Linear Algebra exam problem: is my answer correct?

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This was a problem on a midterm I just took.

Homework Statement



Let f and g be linear transformations such that f: C^n ---> C^m is injective and g: C^m ---> C^n is onto (C = complex numbers).

Prove or disprove with a counterexample: g composed with f is an isomorphism.

The Attempt at a Solution



I claim the proposition is false.

Let g: C^1 --> C^1 be such that g(z) ={ 0 if z=0, z -(1+i) if z does not = 0}.
Let f: C^1 --> C^1 be such that f(z) = z.

Clearly, f is injective. I claim that g is onto, since if y is complex, there exists a complex z such that g(z) = y. Namely, z = y + (1+i). However, since both g(0)=g(1+i)=0, g is not injective.

Hence, g composed with f is not injective, since g(f(0))=g(f(1+i)), and thus cannot be an isomorphism since it is not bijective (not invertible).


How does that look?
 

Answers and Replies

  • #2
Dick
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This was a problem on a midterm I just took.

Homework Statement



Let f and g be linear transformations such that f: C^n ---> C^m is injective and g: C^m ---> C^n is onto (C = complex numbers).

Prove or disprove with a counterexample: g composed with f is an isomorphism.

The Attempt at a Solution



I claim the proposition is false.

Let g: C^1 --> C^1 be such that g(z) ={ 0 if z=0, z -(1+i) if z does not = 0}.
Let f: C^1 --> C^1 be such that f(z) = z.

Clearly, f is injective. I claim that g is onto, since if y is complex, there exists a complex z such that g(z) = y. Namely, z = y + (1+i). However, since both g(0)=g(1+i)=0, g is not injective.

Hence, g composed with f is not injective, since g(f(0))=g(f(1+i)), and thus cannot be an isomorphism since it is not bijective (not invertible).


How does that look?
Pretty bad. There's no value of z such that g(z)=(-(1+i)), so g isn't onto. And g certainly isn't linear.
 
  • #3
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oh damn, yeah that's brutal.

so without providing a proof, is the proposition true or false?
 
  • #4
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In my head I convinced myself that if g isn't injective it ruins the isomorphism but i'm starting to think that's a serious mind-lapse.
 
  • #5
Dick
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If g is onto then m<=n. What does f injective tell you about m and n?
 
  • #6
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g is onto, but in general if we restrict the domain of g to some subset of C^m, there's no reason to suspect that it remains onto as a map to C^n. f isn't necessarily onto, so it's image may be some subset of C^m, and therefore the image of g composed with f will not necessarily be all of C^n. Therefore I suspect the composition is not necessarily an isomorphism. One example you could consider, I think would be if f mapped from C^1 to C^2 where z mapped to (z,0). Then you could consider a g that maps from C^2 to C^1 mapping (z1,z2) to z2. The composition of these will map everything in C^1 to 0.
 
  • #7
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f inj -> ker(f)=0 -> n = m - 0 = m ---> lin trans between vect spaces of same dim is isomorphism.

ugh.
 
  • #8
Dick
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f inj -> ker(f)=0 -> n = m - 0 = m ---> lin trans between vect spaces of same dim is isomorphism.

ugh.
Sure, once you conclude m=n then both f and g are isomorphisms.
 
  • #9
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g is onto, but in general if we restrict the domain of g to some subset of C^m, there's no reason to suspect that it remains onto as a map to C^n. f isn't necessarily onto, so it's image may be some subset of C^m, and therefore the image of g composed with f will not necessarily be all of C^n. Therefore I suspect the composition is not necessarily an isomorphism. One example you could consider, I think would be if f mapped from C^1 to C^2 where z mapped to (z,0). Then you could consider a g that maps from C^2 to C^1 mapping (z1,z2) to z2. The composition of these will map everything in C^1 to 0.
but we can't restrict domain of g, since 1) the domain is given as C^m and still, if we argue that f may restrict g's domain in the composition, see above for proof no.

uhggggg nothing worse than absolutely butchering a pretty simple problem.
 

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