Butkov's book present the theory of linear operators this way:(adsbygoogle = window.adsbygoogle || []).push({});

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.

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# Apparent fallacy in linear operator theory

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