Apparent fallacy in linear operator theory

[neelakash]

Butkov's book present the theory of linear operators this way:

Suppose a linear operator $$\alpha$$ transforms a basis vector
$$\hat{\ e_i}$$ into some vector $$\hat{\ a_i}$$.That is we have

$$\alpha\hat{\ e_i}=\hat{\ a_i}$$......(A)

Now the vectors $$\hat{\ a_i}$$ can be represented by its co-ordinates w.r.t. basis $$\{\hat{\ e_1},\hat{\ e_2}, ...,\hat{\ e_N}}$$.

$$\hat{\ a_i} = \sum\ a_j_i\hat{\ e_j}$$ 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 $$\ a_j_i$$,we can find the co-ordiantes of $$\ y=\alpha\ x$$ from the co-ordiantes of $$\ x$$

$$\ y=\alpha\ x=\alpha\sum [\ x_i\hat{ e_i}]=\sum [\ x_i\hat{ a_i}]$$...(C)

Employing the definition of $$\ a_i$$ as in (B), we obtain

$$\ y= \sum\ x_i\sum\ a_j_i\hat{\ e_j} = \sum[\sum\ a_j_i\ x_i]\hat{\ e_j}$$ in the last term the outer summation is on j.....(D)

From this we could identify that $$\ y=\sum\ y_j\hat{\ e_j}$$...(E)

where $$\ y_j=\sum[\ a_j_i\ x_i}$$...(F)

Last equation shows y and x are column vectors.If they were row vectors, the indices of $$\ a$$ would have interchanged among themselves.

But our very first assumption was $$\hat{ a_i}$$ is a row vector.And y is a linear combination of $$\hat{ a_i}$$.Thus, y should be a row vector!

Can anyone please help me to see where is the fallacy?

-Neel.

 

[jojo12345]

there is a subtle, but important difference between a vector and the matrix of components that represent the vector. Incidentally, the same thing can be said about a linear operator and the matrix of components representing the operator. The vector you express in (B) is a linear combination of the basis vectors you list in the line above (B). The matrix representing this vector is usually written as a Nx1 array, or column vector. Does this help at all?