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)
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