Linear Transformations,Find basis of kernel and range

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Discussion Overview

The discussion revolves around finding the basis of the kernel and range of a linear transformation defined from R4 to R3. Participants explore methods for calculating the kernel and image of the transformation, as well as discussing the implications of dimensionality in this context.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant defines the transformation T and poses the problem of finding the basis of its kernel and range.
  • Another participant suggests calculating the kernel by solving a system of equations derived from the transformation's output being equal to zero.
  • There is a proposal to find the image by solving a system with parameters alpha, beta, and gamma, indicating conditions for solutions.
  • A different viewpoint suggests that to find the image, one should first establish a basis for R4 and then evaluate the transformation on those basis vectors to determine if they are linearly independent.
  • One participant introduces a dimensionality relationship, stating that the sum of the dimensions of the kernel and image equals the dimension of the domain space.

Areas of Agreement / Disagreement

Participants express differing methods for calculating the image of the transformation, with some advocating for a direct approach using basis vectors while others suggest solving systems of equations. There is no consensus on the preferred method, and the discussion remains unresolved.

Contextual Notes

Some participants' approaches depend on the linear independence of the set of transformed basis vectors, which may not be established without further analysis. Additionally, the dimensionality relationship mentioned may require clarification regarding the specific transformation in question.

nehap.2491
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suppose that vectors in R3 are denoted by 1*3 matrices, and define T:R4 to R3 by T9x,y,z,t)=(x-y+z+t,2x-2y+3z+4t,3x-3y+4z+5t).Find basis of kernel and range.
 
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First you'll need to calculate what the kernel and image is of T.

For the kernel, you'll need to solve the following system

[tex]\left\{ \begin{array}{c}<br /> x-y+z+t = 0\\<br /> 2x-2y+3z+4t = 0\\<br /> 3x-3y+4z+5t = 0<br /> \end{array}\right.[/tex]

For the image, you'll have to solve

[tex]\left\{ \begin{array}{c}<br /> x-y+z+t = \alpha\\<br /> 2x-2y+3z+4t = \beta\\<br /> 3x-3y+4z+5t = \gamma<br /> \end{array}\right.[/tex]

where alpha, beta, gamma are parameters. You'll then have to see for which alpha, beta, gamma this system has a solution.
 
Ow, for the image you won't need to do all that stuff, I'm sorry.

You'll first have to find a basis of R4, call this {e1,e2,e3,e4}. Then {T(e1),T(e2),T(e3),T(e4)} is a set which spans the image. If this set is linear independent, then it's a basis. If not, then remove some vectors until it is linear independent...
 
If you are open to it, the equation below (in case f : V -> W) might even help:

dim(ker(f)) + dim(Im(f)) = dim(V)
 
micromass said:
Ow, for the image you won't need to do all that stuff, I'm sorry.

You'll first have to find a basis of R4, call this {e1,e2,e3,e4}. Then {T(e1),T(e2),T(e3),T(e4)} is a set which spans the image. If this set is linear independent, then it's a basis. If not, then remove some vectors until it is linear independent...
Thank you!
 
Outlined said:
If you are open to it, the equation below (in case f : V -> W) might even help:

dim(ker(f)) + dim(Im(f)) = dim(V)
Thank you!
 

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