Surjective and injective linear map

In summary, the conversation discusses a linear transformation T from vector space V to W over a field F. It also mentions bases for V and W, denoted as a and b respectively. The conversation presents two proofs - one for showing that T is surjective if and only if the columns of [T]ba span Fn, and another for proving that T is injective if and only if the columns of [T]ba are linearly independent in Fn. A hint is provided that uses the standard isomorphism between V and Fn and mentions a typo in the second proof.
  • #1
Fernando Revilla
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MHB
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I quote an unsolved question from MHF posted by user jackGee on February 3rd, 2013.

[Let T:V->W Be A Linear Transformation
Where V and W are vector spaces over a Field F
let a={v1,v2,...,vn} be a basis for V and b={w1,w2,...,wm} be a basis for W

a) Prove that T is surjective if and only if the columns of [T]ba span Fn
b) Prove that T is injective if and only if the columns of [T]ba are linearly independent in Fn
P.S. Of course, I meant in the title and instead of an.
 
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  • #2
$(a)\;$ Hint: Use $[T(x)]_B=[T]_{BA}[x]_A=[C_1,\ldots,C_n][x]_A=x_1C_1+\ldots+x_nC_n$ and the standard isomorphism between $V$ and $\mathbb{F}^n$ given by $x\to [x]_A$.

$(b)\;$ There is a typo. It should be $\mathbb{F}^m$ instead of $\mathbb{F}^n$, otherwise does not make sense. Hint: use $\dim (\ker T)=n-\mbox{rank }[T]_{BA}$.
 

1. What is the difference between a surjective and injective linear map?

A surjective linear map is one that maps every element in the domain to at least one element in the codomain. In other words, every element in the codomain has at least one preimage in the domain. An injective linear map, on the other hand, maps each element in the domain to a unique element in the codomain. This means that no two elements in the domain have the same image in the codomain.

2. How can I determine if a linear map is surjective or injective?

To determine if a linear map is surjective, you can check if every element in the codomain has at least one preimage in the domain. This can be done by solving the equation Ax = b, where A is the matrix representation of the linear map and b is an element in the codomain. If the equation has a solution, then the linear map is surjective. To check if a linear map is injective, you can use the kernel of the map. If the kernel only contains the zero vector, then the map is injective.

3. Can a linear map be both surjective and injective?

Yes, a linear map can be both surjective and injective. This type of map is called a bijective linear map. It means that every element in the codomain has a unique preimage in the domain, and vice versa. In other words, the map is both onto and one-to-one.

4. What is the importance of surjective and injective linear maps?

Surjective and injective linear maps are important in linear algebra because they help us understand the relationship between the domain and codomain of a linear map. They also play a crucial role in determining the invertibility of a linear map. A surjective map is invertible if and only if it is also injective. This means that surjective and injective linear maps are essential in solving systems of linear equations and in many other applications.

5. Are surjective and injective linear maps only applicable to vector spaces?

No, surjective and injective linear maps can also be defined for other mathematical structures, such as groups, rings, and modules. In these cases, the maps are called surjective and injective homomorphisms. The same principles apply, where surjective homomorphisms map every element in the domain to at least one element in the codomain, and injective homomorphisms map each element in the domain to a unique element in the codomain.

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