Finding a linear map $\phi:\mathbb{R}^3\mapsto\mathbb{R}^3$ given preimage and image

Question

I've been trying to understand how to solve a question in a Linear Algebra exam I've taken today, but I can't seem to understand where to even start. The question is:

Determine a linear map $\phi:\mathbb{R}^3\mapsto\mathbb{R}^3$ that maps the affine subspace spanned by the plane $x-3y+z-1=0$ to the line $\begin{cases}x+y=0\\y=z-1\end{cases}$

My greatest hurdle seems to be dealing with the fact that the plane and the line are translated. Any help is much appreciated!

Answer 1

The equation $x-3y+z-1=0$ can be written as

$$\langle (x,y,z), (1,-3,1)\rangle = 1. \tag{1}$$

A solution is $(x_0,y_0,z_0) = (1,0,0)$. We can hence rewrite it as

$$\langle (x,y,z) - (x_0,y_0,z_0), (1,-3,1)\rangle = 0$$

This brings to mind the plane through the origin and perpendicular to $(1,-3,1)$,

$$\langle (a,b,c), (1,-3,1)\rangle = 0. \tag{2}$$

Hence, the affine plane defined by $(1)$ is obtained from the plane through the origin defined by $(2)$ via a translation by $(x_0,y_0,z_0)$.

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Similarly, let's consider the line $R$ given by equations $\begin{cases}x+y=0\\y=z-1\end{cases}$. We have two solutions in $(0,0,1)$ and $(1,-1,0)$. The line can hence be parametrized as

\begin{align} r(t) &= (0,0,1)+t\cdot\Big((1,-1,0)-(0,0,1)\Big) \\&= (0,0,1)+t\cdot(1,-1,-1). \end{align}

$R$ is hence obtained from the line $T$ through the origin and with direction $(1,-1,-1)$ via a translation by $(0,0,1)$.

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A linear transformation takes a subspace to a subspace, and an affine space to an affine space. It is hence enough to require that $\phi$ takes $(x_0,y_0,z_0)$ to $(0,0,1)$ and plane $(2)$ to line $R$.

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The plane $(2)$ is spanned by any two linearly independent solutions. If $(a,b,c) = (x,y,z) -$ $(x_0,y_0,z_0)$, we can easily find the linearly independent solutions

$$(a_1,b_1,c_1) = (1,0,-1)\\ (a_2,b_2,c_2) = (3,1,0).$$

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We finally have everything we need. We just need to find a linear transformation $\phi$ such that

• $\phi(1,0,0) = (0,0,1)$: $\phi$ takes an offset of the affine plane to an offset of the affine line
• $\phi(1,0,-1) = \phi(3,1,0) = (1,-1,-1)$: $\phi$ takes the plane through the origin parallel to $(1)$ to the line through the origin parallel to $R$

Notice that $\{(1,0,0), (1,0,-1), (3,1,0)\}$ is a linealy independent set $($and hence a basis for $\mathbb R^3)$, so indeed this uniquely determines $\phi$.

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I calculate $\phi(v) = \pmatrix{0&1&-1\\0&-1&1\\1&-4&2}\,v$. If $v=(x,y,z)$ then

$$\phi(v) = (y-z, -y+z,x-4y+2z)$$

Writing $\phi(v) = (\phi_x, \phi_y, \phi_z)$, notice that $\phi_x+\phi_y=0$.

Moreover, if $x-3y+z-1=0$ $($that is, $v$ lies in the plane$)$, then $\phi_z = (x-3y+z)-y+z =$ $1 + \phi_y$, or $\phi_y = \phi_z - 1$. In other words, $\phi(v)$ satisfies our equation for the line.

Answer 2

I often find working with homogeneous coordinates in a projective space convenient for problems involving flats in $\mathbb R^n$: affine transformations become linear ones and straightforward matrix computations can usually be brought to bear.

In this framework, the equation of the plane says that it is the null space of the matrix $\small{\begin{bmatrix}1&-3&1&-1\end{bmatrix}}$ a basis for which can be found by inspection: it is spanned by $\{(1,0,0,1)^T,(-1,0,1,0)^T,(3,1,0,0)^T\}$. Similarly, the line is the null space of $$\begin{bmatrix}1&1&0&0\\0&1&-1&1\end{bmatrix},$$ which is spanned by $\{(1,-1,0,1)^T,(-1,1,1,0)^T\}$. This gives us all of the pieces that we need to construct the required map.

Affine transformations leave the plane at infinity fixed, so we must take care to map points at infinity only to points at infinity. We also want to leave the origin fixed so that the transformation will be linear. This suggests defining the mappings $$\begin{align}(1,0,0,1)^T &\to (1,-1,0,1)^T \\ (-1,0,1,0)^T &\to (-1,1,1,0)^T \\ (3,1,0,0)^T &\to (-1,1,1,0)^T \\ (0,0,0,1)^T &\to (0,0,0,1)^T \end{align}.$$ The matrix of the transformation of $\mathbb P^3$ that achieves this is then simply $$ \begin{bmatrix}1&-1&-1&0\\-1&1&1&0\\0&1&1&0\\1&0&0&1\end{bmatrix} \begin{bmatrix}1&-1&3&0 \\ 0&0&1&0\\0&1&0&0\\1&0&0&1\end{bmatrix}^{-1} = \begin{bmatrix}1&-4&0&0 \\ -1&4&0&0 \\ 0&1&1&0 \\ 0&0&0&1 \end{bmatrix}.$$ The last row of this matrix verifies that it is indeed an affine transformation and the last column tells us that it is linear, as required. The matrix of the corresponding linear transformation of $\mathbb R^3$ is therefore the upper-left $3\times3$ submatrix. This map has rank 2, as suspected. Other choices of basis for the line and plane might produce different maps.

To verify, we can use the spanning set for the plane to construct a parameterization of it in $\mathbb R^3$, namely $(1,0,0)+\lambda(-1,0,1)+\mu(3,1,0)$. Applying the above map to this point gives $(1-\lambda-\mu, \lambda+\mu-1,\lambda+\mu)$, which clearly satisfies the defining equations of the line.

Answer 3

Hint Find a linear transformations $S$ that maps the given plane $\Pi$ onto the reference plane $\Pi_0 := \{z = 1\}$ and a map $T$ that maps a reference line $\ell_0 := \{x = 0, z = 1\}$ onto the given line $\ell$, so that the line is the image of the plane under the evident linear projection $\pi: (x, y, z) \mapsto (0, y, z)$. Then, by construction, $$\boxed{\phi := T \circ \pi \circ S} \qquad \textrm{(surjectively) maps} \qquad \Pi \stackrel{S}{\mapsto} \Pi_0 \stackrel{\pi}{\mapsto} \ell_0 \stackrel{T}{\mapsto} \ell$$ as desired.

• To find a map $S$, note that if we denote $z' := x - 3 y + z$, then $\Pi = \{z' = 1\}$, so we need to find $S$ such that $\pmatrix{1&-3&1} S = \pmatrix{0&0&1}$, and making the easiest choices gives $$S = \pmatrix{1&0&0\\ 0&1&0\\1&3&1} .$$ The above condition guarantees that $S$ maps $\Pi$ into the reference plane, and $S$ is (visibly) nondegenerate so it maps onto that plane.
• To find a map $T$, note that we can parameterize the reference line via $t \mapsto (0, 0, 1) + t (0, 1, 0)$ and $\ell$ via $t \mapsto (0, 0, 1) + t (1, -1, -1)$, so it suffices to find $T$ that maps $(0, 1, 0)$ to $(1, -1, 1)$ and fixes $(0, 0, 1)$, and thus we may take, e.g., $$T := \pmatrix{0&1&0\\0&-1&0\\0&-1&1} .$$

The choices we made above produced $T, \pi, S$ with many zero entries, making $\phi$ fast to compute by hand: $$\color{#bf0000}{\boxed{\phi = \pmatrix{0&1&0\\0&-1&0\\1&-4&1}}} .$$