How do I show that the mapping of orthogonal complement of $\ker T$ to $\mathrm{range}\ T$ is one to one?

Question

Having trouble figuring this out:

Show that the mapping $T : (\ker T)^{\perp} \to \mathrm{range}\ T$ is one to one.

I have the definitions but am I suppose to set them equal to each other?

Is $T$ a map between Hilbert spaces? – Michael Albanese Mar 17 '13 at 05:37 — Michael Albanese, Mar 17 '13 at 05:37
Yes. T goes from H to H and is a bounded linear operator – User69127 Mar 17 '13 at 05:38 — User69127, Mar 17 '13 at 05:38

score 2 · Accepted Answer · answered Mar 17 '13 at 05:44

2

Hint: for $\,x,y\in\left(\ker T\right)^\perp\,$

$$Tx=Ty\Longrightarrow T(x-y)=0\,\Longleftrightarrow \,x-y\in\ker T\cap(\ker T)^\perp\,\ldots$$

answered Mar 17 '13 at 05:44

DonAntonio

Thanks for the Hint, Don Antonio! I'm confused on your last part. What am I trying to get from that? That this is also the same in the range? – User69127 Mar 17 '13 at 05:56
For any subspace $,W\le V,$ in an inner product space, what is $,W\cap W^\perp,$ ? – DonAntonio Mar 17 '13 at 05:59
a subset of {0} – User69127 Mar 17 '13 at 06:03
No: a subspace of zero, meaning: the trivial subspace $,{0},$ itself! – DonAntonio Mar 17 '13 at 06:04
perfect! so x-y will be in the trivial subspace. Thus, x = y and we are done, right?! – User69127 Mar 17 '13 at 06:08
Así es, or: indeed! – DonAntonio Mar 17 '13 at 06:09
Muchas Gracias, Senor! God bless! – User69127 Mar 17 '13 at 06:10

1 Answers1