- #1
neelakash
- 511
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Butkov's book present the theory of linear operators this way:
Suppose a linear operator [tex]\alpha[/tex] transforms a basis vector
[tex]\hat{\ e_i}[/tex] into some vector [tex]\hat{\ a_i}[/tex].That is we have
[tex]\alpha\hat{\ e_i}=\hat{\ a_i}[/tex]......(A)
Now the vectors [tex]\hat{\ a_i}[/tex] can be represented by its co-ordinates w.r.t. basis [tex]\{\hat{\ e_1},\hat{\ e_2}, ...,\hat{\ e_N}}[/tex].
[tex]\hat{\ a_i} = \sum\ a_j_i\hat{\ e_j}[/tex] where i,j=1,2,3...N and summation over j is implied....(B)
Notice that the in last equation,we have put a row vector(a)=a row vector (e) times a matrix A
Now with the help of the transforming matrix [tex]\ a_j_i[/tex],we can find the co-ordiantes of [tex]\ y=\alpha\ x[/tex] from the co-ordiantes of [tex]\ x[/tex]
[tex]\ y=\alpha\ x=\alpha\sum [\ x_i\hat{ e_i}]=\sum [\ x_i\hat{ a_i}][/tex]...(C)
Employing the definition of [tex]\ a_i[/tex] as in (B), we obtain
[tex]\ y= \sum\ x_i\sum\ a_j_i\hat{\ e_j} = \sum[\sum\ a_j_i\ x_i]\hat{\ e_j}[/tex] in the last term the outer summation is on j.....(D)
From this we could identify that [tex]\ y=\sum\ y_j\hat{\ e_j}[/tex]...(E)
where [tex]\ y_j=\sum[\ a_j_i\ x_i}[/tex]...(F)
Last equation shows y and x are column vectors.If they were row vectors, the indices of [tex]\ a[/tex] would have interchanged among themselves.
But our very first assumption was [tex]\hat{ a_i}[/tex] is a row vector.And y is a linear combination of [tex]\hat{ a_i}[/tex].Thus, y should be a row vector!
Can anyone please help me to see where is the fallacy?
-Neel.
Suppose a linear operator [tex]\alpha[/tex] transforms a basis vector
[tex]\hat{\ e_i}[/tex] into some vector [tex]\hat{\ a_i}[/tex].That is we have
[tex]\alpha\hat{\ e_i}=\hat{\ a_i}[/tex]......(A)
Now the vectors [tex]\hat{\ a_i}[/tex] can be represented by its co-ordinates w.r.t. basis [tex]\{\hat{\ e_1},\hat{\ e_2}, ...,\hat{\ e_N}}[/tex].
[tex]\hat{\ a_i} = \sum\ a_j_i\hat{\ e_j}[/tex] where i,j=1,2,3...N and summation over j is implied....(B)
Notice that the in last equation,we have put a row vector(a)=a row vector (e) times a matrix A
Now with the help of the transforming matrix [tex]\ a_j_i[/tex],we can find the co-ordiantes of [tex]\ y=\alpha\ x[/tex] from the co-ordiantes of [tex]\ x[/tex]
[tex]\ y=\alpha\ x=\alpha\sum [\ x_i\hat{ e_i}]=\sum [\ x_i\hat{ a_i}][/tex]...(C)
Employing the definition of [tex]\ a_i[/tex] as in (B), we obtain
[tex]\ y= \sum\ x_i\sum\ a_j_i\hat{\ e_j} = \sum[\sum\ a_j_i\ x_i]\hat{\ e_j}[/tex] in the last term the outer summation is on j.....(D)
From this we could identify that [tex]\ y=\sum\ y_j\hat{\ e_j}[/tex]...(E)
where [tex]\ y_j=\sum[\ a_j_i\ x_i}[/tex]...(F)
Last equation shows y and x are column vectors.If they were row vectors, the indices of [tex]\ a[/tex] would have interchanged among themselves.
But our very first assumption was [tex]\hat{ a_i}[/tex] is a row vector.And y is a linear combination of [tex]\hat{ a_i}[/tex].Thus, y should be a row vector!
Can anyone please help me to see where is the fallacy?
-Neel.