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Criteria for isomorphism.

  • Thread starter frasifrasi
  • Start date
276
0
Can anyone tell me clearly what the criteria for isomorphism in linear algebra is?

For instance, my book gives the following reason:

Transformation T is not isomoprhic because T((t-1)(t-3)) = T(t^2 - 4t +3) = zero vector.

I don't get why this means T is not an isomorphism. Can anyone explain?

PS. T is a transformation from P_2 to R^(3).

the actual T is: T(f(t)) =

f(1)
f'(2)
f(3)

Thanks.
 

Answers and Replies

229
0
Let V and W be vector spaces. A linear transformation T:V->W is an isomorphism if it's bijective. That's basically the definition.

It's easy to show a linear transformation is injective iff kerT = {0}.

So in your example, T(polynomial) = 0 ,but polynomial != 0, so T is not injective, so it certainly can't be an isomorphism.
 
Last edited:
276
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so, it is not injective if the polynomial is not equal to 0?
How does the statement
"T((t-1)(t-3)) = T(t^2 - 4t +3) = zero vector."
show this?
 
229
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so, it is not injective if the polynomial is not equal to 0?
How does the statement
"T((t-1)(t-3)) = T(t^2 - 4t +3) = zero vector."
show this?
because that implies t^2 - 4t + 3 is in kerT, so kerT != {0}, so T is not injective.
 
276
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Oh, I see; you mean the only solution for the T has to be 0 and in this case there are more?
 
Last edited:
229
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Oh, I see; you mean the only solution for the T has to be 0 and in this case there are more?
i'll show all the steps once more,

We know T(t^2 - 4t + 3) = 0, so t^2 - 4t + 3 is in kerT, and since t^2 - 4t + 3 is not the zero polynomial we know kerT contains a nonzero element, so kerT != {0}, so T is not injective. (this last part is because T is injective iff kerT = {0})

Where exactly did I lose you?
 
276
0
I was jsut asking for confirmation. So, the isomorphism exists only if ker = 0 and if you can write a formula for the inverse, correct? thanks!
 

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