{pmatrix}1&1\\1&1\end{pmatrix}=A$$A$. The matrix $G'=\begin{pmatrix}0&0\\0&1\end{pmatrix}$ works too, and so does $\tfrac{1}{4}A$, since $A^3=4A$$4A$. Note that $AG$ is idempotent but not symmetric.

Solving a system. For $\boldsymbol{b}=(2,2)^{\top}$$(2,2{)}^{\top }$: $AG\boldsymbol{b}=(2,2)^{\top}=\boldsymbol{b}$$\mathbf{b}$, so the system is consistent. With $I-GA=\begin{pmatrix}0&-1\\0&1\end{pmatrix}$$\left(\begin{array}{cc}0& -1\\ 0& 1\end{array}\right)$,

which is the line $x_1+x_2=2$$2$. For $\boldsymbol{b}=(1,0)^{\top}$$(1,0{)}^{\top }$: $AG\boldsymbol{b}=(1,1)^{\top}\neq\boldsymbol{b}$$\mathbf{b}$, so there is no solution.

The figure draws the consistent example with $\boldsymbol{b}=(2,2)^{\top}$$(2,2{)}^{\top }$. The two generalized inverses $G$ and $G'$ give different particular solutions, $G\boldsymbol{b}=(2,0)^{\top}$ and $G'\boldsymbol{b}=(0,2)^{\top}$$(0,2{)}^{\top }$, but both lie on the same line $x_1+x_2=2$$2$. Adding $(I-GA)\boldsymbol{z}=z_2(-1,1)^{\top}$ moves along that line, so the choice of $A^{I}$ changes the starting point, never the solution set.

Prerequisites¶

• Matrix Product
• Rank
• Inverse and Singular Matrices
• Left and Right Inverses

Related Concepts¶

• Moore-Penrose Inverse
• Least Squares
• Affine Subspace
• Idempotent Matrix

Sources¶

• Math V: Advanced Linear Algebra · Week 7

Aliases: g-inverse, generalised inverse, {1}-inverse, inner inverse, 1-inverse Keywords: A^I, A^-, A A^I A = A, consistency condition, x = A^I b + (I - A^I A) z, conditional inverse Symbols: element of, real numbers

Last updated: 22 October 2026 19:04