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$V$ is a finite dimensional vector space over the reals and $W$ is a subspace of $V$. I am to prove that $T(W)$ and $W$ intersect at points other than $0$ for every linear automorphism $T$ defined on $V$ if and only if $\text{dim } W > 1/2 \text{ dim } V$. I tried to use rank nullity theorem but failed to conclude anything from it.

Alejandro Bergasa Alonso
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sxccalmat1100
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1 Answers1

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Suppose $T$ is an automorphism of $V$. Then $\dim T(W) = \dim W$ and by Grassman's formula we have $$ \begin{aligned} \dim T(W) \cap W &= \dim T(W) + \dim W - \underbrace{\dim T(W) \cup W}_{\leq \dim V} \\ &\geq 2 \dim W - \dim V \end{aligned} $$ so $\dim T(W) \cap W \geq 1$ as soon as $\dim W > \frac{1}{2} \dim V$.

Now for the other side of the equivalence, suppose $\dim W \leq \frac{1}{2} \dim V$.

Take $(e_1, \dots, e_k)$ a basis of $W$ and complete it in $(e_1, \dots, e_n)$ a basis of $V$.

We have $k = \dim W \leq \frac{1}{2} \dim V = \frac{n}{2}$ so $2k \leq n$.

Consider the linear application $T : V \rightarrow V$ defined by

$T(e_i) = e_{k+i}$ for $1 \leq i \leq k$,

$T(e_i) = e_{i-k}$ for $k + 1 \leq i \leq 2k$,

$T(e_i) = e_{i}$ for $i > 2k$.

Then $T$ is an automorphism but $T(W) = \text{Vect}(e_{k+1}, \dots, e_{2k})$ while $W = \text{Vect}(e_{1}, \dots, e_{k})$ so $T(W) \cap W = \{0\}$.

DodoDuQuercy
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