An $n\times n$ matrix $A$ with nonegative real entries $a\_{ij}$ is said to be *doubly stochastic* if $\sum\_{i=1}^na\_{ij} = 1$,
for all $j$, and $\sum\_{j=1}^na\_{ij}=1$, for all $i$.


Much is known [1] about the algebraic structure of the semigroup $\Omega \_n$ formed by all doubly stochastic $n\times n$ matrices. For
example, permutation matrices are the only invertible doubly stochastic matrices whose inverse is also doubly stochastic.
On the other hand [3], the idempotent elements in $\Omega \_n$ are precisely the direct sums of $k\times k$ matrices of the form
$$
 \pmatrix{
 1/k & 1/k & \ldots & 1/k \cr
 \vdots & \vdots & \ddots & \vdots\cr
 1/k & 1/k & \ldots & 1/k \cr}
 $$
together with their conjugates by permutation matrices.



> 
> Question: Which doubly stochastic matrices are partial isometries
> (i.e. satisfy the equation $AA^tA = A$)?
> 
> 
> 


See [2] for the characterization of **normal**, partial isometric, doubly stochastic matrices.


[1] *Farahat, H. K.*, [**The semigroup of doubly-stochastic matrices**](http://dx.doi.org/10.1017/S2040618500035401), Proc. Glasg. Math. Assoc. 7, 178-183 (1966). [ZBL0156.26001](https://zbmath.org/?q=an:0156.26001).


[2] *Prasada Rao, P. S. S. N. V.*, On generalized inverses of doubly stochastic matrices, Sankhyā, Ser. A 35, 103-105 (1973). [ZBL0301.15005](https://zbmath.org/?q=an:0301.15005).


[3] *Sinkhorn, R.*, [**Two results concerning doubly stochastic matrices**](http://dx.doi.org/10.2307/2313782), Am. Math. Mon. 75, 632-634 (1968). [ZBL0162.04205](https://zbmath.org/?q=an:0162.04205).



Doubly-stochastic partial-isometric matrices
The following is an attempt to validate the conclusion proposed by
@vidyarthi.



> 
> Theorem: Every
> doubly-stochastic partial-isometric matrix is the product of a
> permutation matrix and a doubly-stochastic projection.
> 
> 
> 


Proof:
Given a doubly-stochastic partial-isometric matrix $A$, one has that $A^tA$ and $AA^t$ are doubly-stochastic
projections, so by Theorem 2 in (*Sinkhorn, R.*, [**Two results concerning doubly stochastic matrices**](http://dx.doi.org/10.2307/2313782), Am. Math. Mon. 75, 632-634 (1968). [ZBL0162.04205](https://zbmath.org/?q=an:0162.04205)) there are permutation matrices $U$ and $V$ such that
$$
 U^tA^tAU = P(k\_1)\oplus P(k\_2)\oplus \cdots \oplus P(k\_n)
 $$
and
$$
 V^tAA^tV = P(l\_1)\oplus P(l\_2)\oplus \cdots \oplus P(l\_m),
 $$
where, for any integer $k$,
$$
 P(k):=
 \pmatrix{
 1/k & 1/k & \ldots & 1/k \cr
 \vdots & \vdots & \ddots & \vdots\cr
 1/k & 1/k & \ldots & 1/k \cr}.
 $$
Replacing $A$ with $V^tAU$, we may assume that $U$ and $V$ coincide with the identity matrix and hence $U$ and $V$ will henceforth
be ommitted.


Set $c(k)=(1/\sqrt k,1/\sqrt k,…,1/\sqrt k) ∈ \mathbb R^k$, so that $c(k)$ is a unit vector spanning the range of $P(k)$.
Moreover the range of the
projection $A^tA$ above admits an orthonormal basis formed by the vectors
$$
 u\_1 = c(k\_1)\oplus 0\_{k\_2}\oplus \cdots \oplus 0\_{k\_n},
 $$
$$
 u\_2 = 0\_{k\_1}\oplus c(k\_2)\oplus \cdots \oplus 0\_{k\_n},
 $$
$$...$$
$$
 u\_n = 0\_{k\_1}\oplus 0\_{k\_2}\oplus \cdots \oplus c(k\_n),
 $$
a similar reasoning yielding a basis $\{v\_1, v\_2, …, v\_m\}$ for the range of $AA^t$. The initial and final
projections of a partial isometry share rank, so $n=m$, and we claim that, up to a permutation of indices $i$, one has
that $k\_i=l\_i$, for all $i$.


Notice that $A$ maps $\hbox{span}\{u\_i\}$ isometrically onto $\hbox{span}\{v\_i\}$ so, for $i\neq j$, one has that
$Au\_i$ and $Au\_j$ are orthogonal vectors. However these vectors have nonnegative coordinates so their support (set of
indices for nonzero coordinates) must be disjoint. By the pigeonhole principle each $u\_i$ must therefore be mapped
under $A$ to a scalar
multiple of some $v\_j$. By positivity and norm preservation these scalars must coincide with 1 so there is a permutation $\sigma $ such that $Au\_i=v\_{\sigma (i)}$, for all $i$.


Observe that, being doubly-stochastic, $A$ leaves invariant the linear functional $\Sigma$ which sums all of the coordinates of a vector. Noticing that $\Sigma(u\_i)=\sqrt{k\_i}$, while $\Sigma(v\_j)=\sqrt{l\_j}$, we deduce that $k\_i=l\_{\sigma (i)}$.


It is now easy to see that there exists a permutation matrix $W$ such that $Wu\_i = v\_{\sigma (i)}$. Letting
$$
 B=W^tA,
 $$
we then have that $Bu\_i=u\_i$, while $B^tB=A^tA$.


It follows that $B$ is a partial isometry coinciding with the identity operator on its initial space, and hence that $B$ coincides with its initial projection $B^tB$. This leads to
$$
 A=WB=WB^tB=WA^tA.
 $$


