Linear algebra - Transformation using a Matrix

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
1 reply · 2K views
BitterX
Messages
35
Reaction score
0

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:
Physics news on Phys.org
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]