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I hope this one is easy. Suppose I have an underdetermined, rectangular matrix $A$ and vector $b$. I want to reason about the subspace where $Ax = b$ and specifically the projection $y:= Tx$. Is there a way to describe the space of $y$'s that satisfy the constraint $Ax = b$? My intuition is this should also be a linearly constrained subspace.

I.e., how do I express constraints that define the space $\{y | \exists x, Ax = b, y = Tx \}$?

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  • $\begingroup$ Okey doke. Reposted at Math.SO. $\endgroup$
    – Bert Huang
    May 1, 2015 at 17:57

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I think the following is right.

I am assuming $L$ and $A$ operate on the same finite dimensional vector space $V$. Write all $x \in V$ as $x=x_0+x_1$ where $x_0 \in W $, where $W=\mathrm{Kernel}(A)$ and $x_1=x-x_0$ is in its orthogonal complement $W^{\perp}$. Then $A$ restricted to $W^{\perp}$, call it $A_r$, is a full rank operator onto some subspace $W'\subset V.$ And one can check that if $Ax=b$ with $b\neq 0,$ then $A_r(x_1)=b.$

If $b=0$ then $\{y: y=T(x_0),x_0 \in W\}$

If $b\neq 0$ then $\{y: y=T(x_0)+T(A_r^{-1}(b)),x_0 \in W\}$

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