A Linear Transformation from Vector Space over Rational Numbers to itself

Ohio State University exam problems and solutions in mathematics

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 transformaiton.

(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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Hint.

  1. For (a), to show that $B$ is linearly independent, consider a linear combination
    \[c_1\cdot 1+c_2\sqrt{2}=0,\]where $c_1, c_2 \in Q$. Use the fact that $\sqrt{2}$ is a irrational number to conclude $c_1=c_2=0$.
  2. You may directly check the linearity, or note that \[T_{\alpha}(x+y\sqrt{2})=\alpha(x+y\sqrt{2}).\]

Proof.

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

Since $V=\{x\cdot 1+y\sqrt{2}\mid x, y \in \Q \}=\Span \{x, y\}$, we see that $B=\{1, \sqrt{2}\}$ spans the vector space $V$.
To show that $B$ is a linearly independent set, consider a linear combination
\[c_1\cdot 1+c_2\sqrt{2}=0,\] where $c_1, c_2 \in \Q$.

We want to show that $c_1=c_2=0$. Assume that this is not the case. Then both $c_1$ and $c_2$ must be nonzero.
Then we have
\[\sqrt{2}=-\frac{c_1}{c_2}.\] (Note that $c_2 \neq 0$.)

Since $c_1, c_2 \in \Q$, the fraction $\frac{c_1}{c_2}$ is also a rational number. However $\sqrt{2}$ is irrational number.

Hence this is a contradiction and we must have $c_1=c_2=0$. Thus $B$ is a linearly independent set, and $B$ is a basis.

(b) Show that $T_{\alpha}$ is a linear transformaiton.

First note that we can simplify the given formula as
\[T_{\alpha}(x+y\sqrt{2})=(a+b\sqrt{2})(x+y\sqrt{2})=\alpha(x+y\sqrt{2}).\] Namely, $T_{\alpha}$ is a multiplication by $\alpha$.

Let $x_1+y_1\sqrt{2}, x_2+y_2\sqrt{2}\in V$, where $x_1,y_1, x_2, y_2 \in \Q$.
Then we have
\begin{align*}
T_{\alpha}\left( (x_1+y_1\sqrt{2})+(x_2+y_2\sqrt{2}) \right) &=
\alpha\left( (x_1+y_1\sqrt{2})+(x_2+y_2\sqrt{2}) \right)\\
&=\alpha (x_1+y_1\sqrt{2})+\alpha (x_2+y_2\sqrt{2})\\
&=T_{\alpha}(x_1+y_1\sqrt{2})+T_{\alpha} (x_2+y_2\sqrt{2}).
\end{align*}

For $x+y\sqrt{2}\in V$ and $c \in \Q$, we also have
\begin{align*}
T_{\alpha}\left(c(x+y\sqrt{2})\right)&=\alpha \left(c(x+y\sqrt{2})\right)\\
&=c \alpha (x+y\sqrt{2})=c T_{\alpha}(x+y\sqrt{2}).
\end{align*}
Therefore $T_{\alpha}$ is a linear transformation.

(c) Find the matrix $T_B$ and compute $\det T_B$.

To find the matrix representation for the linear transformation $T_{\alpha}$ with respect to the basis $B$, we compute $T_{\alpha}(1)$ and $T_{\alpha}(\sqrt{2})$.

We have
\begin{align*}
T_{\alpha}(1)&=\alpha\cdot 1=a+b\sqrt{2}=\begin{bmatrix}
a \\
b
\end{bmatrix}_B\\
T_{\alpha}(\sqrt{2})&=\alpha\cdot \sqrt{2}=2b+a\sqrt{2}=\begin{bmatrix}
2b \\
a
\end{bmatrix}_B.
\end{align*}
Thus the matrix is
\[T_{B}=\begin{bmatrix}
a & 2b\\
b& a
\end{bmatrix}.\] The determinant of this matrix is
\[ \det T_B=a^2-2b^2.\]

Comment.

In a typical first year linear algebra course, we use real numbers $\R$ or complex numbers $\C$ as a base field.
Or just state theorems using a field $K$, for a general field $K$ but most of the examples use $\R$ or $\C$.

In this sense this problem is unorthodox because the field is rational numbers $\Q$.
But the problem is quite typical in Field theory or in Number theory.


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