# Some questions about the pseudoinverse of a matrix

For every mxn-matrix A with real entries, there exist a unique nxm-matrix B, also with real entries, such that

$$ABA = A$$ $$BAB = B$$ $$AB = (AB)^T$$ $$BA = (BA)^T$$

B is called the pseudoinverse of A. There is also a complex version, but I am only interested in the real one. Now my questions :

• If A has rational entries, must B also have rational entries ?
• How can I calculate the pseudoinverse of a matrix with PARI/GP ?
• Is there a simple method to calculate the pseudoinverse by hand for small matrices ?
• Under which condition are the entries of the pseudoinverse integers ?

I know some special properties, for instance, that for invertible square matrices A, the pseudoinverse is simply $A^{-1}$ , or that the pseudoinverse of a zero-matrix is its transposition, but I have not much experience with general pseudoinverses.

• For the fourth question: If $A^+\in\mathbb Z^{n\times m}$ (every entry of $A^+$ is an integer), then from the answer to first question being "yes" we have that $A=(A^+)^+\in\mathbb Q^{m\times n}$ (every entry of $A$ must be rational), which gives us a necessary condition for the entries of the pseudoinverse to be integers. Not sure if there's a nice condition that's both necessary and sufficient. Sep 6 at 7:11

Edit. The old answer is wrong. Here is a corrected one.

For your first question, since $$A$$ has rational entries, it has a rank decomposition $$A=XY$$ such that $$X$$ is a tall matrix with full column rank and rational elements and $$Y$$ is fat matrix with full row rank and rational elements. One may verify, using the four defining properties of Moore-Penrose pseudoinverse, that $$G=Y^+X^+$$ is identical to $$A^+$$. Indeed, \begin{aligned} &AGA=X(YY^+)(X^+X)Y=XY=A,\\ &GAG=Y^+(X^+X)(YY^+)X^+=Y^+X^+=G,\\ &AG=XYY^+X^+=XX^+\text{ is Hermitian},\\ &GA=Y^+X^+XY=Y^+Y\text{ is Hermitian}.\\ \end{aligned} Since $$X^+=(X^TX)^{-1}X^T$$ and $$Y^+=Y^T(YY^T)^{-1}$$, they have rational entries. In turn, so does $$A^+$$.

For your third question, if $$A$$ is at most $$3\times3$$, you may try the formula $$A^+ = \lim_{\delta \searrow 0} (A^\top A + \delta I)^{-1} A^\top = \lim_{\delta \searrow 0} A^\top (A A^\top + \delta I)^{-1}.$$

For your last question, I'm not sure if there are any nice and general sufficient conditions.

• Why can "every matrix $A$ over the rationals can be written as a matrix product of the form $PDQ$, where $P$ and $Q$ are products of elementary matrices with rational entries and $D$ is a rectangular diagonal matrix over the rationals"? Do you have a reference? Sep 6 at 3:43
• @xFioraMstr18 Smith normal form Sep 6 at 5:46
• Why does $A^+=Q^{-1}D^+P^{-1}$? The naive "identity" $(EF)^+=F^+E^+$ does not always hold (Wikipedia). And if $P,Q$ were unitary like in SVD, then we could directly verify the Hermitianness axioms (Axioms 3 and 4) in the axiomatic definition of the pseudoinverse. But that's not necessarily the case here. Sep 6 at 10:04
• @xFioraMstr18 Seems like you are right, but my mother is now under quarantine. I have to take care of her and will review my answer later. Sep 7 at 6:54
• @xFioraMstr18 The answer is fixed now. Thanks for catching my mistake. Sep 14 at 11:38

Regarding the first question: Let $$\mathbb F$$ be a field (not necessarily $$\mathbb R$$ or $$\mathbb C$$), and let $$m,n$$ be nonnegative integers. Given an involutory field automorphism $$a:\mathbb F\to\mathbb F$$ (think complex conjugation), we can define a sort of "conjugate transpose" $$\mathbb F^{m\times n}\to\mathbb F^{n\times m}$$ via $$A\mapsto A^*:=\left(a(A_{j,i})\right)_{i\in n,j\in m}$$. This "conjugate transpose" lets us define what it means for a matrix $$X\in\mathbb F^{n\times m}$$ to be a pseudoinverse of $$A.$$ It can then be shown that every $$A\in\mathbb F^{m\times n}$$ has at most one pseudoinverse $$X\in\mathbb F^{n\times m}$$ (the usual proof carries over). The topic of existence is more delicate; it can be shown that an arbitrary $$A\in\mathbb F^{m\times n}$$ has a pseudoinverse belonging to $$\mathbb F^{n\times m}$$ if and only if we have the equalities $$\text{rank}(A^*A)=\text{rank }A=\text{rank}(AA^*)$$ [Bot13].

Now, consider the case where we have two fields $$\mathbb E,\mathbb F$$, and an involutory field automorphism $$a:\mathbb F\to\mathbb F,$$ where $$\mathbb E\subseteq\mathbb F$$ and $$a|_{\mathbb E}:\mathbb E\to\mathbb E$$ has range $$\text{range}(a|_{\mathbb E})\subseteq\mathbb E.$$ It is not difficult to see that $$a|_{\mathbb E}$$ will be an involutory field automorphism on $$\mathbb E$$, so, given two nonnegative integers $$m,n$$, we can apply the above theory to $$\mathbb E^{m\times n}:$$ Any $$A\in\mathbb E^{m\times n}$$ has a pseudoinverse belonging to $$\mathbb E^{n\times m}$$ iff $$\text{rank}_{\mathbb E}(A^*A)=\text{rank}_{\mathbb E}A=\text{rank}_{\mathbb E}(AA^*).$$ This string of equalities is in turn equivalent to $$\text{rank}_{\mathbb F}(A^*A)=\text{rank}_{\mathbb F}A=\text{rank}_{\mathbb F}(AA^*)$$ (rank does not change under field extension). This second string of equalities is equivalent to $$A$$ having a pseudoinverse belonging to $$\mathbb F^{n\times m}$$. Thus, $$A$$ has a pseudoinverse belonging to $$\mathbb F^{n\times m}$$ iff $$A$$ has a pseudoinverse belonging to $$\mathbb E^{n\times m}$$. (One direction is almost obvious, from the set inclusion $$\mathbb E^{n\times m}\subseteq \mathbb F^{n\times m}$$, but the converse might have been not obvious.)

Now, let us specialize to the case where $$\mathbb F=\mathbb C$$, $$a=(\bar z)_{z\in\mathbb C}$$ is complex conjugation, and $$\mathbb E$$ is any subfield of $$\mathbb C$$ such that $$a|_{\mathbb E}:\mathbb E\to\mathbb E$$. (Examples: $$\mathbb C$$ itself; the algebraic numbers $$\mathbb A$$; or any subfield of $$\mathbb R$$, such as $$\mathbb R$$ itself, $$\mathbb Q$$, or $$\mathbb Q(\sqrt2)$$.) We know that every $$A\in\mathbb E^{m\times n}\subseteq\mathbb C^{m\times n}$$ has a pseudoinverse, which we denote $$A^+$$, belonging to $$\mathbb C^{n\times m}.$$ Thus, from the above paragraph, $$A$$ has a pseudoinverse in $$\mathbb E^{n\times m}\subseteq\mathbb C^{n\times m}$$, and so from uniqueness in $$\mathbb C^{n\times m}$$ we have $$A^+\in\mathbb E^{n\times m}$$. Thus, the answer is yes.

In fact, for any $$A\in\mathbb Z^{m\times n}$$, the least common denominator of the image of $$A^+$$, $$\ell:=\min\{d\in\mathbb Z_{\ge1}:\forall i\in m~\forall j\in n\quad d(A^+)_{i,j}\in\mathbb Z\}$$ will divide (in $$\mathbb Z$$) $$(\text{vol}A)^2$$, where we define the volume $$\text{vol}~A:=\sqrt{\sum_{I,J}\det(A|_{I\times J})^2},$$ where the summation ranges over all cardinality-$$r$$ subsets $$I,J\subseteq A$$, where $$r=\text{rank}~A$$ [BhR02 p.81 Theorem 6.7, BouVre20]. (If $$A$$ is square and invertible, then $$\text{vol}~A=|\det A|$$.) In fact, based on a few examples I computed (with noninvertible integer-valued matrices whose nonzero entries are relatively prime), it seems we "often" have $$\ell=(\text{vol}~A)^2$$. For instance, in the example $$[1;2]^+=5^{-1}[1,2]$$ we have $$\ell=5$$.

Regarding the fourth question: We also have a necessary and sufficient condition when the entries of $$A$$ itself are integers: if $$A\in\mathbb Z^{m\times n}$$, then $$A^+\in\mathbb Z^{n\times m}$$ iff $$\text{vol}~A=1$$ [BhR02 Theorem 6.7]. As for arbitrary $$A\in\mathbb Q^{m\times n}$$, I'm unsure.

[BhR02] Bhaskara Rao, K. (2002). Theory of generalized inverses over commutative rings. CRC Press.
[Bot13] Botha, J. D. (2013). Matrices over finite fields. In: Hogben, L. (ed.). Handbook of Linear Algebra (2nd ed.). CRC Press. Section 31.5.
[BouVre20] Bouman, N. J., and de Vreede, N. (2020). A practical approach to the secure computation of the Moore–Penrose pseudoinverse over the rationals. In International Conference on Applied Cryptography and Network Security (pp. 398-417). Springer, Cham.

• (The material describing arbitrary fields is rather inessential; the main point here is the equivalence with $\text{rank}(A^*A)=\text{rank }A=\text{rank}(AA^*)$.) Sep 6 at 12:54
• And actually this definition of pseudoinverse carries over from the setting of an arbitrary field to the setting of an arbitrary unital ring [BhR02 Section 3.1]. Sep 6 at 14:41
• @xFioraMstr18 some care is needed, however; we might want to assume the underlying ring has invariant basis number (e.g. if the underlying ring is commutative and nonzero) and assume all submodules of a module are free (e.g. if the underlying ring is a principal-ideal domain), so that we can define the rank of a matrix (cf. Hungerford's Algebra, Section VII.2). All in all, everything (the definition and proofs) seem to work for an arbitrary principal-ideal domain. Sep 6 at 15:26
• @xFioraMstr18 the result doesn't hold for arbitrary principal-ideal domains even; consider the case where the two principal-ideal domains are $\mathbb Z$ and $\mathbb C$. Sep 7 at 1:17
• @xFioraMstr18 It seems the result in [Bot13] fails to hold for arbitrary principal-ideal domains. The proof that he references relies on 1. rank factorization, which no longer holds once we ditch a field for a PID ("Over a commutative ring $R$ if every [finite?] matrix has a rank factorization show that $R$ is a field" [BhR02 Exercise 3.3(b)]), 2. the fact that over a field, any full-rank finite square matrix will be invertible (not generally true for PIDs). Sep 7 at 2:00

This is not a proper definition of the pseudoinverse. There are two common definitions, the first concerning matrices where the Singular Value Decomposition is used, and the more general (which I prefer) in Hilbert spaces where we consider the minimum of $$\left\|Ax-y \right\|$$ over $$X$$ and then we define as $$A^{+}y$$ the minimizing $$x$$ with the minimum norm. All the condition you use are obtained as propositions by this definition of the pseudoinverse!