The row extraction (as you call it) is basically a projection $P$, which acts from the right on the coefficient matrix

$$ A_{S_i} = A \cdot P_{S_i} $$

where $P_{S_i}$ is a matrix of dimension $N \times |S_i|$, which contains unit vectors if the corresponding row is included.

$$ P_{S_i} = \begin{pmatrix} e_{s_1}, e_{s_2}, \ldots , e_{s_{|S_i|}} \end{pmatrix} $$

With this, the task is to solve the systems

$$ (A P_{S_i}) x = y $$

for which you can you can successively solve the two problems $$ A \underbrace{(P_{S_i} x)}_{=z} = y $$ and $$ P_{S_i} x = z $$

By decomposing the projection as 1 = P + (1 - P) and using the fact that a projection is idempotent $PP=P$, the second equation can be decomposed into two equations (i'll drop the subscript)

\begin{align} Px &= Pz\,,\\ 0 &= (1-P)z \end{align}

That is, you solve for the unknown x within the given subspace, and you can't solve for $z$ outside the projected space. An exact solution therefore only exists if $0 = (1-P)z$ exactly holds. In the least squares case, you should ignore this equation, and use $x = Pz$.

The row extraction (as you call it) is basically a projection $P$, which acts from the right on the coefficient matrix

$$ A_{S_i} = A \cdot P_{S_i} $$

where $P_{S_i}$ is a matrix of dimension $N \times |S_i|$, which contains unit vectors if the corresponding row is included.

$$ P_{S_i} = \begin{pmatrix} e_{s_1}, e_{s_2}, \ldots , e_{s_{|S_i|}} \end{pmatrix} $$

With this, the task is to solve the systems

$$ (A P_{S_i}) x = y $$

for which you can you can successively solve the two problems $$ A \underbrace{(P_{S_i} x)}_{=z} = y $$ and $$ P_{S_i} x = z $$

For the first equation, you can decompose the system matrix $A$ once by a QR or SVD decomposition.

By decomposing the projection as 1 = P + (1 - P) and using the fact that a projection is idempotent $PP=P$, the second equation can be decomposed into two equations (i'll drop the subscript)

\begin{align} Px &= Pz\,,\\ 0 &= (1-P)z \end{align}

That is, you solve for the unknown x within the given subspace, and you can't solve for $z$ outside the projected space. An exact solution therefore only exists if $0 = (1-P)z$ exactly holds. In the least squares case, you should ignore this equation, and use $x = Pz$.

The row extraction (as you call it) is basically a projection $P$, which acts from the right on the coefficient matrix

$$ A_{S_i} = A \cdot P_{S_i} $$

where $P_{S_i}$ is a matrix of dimension $N \times |S_i|$, which contains unit vectors if the corresponding row is included.

$$ P_{S_i} = \begin{pmatrix} e_{s_1}, e_{s_2}, \ldots , e_{s_{|S_i|}} \end{pmatrix} $$

With this, you can successively solve the two problems $$ A \underbrace{(P_{S_i} x)}_{=z} = y $$ and $$ P_{S_i} x = z $$

By decomposing the projection as 1 = P + (1 - P) and using the fact that a projection is idempotent $PP=P$, one gets two equations (i'll drop the subscript)

\begin{align} Px &= Pz\,,\\ 0 &= (1-P)z \end{align}

That is, you solve for the unknown x within the given subspace, and you can't solve for $z$ outside the projected space. An exact solution therefore only exists if $0 = (1-P)z$ exactly holds. In the least squares case, you should ignore this equation, and use $x = Pz$.

The row extraction (as you call it) is basically a projection $P$, which acts from the right on the coefficient matrix

$$ A_{S_i} = A \cdot P_{S_i} $$

where $P_{S_i}$ is a matrix of dimension $N \times |S_i|$, which contains unit vectors if the corresponding row is included.

$$ P_{S_i} = \begin{pmatrix} e_{s_1}, e_{s_2}, \ldots , e_{s_{|S_i|}} \end{pmatrix} $$

With this, the task is to solve the systems

$$ (A P_{S_i}) x = y $$

for which you can you can successively solve the two problems $$ A \underbrace{(P_{S_i} x)}_{=z} = y $$ and $$ P_{S_i} x = z $$

By decomposing the projection as 1 = P + (1 - P) and using the fact that a projection is idempotent $PP=P$, the second equation can be decomposed into two equations (i'll drop the subscript)

\begin{align} Px &= Pz\,,\\ 0 &= (1-P)z \end{align}

That is, you solve for the unknown x within the given subspace, and you can't solve for $z$ outside the projected space. An exact solution therefore only exists if $0 = (1-P)z$ exactly holds. In the least squares case, you should ignore this equation, and use $x = Pz$.