Linear Algebra Basis Problem

  • #1
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 Im 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 Im 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
 

Answers and Replies

  • #2
351
1
Well xn is linearly independent and it has the same dimension as V/ker T. So it's a basis.
 
  • #3
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
124
0
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:

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