Linear algebra - Transformation using a Matrix

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SUMMARY

The discussion focuses on solving a linear algebra problem involving a linear transformation T: R^3 → R^3, represented by the matrix [T]_B in the basis B = {(1,1,1), (1,1,0), (1,0,1)}. The user successfully determines the values of a, b, and c in the matrix as a = 1, b = 1, and c = 1. Additionally, the user confirms the representation of the vector (1,0,0) in the basis B, validating the calculations through matrix multiplication.

PREREQUISITES
  • Understanding of linear transformations in R^3
  • Familiarity with matrix representation of linear maps
  • Knowledge of basis and kernel concepts in linear algebra
  • Ability to perform matrix multiplication and solve linear equations
NEXT STEPS
  • Study the properties of linear transformations in R^n
  • Learn about the Rank-Nullity Theorem in linear algebra
  • Explore the concept of change of basis and its applications
  • Investigate the implications of kernel and image in linear mappings
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Students and educators in mathematics, particularly those studying linear algebra, as well as anyone looking to deepen their understanding of linear transformations and matrix representations.

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Homework Statement



let [itex]T:R^3 \rightarrow R^3[/itex]
be a Linear map,
and let [itex]B=\left \{ (1,1,1),(1,1,0),(1,0,1) \right \}[/itex]
be a Basis

and [itex](1,0,0)\in kerT[/itex][itex][T]_B=\begin{pmatrix}<br /> a & 0 & b\\ <br /> 3 & 2a & 1\\ <br /> 2c& b & a<br /> \end{pmatrix}[/itex]

a. find a,b,c
b. find a Basis for ImT

Homework Equations


The Attempt at a Solution



as for a.

I think that what we need to do is find a general vector (x,y,z)
and express (1,0,0) thorugh it
if we multiply the matrix by what we got we will have to get (0,0,0) (because (1,0,0) is in the kernel)
but I'm not sure how to express (1,0,0).
I think the vector (1,0,0) in the basis B is -1(1,1,1)+1(1,1,0)+1(1,0,1) and so is
(-x+y+z,0,0)

and for the matrix we have
-a+0+b=0
-3+2a+1=0
-2c+b+a=0

from the second equation:
a= 1
from the first
b=1
from the third
c=1

Am I right?
 
Last edited:
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yeah look good calling the natural basis E, (1,0,0) in the B basis is as follows
[tex](1,0,0)^T_E = (-1,1,1)^T_B[/tex]

And as a check in E
[tex]-1\textbf{b}_1+1\textbf{b}_2+1\textbf{b}_3 = -1(1,1,1)^T+1(1,1,0)^T+1(1,0,1)^T=(-1+1+1,-1+1+0,-1+0+1)^T =(1,0,0)^T[/tex]
 

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