How would I solve this algebra problem?

[Dari]

Prove that a homomorphism is surjective iff its rank is the dimension of its codomain.

 

[JohnCU]

cookie monster

 

[Kirby]

Prove that a homomorphism is surjective iff its rank is the dimension of its codomain.

Take your bigotry elsewhere. 😡

 

[drinkmorejava]

Liar, that's not algebra, that's linear algebra.

 

[GodlessAstronomer]

Sounds like more discrete math to me, but it's a little over my head so I'm probably wrong.

 

[Dari]

Liar, that's not algebra, that's linear algebra.

ok. So how do I prove it?

 

[Kirby]

Do your own homework.

 

[Maximilian]

http://www.axesandalleys.com/Index/aa024/algae-bra.jpg

Just unhook it, simple as 😛

 

[Sumguy]

Because the determinant is 42

 

[simpletron]

Prove that a homomorphism is surjective iff its rank is the dimension of its codomain.

try proving the contrapositive:

if its rank is not the dimension of its codomain then the homomorphism is not surjective

 

[Dari]

1.One direction is obvious: if the homomorphism is onto then its range is the codomain and so its rank equals the dimension of its codomain. For the other direction assume that the map's rank equals the dimension of the codomain. Then the map's range is a subspace of the codomain, and has dimension equal to the dimension of the codomain. Therefore, the map's range must equal the codomain, and the map is onto. (The "therefore" is because there is a linearly independent subset of the range that is of size equal to the dimension of the codomain, but any such linearly independent subset of the codomain must be a basis for the codomain, and so the range equals the codomain.)