# Finding basis of inverse image

Let $\psi$ be a linear transformation such that$$\psi ([x_1,x_2,x_3,x_4])=[x_1+x_3+x_4, -x_2-x_4,x_1+x_2+x_3+2x_4].$$ Find basis of inverse image $\psi^{-1}(W)$ of subspace $W=span([1,1,1],[3,2,1])\subset \mathbb{R}^3$ How to approach this?

• I am sorry, my mistake Dec 18, 2014 at 18:58
• Since I just spent a bunch of time earlier walking you through another problem rather like this, I'm going to suggest that the way to approach this is to get to know your definitions very well. Calling that span $S$, is it true that each of the spanning vectors for $S$ is in the image of $\psi$? What are their pre-images under $\psi$? How are these preimages related to the preimage of all of $S$ under $\psi$? What's the dimension of the preimage of $W$? Dec 18, 2014 at 19:16

I suggest that you find a basis for $\psi(\mathbb{R}^4) = T$. This could be done by simply calculating $\psi(e_1), \dots \psi (e_4)$ and then removing one by one those of them which breaks the linear independence. When you have the basis of T, it's time to calculate $T \cap W$. One way to do this is by solving several systems of equations. Namely you can find basis $t_1, \dots, t_k$ for $T^{\perp}$ and basis $w_1, \dots w_s$ for $W^{\perp}$ and then basis $b_1, \dots, b_m$ for $(\operatorname{span}(t_1, \dots, t_k, w_1, \dots, w_s))^{\perp}$, which will be just the basis of $T \cap W$. Now you can be sure that all of the $b_i \in \operatorname{Im}(\psi)$, so you can use the definition of $\psi$ to form a system of equations for every single one of them : $\psi(x_{1_{b_i}}, x_{2_{b_i}}, x_{3_{b_i}}, x_{4_{b_i}}) = b_i$ find the corresponding vectors and take their union for the final result. Ask me if you have questions. Not sure if I explained it very well... (However did my best :))