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Given linear map $A:\mathbb{R}^2\to \mathbb{R}^4$ defined as $$A = \begin{pmatrix} 1 & 1 \\ 1 & -1 \\ 0 & 2 \\ 3 & 1 \end{pmatrix}$$

and linear manifold $ W \subset \mathbb{R}^4$ $$ W = \begin{pmatrix} 1 \\ -2 \\ 0 \\ 0 \end{pmatrix} + \left[ \begin{pmatrix} 1 \\ -2 \\ 1 \\ 0 \end{pmatrix} , \begin{pmatrix} 1 \\ -1 \\ 0 \\ 1 \end{pmatrix} \right]_\lambda$$

($\left[ \dots \right]_\lambda$ denotes linear span). How do I find $A^{-1}$(W)?

All I know is that generally $A^{-1}(W) \neq A^{-1}(a) + A^{-1}(P)$ ($P$ is the subspace spanned by those 2 vectors). The result has to be manifold in $\mathbb{R}^2$ right?

Are non-parametric equations of W needed here?

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Be careful with the notation $A^{-1}$. The linear map $A$ does not have an inverse and so $A^{-1}$ is not well-defined. I assume by $A^{-1}(W)$ you mean the pre-image of $W$, i.e. the set of points that are sent to $W$. –  Fly by Night Jan 6 '13 at 0:53
    
I changed every occurrence of your $\mathbb{A}$ to ordinary $A$, as blackboard bold face is usually used to denote a set rather than a mapping. –  user1551 Jan 6 '13 at 8:00
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2 Answers

up vote 1 down vote accepted

There are most likely many different ways of doing this. Your affine subspace $W$ can be parametrised by ${\bf X} : \mathbb{R}^2 \to \mathbb{R}^4$, given by:

$${\bf X}(\lambda,\mu) = (1+\lambda+\mu,-2-2\lambda-\mu,\lambda,\mu) \, . $$

If we use $(x,y,z,t)$ as coordinates for $\mathbb{R}^4$ then $W$ can also be given by the simultaneous equations:

$$\begin{array}{ccc} x & = & 1+z+t \, , \\ y & = & -2 - 2z - t \, . \end{array}$$

If we use $(u,v)$ as coordinates for $\mathbb{R}^2$ then $A : (u,v) \mapsto (u+v,u-v,2v,3u+v).$ We see that $A(u,v)$ belongs to $W$ if $x=u+v,$ $y=u-v$, $z=2v$ and $t=3u+v$ satisfy the simultaneous equations $x=1+z+t$ and $y=-2-2z-t.$ From this we get:

$$\begin{array}{ccc} u+v &=& 1+3u + 3v \, , \\ u-v &=& -2-3u-5v \, . \end{array}$$

Thus, the set of point $(u,v) \in \mathbb{R}^2$ for which $A(u,v) \in W$ satisfy:

$$2u+2v+1=0 \, . $$

Note that $A^{-1}(W)$ is just shorthand for $\{(u,v) \in \mathbb{R}^2 : A(u,v) \in W \}.$

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Thanks for your quick and very informative answer! This is exactly what I needed! –  NumberFour Jan 6 '13 at 1:05
    
My pleasure. I'm really pleased I could help! –  Fly by Night Jan 6 '13 at 1:05
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$A^{-1}(W)=A^{-1}\left((\operatorname{range}A)\cap W\right)$ and in turn it is equal to the set of vectors $(x,y)^T\in\mathbb{R}^2$ such that $$ \begin{bmatrix} 1& 1& 1& 1\\ 1&-1&-2&-1\\ 0& 2& 1& 0\\ 3& 1& 0& 1 \end{bmatrix} \begin{bmatrix}x\\y\\z\\w\end{bmatrix} =\begin{bmatrix}1\\-2\\0\\0\end{bmatrix}. $$ It is not hard to see that $(-\frac14, -\frac14, \frac12, 1)^T$ is a solution to the above equation and the kernel of the $4\times 4$ matrix on the left is $\operatorname{span}\{(1, -1, 2, -2)^T\}$. Therefore $A^{-1}(W)$ is the line $(-\frac14, -\frac14)^T+s(1, -1)^T$, or $2x+2y+1=0$.

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