Quod Erat Demonstrandum

2008/05/10

Filed under: University Mathematics — johnmayhk @ 5:49 下午

Let $V$ be a finite vector space over $\mathbb{Q}$, let $F: V \rightarrow V$ be a linear transformation. Suppose $x,y,z \in V$ such that

1. $F(x) = y$
2. $F(y) = z$
3. $F(z) = x+y$

Suppose $x$ is non-zero. Show that $x,y,z$ are linearly independent.

2 則迴響 »

1. i am a maths undergraduate in the UK
多多指教！

迴響 由 雅典娜 — 2008/05/11 @ 2:05 上午 | 回應

2. Nice to meet you 雅典娜! It is great to know someone study math in different countries, share something about the studying in UK with us if you want to.

Solution to the question

Firstly, we claim that $x, y$ are independent, otherwise, $y = ax$ for some non-zero $a \in \mathbb{Q}$. Apply $F$, $F(y) = F(ax)$, yield $z = ay = a^2x$.

From (3), $F(z) = x + y$
$\Rightarrow F(a^2x) = x + ax$
$\Rightarrow a^2F(x) = x(1 + a)$
$\Rightarrow a^2y = x(1 + a)$
$\Rightarrow a^3x = x(1 + a)$

Since $x \ne 0$, yield
$a^3 - a - 1 = 0$

But there should be no rational root for the equation above, hence $a$ does not exist and therefore $x, y$ are independent.

Now, similarly, we claim that $z$ is independent of $x$ and $y$, otherwise, let $z = bx + cy$ for some $b, c \in \mathbb{Q}$. Apply $F$, yield

$F(z) = bF(x) + cF(y)$
$\Rightarrow x + y = by + cz$
$\Rightarrow z = \frac{1}{c}x + \frac{1 - b}{c}y$ (Note: $c \ne 0$ otherwise $x, y$ will not be independent)

Hence $\frac{1}{c}x + \frac{1 - b}{c}y = bx + cy$
By the independence of $x$ and $y$, we have

$\frac{1}{c} = b$
$\frac{1 - b}{c} = c$

Eliminate $b$; thus $c^3 - c + 1 = 0$ and there is no rational $c$ satisfying the equation, hence $x,y,z$ are independent.

I forget the source of this question, sorry.

迴響 由 johnmayhk — 2008/05/13 @ 10:46 上午 | 回應