I can not explain a difference between to calculations regarding linear functionals

  • #1
Hello,
Problem, let [itex]B=[/itex]{[itex]a_1,a_2,a_3[/itex]} be a basis for [itex]C^3[/itex] defined by [itex]a_1=(1,0,-1)[/itex] [itex]a_2=(1,1,1)[/itex] [itex]a_3=(2,2,0)[/itex]
Find the dual basis of [itex]B[/itex].

My Solution. Let [itex]W_1[/itex] be the subspace generated by [itex]a_2=(1,1,1)[/itex] [itex]a_3=(2,2,0)[/itex], lets find [itex][/itex] [itex]W*[/itex], where [itex]W*[/itex] is the set of linear anihilator of [itex]W_1[/itex]. Consider the system:

\begin{align*}
\begin{bmatrix}1 & 1 & 1 \\ 2 & 2 & 0 \\\end{bmatrix} &\begin{bmatrix} x_1 \\x_2\\x_3 \\\end{bmatrix}
=0
\end{align*}
solving this will yeild [itex]x_2=-x_1[/itex] and [itex]x_3=0[/itex]
so from this, by assinging [itex]x_1=1[/itex] ,we find out that the map [itex]f_1=x_1-x_2[/itex].
Similarly,




Let [itex]W_2[/itex] be the subspace generated by [itex]a_2=(1,0,-1)[/itex] [itex]a_3=(2,2,0)[/itex], lets find [itex][/itex] [itex]W*[/itex], where [itex]W*[/itex] is the set of linear anihilator of [itex]W_2[/itex]. Consider the system:

\begin{align*}
\begin{bmatrix}1 & 0 & -1 \\ 2 & 2 & 0 \\\end{bmatrix} &\begin{bmatrix} x_1 \\x_2\\x_3 \\\end{bmatrix}
=0
\end{align*}
solving this will yeild [itex]x_2=-x_1[/itex] and [itex]x_3=x_1[/itex]
so from this, by assinging [itex]x_1=1[/itex] ,we find out that the map [itex]f_2=x_1-x_2+x_3[/itex].

Finally,

Let [itex]W_3[/itex] be the subspace generated by [itex]a_2=(1,1,1)[/itex] [itex]a_3=(1,0,-1)[/itex], lets find [itex][/itex] [itex]W*[/itex], where [itex]W*[/itex] is the set of linear anihilator of [itex]W_3[/itex]. Consider the system:

\begin{align*}
\begin{bmatrix}1 & 1 & 1 \\ 1 & 0 & -1 \\\end{bmatrix} &\begin{bmatrix} x_1 \\x_2\\x_3 \\\end{bmatrix}
=0
\end{align*}
solving this will yeild [itex]x_2=-2x_1[/itex] and [itex]x_3=x_1[/itex]
so from this, by assinging [itex]x_1=1[/itex] ,we find out that the map [itex]f_3=x_1-2x_2+x_3[/itex].
So far so Good. The issue is that the last map [itex]f_3[/itex]does not agree with the solution I have, but every other thing is the same.Note that I have the final solution with a diffrenet method of solution than mine. In the solution, I got that [itex]f_3=-1/2x_1+x_2-1/2x_3[/itex], I know that I assigned [itex]x_1=1[/itex] when I was trying to find [itex]f_3[/itex], and I could get the same answer if I put [itex]x_1=-1/2[/itex].
This significantly affect the nature of the map. i.e.
let [itex](5,4,2)\in C^3[/itex], then
[itex](5,4,2)=1(1,0,-1)+3(1,1,1)+1/2(2,2,0)[/itex]
so we have [itex]c_1=1, c_2=3, c_3=1/2[/itex]. while
[itex]f_1(5,4,2)=1[/itex],
[itex]f_2(5,4,2)=3[/itex],
[itex]f_3(5,4,2)=-1[/itex]
so note that [itex]c_3 \ \ does \ \ not \ \ equal \ \ f_3(5,4,2)[/itex]
While if I used the map [itex]f_3=-1/2x_1+x_2-1/2x_3[/itex] we get that [itex]f_3(5,4,2)=1/2[/itex]. The equality should ocuur since [itex]f_3[/itex]is a vector in the dual which determine the scalar [itex]c_3[/itex].
So my explanation is that I should consider also another equation to each system saying that [itex]c_1x_1+ c_2x_2 + c_2x_3=1[/itex] for each system depending on our choice of the vectors to determine [itex]c_1,c_2,c_3[/itex].
My question, Is this method always work when we find the Dual basis to a given Basis?.
Because I figured this method by myself using a similar method to find anihilator space along with the fact that [itex]dim \ \ W + dim \ \ of \ \ annihilator \ \ space= dim \ \ V[/itex]
, where [itex]W\subset V[/itex]
 

Answers and Replies

  • #2
fresh_42
Mentor
Insights Author
2021 Award
15,940
14,371
You could just have taken the suitably normed perpendicular vectors of your basis. Should have been easier to find, and in the end to cross check by a matrix multiplication.
 

Related Threads on I can not explain a difference between to calculations regarding linear functionals

Replies
7
Views
9K
Replies
2
Views
16K
  • Last Post
Replies
13
Views
3K
Replies
8
Views
5K
Replies
1
Views
3K
Replies
4
Views
1K
Replies
1
Views
2K
Top