Exercise 14.4.4

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

Let $V$ be a vector space of dimension $2$ over a field $F$, and let $T : V 	o V$ be a linear map that is not a multiple of the identity. Also assume that $T$ is an isomorphism. Prove that there is $v \in V$ such that $v$ and $T(v)$ form a basis of $V$ over $F$.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof} Consider a basis ${\cal B} = (e,f)$ of $V$.

Reasoning by contradiction, suppose that for all $v \in V$, the vectors $v,T(v)$ are linearly dependent. Then, since $e
e 0,f 
e 0, e+f 
e 0$, there are some $\lambda, \mu,
u \in F$ such that
$$T(e) = \lambda e,\qquad T(f) = \mu f, \qquad T(e+f) = 
u (e+f).$$
Since $T$ is linear,
$$
u e + 
u f = T(e+f) = T(e) + T(f) = \lambda e + \mu f;$$
But $(e,f)$ is a basis, therefore $\lambda = \mu = 
u$, so that the matrix of $T$ in the basis $\cal B$ is
$
\begin{pmatrix}
\lambda & 0 \
0 &\lambda
\end{pmatrix}
= \lambda I_2
$, and $T$ would be a multiple of the identity, which is in contradiction with the hypothesis.

This proves that there is some $v \in V$ such that $v,T(v)$ are linearly independent. Since the dimension of $V$ is $2$, $v$ and $T(v)$ form a basis of $V$ over $F$.
\end{proof}

Note: The hypothesis ``$\,T$ is an isomorphism'' was useless.

Exercise 14.4.4

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Exercise 14.4.4

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