0
$\begingroup$

V vectorspace, $\tau: V \rightarrow V$, $\tau$ $\mathbb{K}$-linear and bijective. Proof $\tau^{-1}: V \rightarrow V$

So basically $\tau$ takes an inputvector of V and maps it onto an outputvector in the same vectorspace. Now, intuitively $\tau^{-1}$ will an inputvector of V and map it back into the same Vectorspace.

In my attempt, I try to proof $\tau^{-1}$ is $\mathbb{K}$- linear:

$\tau(\lambda u) = \alpha w$

$\tau(\mu v) = \beta x$,

$\tau^{-1}(\alpha w) = \lambda u$

$\tau^{-1}(\beta x) =\mu v$

$\tau^{-1}(\alpha w) + \tau^{-1}(\beta x) = \lambda u + \mu v \Leftrightarrow$

$\alpha + \beta x = \tau(\lambda u) + \tau (\mu v) \Leftrightarrow$

$\alpha w + \beta x = \tau(\lambda u + \mu v) \Leftrightarrow$

$\tau^{-1}(\alpha w+ \beta x) = \lambda u + \mu v$

So all in all: $\tau^{-1}(\alpha w) + \tau^{-1}(\beta x) = \lambda u + \mu v = \tau^{-1}(\alpha w+ \beta x) \Rightarrow \tau^{-1}(\alpha w) + \tau^{-1}(\beta x) = \tau^{-1}(\alpha w+ \beta x)$

  • 1
    Yes, you are fine, although the working is elaborate, but that's good if you are a beginner. – 2017-01-28
  • 0
    @астонвіллаолофмэллбэрг Thanks for your verification! – 2017-01-28
  • 0
    Thou art welcome. – 2017-01-28

0 Answers 0