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May 19th, 2009, 07:19 AM   #1
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linear algebra- vector spaces

i really need help with this question:

Let V be a finite-dimensional vector space. suppose that T : V-->V is a linear operator. Show that T is injective if and only if T is surjective.

and are there any results for sets and functions?
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May 20th, 2009, 07:53 AM   #2
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Re: linear algebra- vector spaces

Use the rank-nullity theorem - i.e. the one that says



You can find by noting that if is injective, then and using the properties of a linear map.
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May 20th, 2009, 09:23 AM   #3
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Re: linear algebra- vector spaces

sorri could u possibly expand you reply im not getting you

cheers
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May 20th, 2009, 09:43 AM   #4
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Re: linear algebra- vector spaces

The rank-nullity theorem states that for a linear map the dimension of the kernel of and the dimension of the image of sum to the dimension of

The kernel of is defined to be the space of vectors such that
The image of is the space of vectors such that there exists an such that

Clearly, a linear operator is surjective iff

You need to show that for we have iff is injective, using the rank-nullity theorem.

To find the kernel of an injective map, note that if is injective, then for all or If you let you therefore have that is injective iff for all

Then use the properties of a linear operator to show that Once this is established, you get the desired result by a simple application of the theorem.
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May 26th, 2009, 01:48 AM   #5
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Re: linear algebra- vector spaces

thankyou so much for your help

it was much appreciated
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