# All Linear Transformations that Take the Line $y=x$ to the Line $y=-x$

## Problem 454

Determine all linear transformations of the $2$-dimensional $x$-$y$ plane $\R^2$ that take the line $y=x$ to the line $y=-x$.

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## Solution.

Let $T:\R^2 \to \R^2$ be a linear transformation that maps the line $y=x$ to the line $y=-x$.

Note that the linear transformation $T$ is completely determined if the values of $T$ on basis vectors of the vector space $\R^2$ are known.

Let

\[B=\left\{\, \begin{bmatrix}

1 \\

0

\end{bmatrix}, \begin{bmatrix}

1 \\

1

\end{bmatrix} \,\right\}\]
be a basis of $\R^2$.

The reason of this choice is as follows.

Since we know that $T$ takes the line $y=x$ to the line $y=-x$, the vector $\begin{bmatrix}

1 \\

1

\end{bmatrix}$ is mapped into some point on the line $y=-x$.

This is how we chose the vector $\begin{bmatrix}

1 \\

1

\end{bmatrix}$. The other basis vector could be any vector that is not a multiple of $\begin{bmatrix}

1 \\

1

\end{bmatrix}$, and we just chose the simple vector $\begin{bmatrix}

1 \\

0

\end{bmatrix}$.

Let

\[T\left(\, \begin{bmatrix}

1 \\

0

\end{bmatrix} \,\right)=\begin{bmatrix}

a \\

b

\end{bmatrix}\]
for some $a,b \in \R$.

Since we know that the vector $T\left(\, \begin{bmatrix}

1 \\

1

\end{bmatrix} \,\right)$ is on the line $y=-x$, let

\[T\left(\, \begin{bmatrix}

1 \\

1

\end{bmatrix} \,\right)=\begin{bmatrix}

c \\

-c

\end{bmatrix}\]
for some $c\in \R$.

We now find a formula for this linear transformation $T$.

Let $\begin{bmatrix}

x \\

y

\end{bmatrix}$ be an arbitrary vector in the plane $\R^2$.

We express this vector as a linear combination of basis vectors:

\[\begin{bmatrix}

x \\

y

\end{bmatrix}=(x-y)\begin{bmatrix}

1 \\

0

\end{bmatrix}+y\begin{bmatrix}

1 \\

1

\end{bmatrix}.\]

Then we have

\begin{align*}

&T\left(\, \begin{bmatrix}

x \\

y

\end{bmatrix} \,\right)\\

&=T\left(\, (x-y)\begin{bmatrix}

1 \\

0

\end{bmatrix}+y\begin{bmatrix}

1 \\

1

\end{bmatrix} \,\right)\\

&=(x-y)T\left(\, \begin{bmatrix}

1 \\

0

\end{bmatrix}\,\right)+yT\left(\, \begin{bmatrix}

1 \\

1

\end{bmatrix} \,\right) && \text{since $T$ is a linear transformation}\\

&=(x-y)\begin{bmatrix}

a \\

b

\end{bmatrix}+y\begin{bmatrix}

c \\

-c

\end{bmatrix}\\

&=\begin{bmatrix}

ax+(c-a)y \\

bx-(c+b)y

\end{bmatrix}.

\end{align*}

We conclude that any linear transformation $T:\R^2\to \R^2$ that takes the line $y=x$ to the line $y=-x$ is of the form

x \\

y

\end{bmatrix} \,\right)=\begin{bmatrix}

ax+(c-a)y \\

bx-(c+b)y

\end{bmatrix}\]

for some $a, b, c\in \R$.

## Remark.

Remark that if $c=0$, then all the points on the line $y=x$ are mapped into the origin, which is on the line $y=-x$.

If we want to avoid this degenerate case, we need to assume that $c\neq 0$.

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