Return to Answer

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, Am. Math. Mon. 75, 632-634 (1968). ZBL0162.04205) 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, upon permuting the $v_j$, we may assume that $Au_i=v_i$, for all $i$.

Observe that the fact that $A$ is doubly-stochastic means that $A$ leaves invariant the linear functional $\Sigma$ which sums all of the coordinates of a vector. Noticing that $\Sigma(u_i)=\sqrt{k_i}$, while $\Sigma(v_i)=\sqrt{l_i}$, we deduce that $k_i=l_i$.

We then conclude that $A^tA$ coincides with $AA^t$, and hence also that $A$ admits a direct sum decomposition into blocks with the same sizes as both $A^tA$ and $AA^t$. Dealing with each block separately, we may assume that $n=m=1$, and that $k_1=l_1=n$, and a moments' thought will reveal that $A=P(n)$.

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, Am. Math. Mon. 75, 632-634 (1968). ZBL0162.04205) 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 $$ A=WB=WB^tB=WA^tA. $$

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 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, upon permuting the $v_j$, we may assume that $Au_i=v_i$, for all $i$.

Observe that the fact that $A$ is doubly-stochastic means that $A$ leaves invariant the linear functional $\Sigma$ which sums all of the coordinates of a vector. Noticing that $\Sigma(u_i)=\sqrt{k_i}$, while $\Sigma(v_i)=\sqrt{l_i}$, we deduce that $k_i=l_i$.

We then conclude that $A^tA$ coincides with $AA^t$, and hence also that $A$ admits a direct sum decomposition into blocks with the same sizes as both $A^tA$ and $AA^t$. Dealing with each block separately, we may assume that $n=m=1$, and that $k_1=l_1=n$, and a moments' thought will reveal that $A=P(n)$.

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, Am. Math. Mon. 75, 632-634 (1968). ZBL0162.04205) 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, upon permuting the $v_j$, we may assume that $Au_i=v_i$, for all $i$.

Observe that the fact that $A$ is doubly-stochastic means that $A$ leaves invariant the linear functional $\Sigma$ which sums all of the coordinates of a vector. Noticing that $\Sigma(u_i)=\sqrt{k_i}$, while $\Sigma(v_i)=\sqrt{l_i}$, we deduce that $k_i=l_i$.

We then conclude that $A^tA$ coincides with $AA^t$, and hence also that $A$ admits a direct sum decomposition into blocks with the same sizes as both $A^tA$ and $AA^t$. Dealing with each block separately, we may assume that $n=m=1$, and that $k_1=l_1=n$, and a moments' thought will reveal that $A=P(n)$.

The following is an attempt to validate the conclusion proposed by @vidyarthi.

Theorem: Up to conjugation by a permutation matrix, 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 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, upon permuting the $v_j$, we may assume that $Au_i=v_i$, for all $i$.

Observe that the fact that $A$ is doubly-stochastic means that $A$ leaves invariant the linear functional $\Sigma$ which sums all of the coordinates of a vector. Noticing that $\Sigma(u_i)=\sqrt{k_i}$, while $\Sigma(v_i)=\sqrt{l_i}$, we deduce that $k_i=l_i$.

We then conclude that $A^tA$ coincides with $AA^t$, and hence also that $A$ admits a direct sum decomposition into blocks with the same sizes as both $A^tA$ and $AA^t$. Dealing with each block separately, we may assume that $n=m=1$, and that $k_1=l_1=n$, and a moments' thought will reveal that $A=P(n)$.

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 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, upon permuting the $v_j$, we may assume that $Au_i=v_i$, for all $i$.

Observe that the fact that $A$ is doubly-stochastic means that $A$ leaves invariant the linear functional $\Sigma$ which sums all of the coordinates of a vector. Noticing that $\Sigma(u_i)=\sqrt{k_i}$, while $\Sigma(v_i)=\sqrt{l_i}$, we deduce that $k_i=l_i$.

We then conclude that $A^tA$ coincides with $AA^t$, and hence also that $A$ admits a direct sum decomposition into blocks with the same sizes as both $A^tA$ and $AA^t$. Dealing with each block separately, we may assume that $n=m=1$, and that $k_1=l_1=n$, and a moments' thought will reveal that $A=P(n)$.