Characterization and Uniqueness of the Moore–Penrose Inverse

Why Four Equations?

For a nonsingular square matrix, the inverse is characterized by two equations: multiplying on either side gives the identity. A matrix can be rectangular or singular, however, so it may have no ordinary inverse. The Moore–Penrose inverse extends the useful idea of an inverse to every real matrix. Its four equations impose exactly the conditions needed to make the resulting matrix unique.

The focus here is not on treating the pseudoinverse as a computational shortcut. It is on understanding its defining equations: why a solution exists, why no two different matrices can satisfy them, and how to check the equations without relying on a remembered formula.

Definition: Let $A$ be a real $m\times n$ matrix. The Moore–Penrose inverse of $A$, denoted $A^+$, is an $n\times m$ matrix satisfying $$ AA^+A=A,\qquad A^+AA^+=A^+,\qquad (AA^+)^T=AA^+,\qquad (A^+A)^T=A^+A. $$ The first two equations are called the reflexive equations. The last two require the products $AA^+$ and $A^+A$ to be symmetric.

The order of the factors matters. For $A$ of size $m\times n$, $AA^+$ is $m\times m$, while $A^+A$ is $n\times n$. Both products must be symmetric, but they need not be the same matrix. In the equations above, the transpose conditions are separate requirements, not consequences of the first two equations alone.

Existence from a Rank Factorization

The four equations do not merely describe a matrix if one happens to exist: every real matrix has such a matrix. The rank-factorization construction makes this existence statement explicit. If $A$ has rank $r>0$, write $A=FG$, where $F$ is $m\times r$ with full column rank and $G$ is $r\times n$ with full row rank. The matrices $F^TF$ and $GG^T$ are invertible: for example, $u^TF^TFu=\|Fu\|^2>0$ for every nonzero $u$, since $F$ has full column rank.

Theorem: For a rank factorization $A=FG$ with $r>0$, the matrix $$ B=G^T(GG^T)^{-1}(F^TF)^{-1}F^T $$ satisfies the four Moore–Penrose equations. If $A=0$, the zero matrix of size $n\times m$ satisfies them.

To verify the claim, define the two products $P=AB$ and $Q=BA$. Direct multiplication gives $$ P=F(F^TF)^{-1}F^T,\qquad Q=G^T(GG^T)^{-1}G. $$ Both are symmetric. Also $P^2=P$ and $Q^2=Q$: in the first calculation the middle factor $F^TF$ cancels its inverse, and in the second the middle factor $GG^T$ does so. Thus $P$ and $Q$ are symmetric idempotent matrices.

Since $PF=F$, we have $ABA=PA=PF G=FG=A$. Since $QG^T=G^T$, the displayed formula for $B$ gives $QB=B$, so $BAB=QB=B$. The symmetry of $P=AB$ and $Q=BA$ supplies the remaining two equations. If $A=0$, taking $B=0$ makes all four products zero, as required.

Why the Four Equations Give Only One Matrix

The uniqueness proof uses a simple fact about symmetric idempotent matrices. Here, idempotent means $P^2=P$. We prove the fact in two forms so that it can be applied to both products in the Moore–Penrose equations.

Lemma: Two symmetric idempotent matrices of the same size are equal if they have the same range, and they are also equal if they have the same null space.

For the range claim, let $P$ and $Q$ have the same range. Each idempotent acts as the identity on its range: if $z=Pw$, then $Pz=P^2w=Pw=z$. Therefore $PQ=Q$, because $Qv$ is in the common range, and $QP=P$, because $Pv$ is in that range. Transposing $QP=P$, and using symmetry, gives $PQ=P$. Hence $P=Q$.

For the null-space claim, suppose $\operatorname{Null}(P)=\operatorname{Null}(Q)$. For any $v$, $(I-Q)v$ is in $\operatorname{Null}(Q)$, because $Q(I-Q)=0$; it is therefore in $\operatorname{Null}(P)$, giving $P(I-Q)=0$. Transposing this equality gives $(I-Q)P=0$, or $QP=P$. Reversing the roles of $P$ and $Q$ gives $PQ=Q$. Transposing $QP=P$ also gives $PQ=P$. Thus $P=Q$.

Theorem: For each real $m\times n$ matrix $A$, exactly one $n\times m$ matrix satisfies the four Moore–Penrose equations. This unique matrix is $A^+$.

Existence follows from the rank-factorization construction above (including its zero-matrix case). For uniqueness, suppose $X$ and $Y$ are two $n\times m$ matrices satisfying all four equations for the same $A$.

First consider $AX$ and $AY$. Each is symmetric by the third and fourth equations, respectively. Each is idempotent, since $(AX)^2=AXAX=(AXA)X=AX$, and the same calculation applies to $AY$. Both have range equal to the range of $A$. Indeed, their ranges are contained in the range of $A$, while $A=AXA$ implies that every vector $Av$ is also $AX(Av)$, in the range of $AX$; the corresponding equation for $Y$ gives the same conclusion. By the lemma’s range claim, $AX=AY$.

Next consider $XA$ and $YA$. They are symmetric by the last two equations, and idempotent because $XAX=X$ and $YAY=Y$. Their null spaces both equal $\operatorname{Null}(A)$. If $Av=0$, then $XAv=0$. Conversely, if $XAv=0$, applying $A$ gives $AXAv=0$, and $AXA=A$ then gives $Av=0$. The same reasoning, using $AYA=A$, works for $YA$. The lemma’s null-space claim now gives $XA=YA$.

Finally, use the two equalities just obtained: $$ X=XAX=(XA)X=(YA)X=Y(AX)=Y(AY)=YAY=Y. $$ Thus any two matrices satisfying the four equations must coincide. Together with existence, this proves the theorem. $\square$

Worked Verifications

In each example, verify all four equations. Checking only $AA^+A=A$ is not enough: a candidate must also satisfy the second reflexive equation and both symmetry conditions.

Worked Example: A Rectangular Matrix with Full Row Rank

Let $$ A=\begin{pmatrix}1&0&0\\0&2&0\end{pmatrix},\qquad B=\begin{pmatrix}1&0\\0&\tfrac12\\0&0\end{pmatrix}. $$ Here $A$ is $2\times3$ and $B$ is $3\times2$, so the products in the four equations are defined.

First, $$ AB=\begin{pmatrix}1&0\\0&1\end{pmatrix},\qquad BA=\begin{pmatrix}1&0&0\\0&1&0\\0&0&0\end{pmatrix}. $$ Both products are symmetric. Since $AB=I_2$, $ABA=(AB)A=A$. Since $BA$ leaves $B$ unchanged, direct multiplication gives $BAB=(BA)B=B$. Thus $B=A^+$ by uniqueness.

Worked Example: A Rank-One Matrix

Let $A=\begin{pmatrix}1&1\\1&1\end{pmatrix}$ and propose $B=A/4$. Since $A^2=2A$, we obtain $$ AB=BA=\frac{A^2}{4}=\frac{A}{2}. $$ This product is symmetric. Also, $$ ABA=\frac{A}{2}A=\frac{A^2}{2}=A,\qquad BAB=\frac{A}{4}\frac{A}{2}=\frac{A^2}{8}=\frac{A}{4}=B. $$ All four equations hold, so $A^+=A/4$. The matrix $A$ is singular, but that presents no obstacle to the defining equations.

Worked Example: A Singular Nonsymmetric Matrix

Consider $$ A=\begin{pmatrix}0&2\\0&0\end{pmatrix},\qquad B=\begin{pmatrix}0&0\\\tfrac12&0\end{pmatrix}. $$ The products are $$ AB=\begin{pmatrix}1&0\\0&0\end{pmatrix},\qquad BA=\begin{pmatrix}0&0\\0&1\end{pmatrix}. $$ Both are symmetric. Multiplying $AB$ by $A$ gives $ABA=A$, while multiplying $BA$ by $B$ gives $BAB=B$. Therefore $B=A^+$. Notice that $A$ itself is not symmetric; the equations require symmetry of $AB$ and $BA$, not symmetry of $A$.

Common Pitfalls and Why the Characterization Matters

The defining equations are useful because they turn an abstract pseudoinverse into a checkable algebraic claim. A proposed answer can be verified directly, and the uniqueness theorem means that a successful verification identifies the Moore–Penrose inverse without needing to compare it with a separate formula. The equations also make clear that the two sides of a rectangular matrix play different dimensional roles.

Checking only one equation. A matrix that satisfies $ABA=A$ need not satisfy the other three. Verify every condition.
Using the wrong product dimensions. For $A$ of size $m\times n$, $AB$ is $m\times m$ and $BA$ is $n\times n$. Keep the order and sizes visible while multiplying.
Assuming $AB$ and $BA$ are equal. They can even have different sizes. Each must be symmetric, but equality is not required.
Assuming the original matrix must be symmetric. The symmetry conditions apply to $AA^+$ and $A^+A$, not to $A$ itself.

When checking a product, calculate entries from the indicated row and column rather than inferring them from a hoped-for identity. For instance, a zero row or column in one factor can force a product entry to be zero; every such entry should be recomputed before claiming that a product is an identity matrix.

Check Your Understanding

Use the four equations and the uniqueness argument to answer these questions.

If $A$ is $4\times2$, what are the dimensions of $A^+$, $AA^+$, and $A^+A$?
Why does $AXA=A$ imply that the range of $AX$ contains the range of $A$?
In the uniqueness proof, why does $\operatorname{Null}(XA)=\operatorname{Null}(A)$ follow from $XAX=X$ and $AXA=A$?
For the rank-one example, calculate $A^2$ and use it to verify $BAB=B$.
Why is checking that $AB$ and $BA$ are symmetric insufficient unless the two reflexive equations are checked as well?