Proving (x_k+1, . . . , x_n) forms a basis for V/kerT

In summary, the conversation discusses the proof of ker T \subseteq V and I am T \cong V/kerT, and the basis of V/kerT is shown to be (x_k+1, . . . , x_n). The problem is understanding how (x_k+1, . . . , x_n) forms a basis for V/kerT, but it is clarified that it follows from the definition of basis.
  • #1
HyperbolicMan
14
0
Hi, I was working through this proof in my linear al textbook and there's this one step I can't get past. Any help would be appreciated.

Homework Statement



Let V be a finite dimensional vector space, and let T be a linear map defined on V.

ker T [tex]\subseteq[/tex] V and I am T [tex]\cong[/tex] V/kerT

Let (y_1, . . . , y_k) be a basis of ker T. Augment this list by (x_k+1, . . . , x_n) to a basis for V: (y_1, . . . , y_k, x_k+1, . . . , x_n).

Now here's the part that's getting me:

"Obviously, (x_k+1, . . . , x_n) forms a basis of V/kerT"

Homework Equations



ker T [tex]\subseteq[/tex] V and I am T [tex]\cong[/tex] V/kerT

The Attempt at a Solution



I believe that span(x_k+1, . . . , x_n) is isomorphic to V/kerT (they have the same dimension), but I don't see how (x_k+1, . . . , x_n) actually forms a basis for V/kerT
 
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  • #2
Well xn is linearly independent and it has the same dimension as V/ker T. So it's a basis.
 
  • #3
hgfalling said:
Well xn is linearly independent and it has the same dimension as V/ker T. So it's a basis.

yes, but I think this only shows that (x_k+1, . . . , x_n) is basis for span(x_k+1, . . . , x_n), not for V/kerT.
 
  • #4
Follows straight from the definition of basis:
  1. [tex]x_{k+1} + \ker{T}, \dots , x_{n} + \ker{T}[/tex] are linearly independent in [tex]V / \ker{T}[/tex];
  2. [tex]x_{k+1} + \ker{T}, \dots , x_{n} + \ker{T}[/tex] span whole [tex]V / \ker{T}[/tex].

Tell me which part you have problems with.
 
Last edited:

Related to Proving (x_k+1, . . . , x_n) forms a basis for V/kerT

1. What is a basis in linear algebra?

A basis in linear algebra is a set of linearly independent vectors that span the entire vector space. This means that any vector in the vector space can be expressed as a linear combination of the basis vectors.

2. What is the importance of a basis in linear algebra?

A basis is important in linear algebra because it provides a way to represent and manipulate vectors in a vector space. It also allows for a more efficient and concise way to solve linear systems of equations.

3. How do you find a basis for a given vector space?

To find a basis for a vector space, you need to first determine the linearly independent vectors in the space. This can be done by using elimination techniques or by finding the null space of a matrix. Once you have the linearly independent vectors, you can use them as the basis for the vector space.

4. Can a vector space have more than one basis?

Yes, a vector space can have multiple bases. This is because there are often many different sets of linearly independent vectors that can span a given vector space. However, all bases for a particular vector space will have the same number of vectors, known as the dimension of the vector space.

5. How is a basis used in solving systems of linear equations?

A basis is used in solving systems of linear equations by providing a set of vectors that can be used to represent the variables in the system. This allows for a more organized and systematic approach to solving the equations, often using techniques such as Gaussian elimination or matrix operations.

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