# Linear Algebra - Hyperplane and a linear mapping

I'm doing some problems in Linear Algebra and this is the one I just have no idea how to start with. We are given a linear transformation $f \in L(\mathbb R^{6\times1} , \mathbb R^{4\times1})$ with its coordinate matrix:

$$\begin{pmatrix} -1 & -2 & -2 & -1 & -5 & 0 \\ 1 & 1 & 1 & 1 & 3 & 0 \\ 1 & 3 & 2 & 2 & 5 &1 \\ 1 & 2 &1 & 1 &3 &0 \\ \end{pmatrix}$$

And besides that, a hyperplane $\mathbf H \subset \mathbb R^{4\times1}$ with the equation $y_1 + y_2 + y_3 + y_4 = 0$ We are supposed to find a linear equation with 6 unknown variables $x_1, ... , x_6$ that have a solution set matching with $f^{-1}(\mathbf H)$. We are also supposed to find a basis of the subspace $f^{-1}(\mathbf H)$.

I'm not very clear with the fact that they can build an inverse of a function that isn't bijective (or am I wrong for concluding that just because I can't build an inverse of the coordinate matrix)?

There was a tip given: we should use a transpose $f^T$ for solving the fist part of the question.

Any kind of help/tip is welcome, at least how to approach or start the problem!

• If $y=f(x)$ then $y_1+y_2+y_3+y_4=g(y)$ for $g=(1,1,1,1)\in L(\mathbb R^4,\mathbb R)$. So, use $g\circ f$ for constructing your equation. – Sergei Golovan Dec 25 '16 at 20:49
• @SergeiGolovan Thanks a lot! I still have a question though. I calculated the $g \circ f$ simply by multiplying the two matrices, and got something that looks like the answer I was looking for, but I still don't understand what am I supposed to do with the transpose? – Milena Dec 25 '16 at 21:02
• I don't know what is supposed to be done with the transposed matrix, sorry. – Sergei Golovan Dec 25 '16 at 21:05

Remember the handy-dandy formula $$A^\top z\cdot x = z\cdot Ax \quad\text{for A\in L(\Bbb R^n,\Bbb R^m), x\in\Bbb R^n, z\in\Bbb R^m}.$$ You are given the hyperplane $\mathbf H = \{y\in\Bbb R^4: z\cdot y=0\}$, where $z=\begin{bmatrix} 1\\1\\1\\1 \end{bmatrix}$. Then $f^{-1}(\mathbf H) = \{x\in\Bbb R^6: z\cdot f(x) = 0\} = \{x\in\Bbb R^6: f^\top(z)\cdot x = 0\}$.