## A Linear Transformation from Vector Space over Rational Numbers to itself

## Problem 75

Let $\Q$ denote the set of rational numbers (i.e., fractions of integers). Let $V$ denote the set of the form $x+y \sqrt{2}$ where $x,y \in \Q$. You may take for granted that the set $V$ is a vector space over the field $\Q$.

**(a)** Show that $B=\{1, \sqrt{2}\}$ is a basis for the vector space $V$ over $\Q$.

**(b)** Let $\alpha=a+b\sqrt{2} \in V$, and let $T_{\alpha}: V \to V$ be the map defined by

\[ T_{\alpha}(x+y\sqrt{2}):=(ax+2by)+(ay+bx)\sqrt{2}\in V\]
for any $x+y\sqrt{2} \in V$.

Show that $T_{\alpha}$ is a linear transformation.

**(c)** Let $\begin{bmatrix}

x \\

y

\end{bmatrix}_B=x+y \sqrt{2}$.

Find the matrix $T_B$ such that

\[ T_{\alpha} (x+y \sqrt{2})=\left( T_B\begin{bmatrix}

x \\

y

\end{bmatrix}\right)_B,\]
and compute $\det T_B$.

(*The Ohio State University, Linear Algebra Exam*)