EP matrices

Motivation: We know that when $A$ is nonsingular, then $AA^{-1}=A^{-1}A=I$. Consider $A=\begin{pmatrix}1 &1\\0 &0 \end{pmatrix}$, then it is easy to check that the Moore-Penrose inverse of this matrix is given by $A^{\dagger}=\begin{pmatrix}1/2 &0\\1/2 &0 \end{pmatrix}$. Now, $AA^{\dagger}=\begin{pmatrix}1 &0\\0 &0 \end{pmatrix}\neq \begin{pmatrix}1/2 &1/2\\1/2 &1/2 \end{pmatrix}=A^{\dagger}A$. Further,  Let $B=\begin{pmatrix}1 &0 &1\\1 &0 &1\\ 0 &1 &0\end{pmatrix}$. Then, $B^{\dagger}=\begin{pmatrix}0.25 &0.25 &0\\ 0 &0 &1\\ 0.25 &0.25 &0\end{pmatrix}\neq \begin{pmatrix}1 &-1 &1\\1 &-1 &1\\-1 &2 &-1\end{pmatrix}=B^{\#}$, where $B^{\dagger}$ is the Moore-Penrose inverse and $B^{\#}$ is the Group inverse of $B$.

Observe that when $AA^{\dagger}=A^{\dagger}A$, then from the third equation (recall, $AX=XA$) in the definition of the Group inverse, $A^{\dagger}=A^{\#}$.

It is natural from the above examples to search for the class of matrices for which $AA^{\dagger}=A^{\dagger}A$, i.e., the product of $A$ and $A^{\dagger}$ commutes or equivalently, $A^{\dagger}=A^{\#}$.

Problem statement: Is there a class of matrices for which the product of matrices $A$ and $A^{\dagger}$ commutes? 

or equivalently, 

for which class of matrices the group inverse and the Moore– Penrose inverse are the same?

Formulation: We know that $AA^{\dagger}=P_{R(A)}$, an orthogonal projection onto $R(A)$ and $A^{\dagger}A=P_{R(A^*)}$, an orthogonal projection onto $R(A^*)$. So, if we want $AA^{\dagger}=A^{\dagger}A$, we must have $P_{R(A)}=P_{R(A^*)}$ or equivalently,  $R(A)=R(A^*)$.

Definition: A square matrix $A$ is called EP (also called Range-Hermitian) matrix if $R(A)=R(A^T).$

NOTE: 1. When $A$ is Symmetric, i.e., $A=A^T$, then it is obvious that $R(A)=R(A^T)$. But, not all EP matrices are symmetric as the following examples shows:

Ex: Consider $A=\begin{pmatrix}1 &2\\1&1\end{pmatrix}$, then $A\neq A^T$ but $R(A)=R(A^T)$. So, all nonsingular nonsymmetric square matrices are contained in the class of  EP matrices.

2. When $A$ is normal, i.e., $AA^T=A^TA$, then clearly $R(A)=R(A^T)$. So, every Normal matrices are EP but converse is not true:

Ex:  Consider the same matrix $A=\begin{pmatrix}1 &2\\1&1\end{pmatrix}$, then $AA^T=\begin{pmatrix}5 &3\\3&2\end{pmatrix}\neq \begin{pmatrix}2 &3\\3&5\end{pmatrix}=A^TA$ but $R(A)=R(A^T)$ (since $A$ is nonsingular the columns of $A$ and $A^T$ span same space $\mathbb{R}^2$). 

3. EP matrices have index 1.

Proof: Let $A$ be an EP matrix, then $R(A)=R(A^T)$. But, $R(A^T)=N(A)^{\perp}$, so $R(A)=N(A)^{\perp}$ and thus, $R(A)\cap N(A)=\{0\}$.

Hence, we can draw the following conclusion:

 set of  Normal matrices $\subset$ set of  Symmetric matrices $\subset$ set of  EP matrices $\subset$ set of matrices with index 1.

Some results: 

Theorem 1. $A$ is an EP matrix if and only if the Moore-Penrose inverse of $A$ is an EP matrix.

proof. Let $A$ be an EP matrix. So, $AA^{\dagger}=A^{\dagger}A$, but we know that $(A^{\dagger})^{\dagger}=A$. Hence, $A^{\dagger}$ is EP, i.e., $A^{\dagger}(A^{\dagger})^{\dagger}=(A^{\dagger})^{\dagger}A^{\dagger}$,  Hence proved. Conversely, we have $A^{\dagger}(A^{\dagger})^{\dagger}=(A^{\dagger})^{\dagger}A^{\dagger}$ and using the property $(A^{\dagger})^{\dagger}=A$, we get $AA^{\dagger}=A^{\dagger}A$ which shows that $A$ is an EP matrix.

Theorem 2. $A$ is normal if and only if $A$ is EP matrix and $AA^TA^2=A^2A^TA$.

proof. Let $A$ be an EP and $AA^TA^2=A^2A^TA$, then we will show that $A$ is normal, i.e., $AA^T=A^TA$. The other part is trivial. Pre-multiplying and post-multiplying $AA^TA^2=A^2A^TA$ by $A^{\dagger}$, we get $A^{\dagger}AA^T(A^2A^{\dagger})=(A^{\dagger}A^2)A^TAA^{\dagger}$, but $A^2A^{\dagger}=AA^{\dagger}A=A$ and $A^{\dagger}A^2=AA^{\dagger}A=A$ (since $A~\&~A^{\dagger}$ commutes). So, $A^{\dagger}AA^TA=AA^TAA^{\dagger}$. Further, $A^{\dagger}AA^T=A^T$ as $R(A^T)\subseteq R(A^{\dagger})$ and $A^TAA^{\dagger}=A^T$ as $N(A^\dagger)\subseteq N(A^T)$. Hence, $A^TA=AA^T$ and thus $A$ is normal.

Theorem 3. For any two matrices $A$ and $B$ such that $AB$ exists, the reverse order law

                                                $(AB)^{\dagger}=B^{\dagger}A^{\dagger}$

 holds if and only if $A^TABB^T$ is EP.

proof. Let $W=A^TABB^T$ be an EP matrix. We prove this results by showing that '$W$ is an EP' if and only if '$R(W)=R(A^TAB)\subset R(B)$' and $R(W^T)=R(BB^TA^T)\subset R(B^T)$'. Then, utilizing Theorem 1 (in the Moore-Penrose inverse post), we get the proof. Clearly, $R(W)\subseteq R(A^TAB)$. Now, $W(B^T)^{\dagger}=A^TABB^T(B^T)^{\dagger}=A^TA(B^{\dagger}BB^T)^T=A^TAB$, so $R(A^TAB)\subseteq R(W)$. Thus, $R(W)=R(A^TAB)$. Using similar arguments, we can show that $W^TA^{\dagger}=BB^TA^T$ and then it follows that $R(W^T)=R(BB^TA^T)$. But, $R(W)=R(W^T)$. So, $R(W)=R(BB^TA^T)\subset R(B)$ and $R(W^T)=R(A^TAB)\subset R(A^T)$.

Conversely, assume that $R(W)\subset R(B)$ and $R(W^T)\subset R(A^T)$ and let $F=A^TA$ and $G=BB^T$, then by Theorem 1 (in the group inverse post), $R(C)=R(A^T)\cap R(B)$. Interchanging the roles of $F$ $\&$ $G$ and applying Theorem 1 (in the group inverse post) again, we get $R(W^T)=R(A^T)\cap R(B)$. Thus, $R(W)=R(W^T)$. Hence, we have proved that $R(W)=R(W^T)$ if and only if $R(W)\subset R(B)$ and $R(W^T)\subset R(B^T)$. Now directly using the conclusion of Theorem 1 (in the Moore-Penrose inverse post), we get our conclusion.

Theorem 4. Let $A$ and $B$ be two EP matrices. Then, $C=A+B$ is an EP matrix if any one of the following equivalent conditions hold

(i) $N(C)\subseteq N(A)$ and $N(C)\subseteq N(B)$.

(ii) rank $\begin{bmatrix} A\\B \end{bmatrix}$=rank$(C)$.

proof. First we prove that (i) and (ii) are equivalent. Assume that (i) holds, then $N(C)\subseteq N(A)\cap N(B)$. Since $C=A+B$, then $N(A)\cap N(B)\subseteq N(A+B)=N(C)$. Thus, $N(C)=N(A)\cap N(B)=N\left(\begin{bmatrix}A\\B\end{bmatrix}\right)$ and by Rank-Nullity Theorem, $rank(C)=rank\begin{bmatrix}A\\B\end{bmatrix}$. 

Conversely, observe that $N\left(\begin{bmatrix}A\\B\end{bmatrix}\right)=N(A)\cap N(B)\subset N(C)$ and if (ii) holds, then clearly, $N(C)=N(A)\cap N(B)$ and hence $N(C)\subseteq N(A)$ and $N(C)\subseteq N(B).$ So, (i) $\&$ (ii) are equivalent.

Now, if  $A$ and $B$ are EP, i.e., $N(A)=N(A^T)$ and $N(B)=N(B^T)$, then (i) implies that $N(C)\subseteq N(A)\cap N(B)=N(A^T)\cap N(B^T)\subseteq N(C^T)$ (since $N(A^T)\subseteq N(C^T)$ and $N(B^T)\subset N(C^T)$ by (i)) and $rank(C)=rank(C^T)$. Hence, $N(C)=N(C^T)$, i.e., $C$ is EP.

Computing the Moore-Penrose inverse of a matrix is considered to be infeasible. Very recently, Ignacio Bajo [3] proposed a method for computing the Moore-Penrose inverse using polynomial in matrices. The following result provides a characterization using an EP matrix of all those matrices whose Moore-Penrose inverse is a polynomial in $A^T$. The proof of this result will be added later.

Theorem 5. Let $A$ be an EP matrix. Then, there exists a polynomial $p(x)$ such that $A^{\dagger}=p(A^T)$ if and only if $A$ is normal.

SOURCES:

[1] Ben-Israel, A.; Greville, T. N. E., Generalized Inverses. Theory and Applications, Springer-Verlag, New York, 2003.

[2] Meenakshi, A. R., On sums of EP matrices. Houston Journal of Mathematics, 9 (1983).

[3] Bajo, I., Computing Moore-Penrose inverses with polynomials in matrices. https://doi.org/10.1080/00029890.2021.1886840