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

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Homework Help Overview

The discussion revolves around proving that a set of vectors, specifically (x_k+1, . . . , x_n), forms a basis for the quotient space V/kerT, where V is a finite-dimensional vector space and T is a linear map. Participants are exploring the implications of linear independence and dimensionality in relation to the basis of the kernel of T.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are examining the relationship between the linear independence of the vectors (x_k+1, . . . , x_n) and their ability to form a basis for V/kerT. There is a focus on the dimensions of the spaces involved and the definitions of basis and span.

Discussion Status

Some participants have provided insights regarding the linear independence of the vectors and their dimensionality in relation to V/kerT. However, there remains a lack of consensus on whether these properties are sufficient to conclude that (x_k+1, . . . , x_n) forms a basis for V/kerT.

Contextual Notes

Participants are working under the constraints of definitions related to vector spaces and linear maps, specifically regarding the kernel and quotient spaces. There is an ongoing examination of the assumptions made in the proof process.

HyperbolicMan
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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 \subseteq V and I am T \cong 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 \subseteq V and I am T \cong 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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Well xn is linearly independent and it has the same dimension as V/ker T. So it's a basis.
 
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.
 
Follows straight from the definition of basis:
  1. x_{k+1} + \ker{T}, \dots , x_{n} + \ker{T} are linearly independent in V / \ker{T};
  2. x_{k+1} + \ker{T}, \dots , x_{n} + \ker{T} span whole V / \ker{T}.

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

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