From my understanding, this question asks me is there any formula which can make T ( x

Brennen Fisher

Brennen Fisher

Answered question

2022-05-27

From my understanding, this question asks me is there any formula which can make T ( x ) = x 2 a non-linear transformation into a matrix transformation ? What am I supposed to answer to this question?

Answer & Explanation

sag2y8s

sag2y8s

Beginner2022-05-28Added 10 answers

I think that the question is rather poor. There is no linear transformation S such that S ( x ) = x 2 , so there is no linear transformation that agrees with T everywhere.
On the other hand, there is a "natural" way to obtain a linear transformation S from T ( x ) = x 2 , and that would be to take the derivative. The derivative S ( x ) = T ( x ) = 2 x is the linear map which scales vectors by
More generally, given a "non-linear" (but still differentiable) map T : R n R m , written in the standard bases as
T ( x 1 , , x n ) = ( T 1 ( x 1 , , x n ) , , T m ( x 1 , , x n ) ) ,
the derivative D T p at any point p = ( p 1 , , p n ) is a linear map R n R m . Moreover, in the standard bases, D T p can be identified with the m × n Jacobian matrix of partial derivatives:
D T p = ( T 1 x 1 ( p ) T 1 x 2 ( p ) T 1 x n ( p ) T 2 x 1 ( p ) T 2 x 2 ( p ) T 2 x n ( p ) T m x 1 ( p ) T m x 2 ( p ) T m x n ( p ) ) .
The upshot is that as long as we look at points q = ( q 1 , , q n ) that are "very small," then the affine transformation q T ( p ) + D T p ( q ) closely approximates the (in general nonlinear) map q T ( p + q )
So, for the example T : R R defined by T ( x ) = x 2 , we get a linear map at every point p defined as D T p which we typically just write as T ( p ) which is really just a 1 × 1 matrix T ( p ) = ( 2 p ) . As a linear map acting on 1 × 1 column vectors (q), we have
D T p ( q ) = T ( p ) ( q ) = ( 2 p ) ( q ) = ( 2 p q ) .
Hence the end result is that "vectors" (q) are mapped linearly to (2pq) by the derivative, and if we identify 1 × 1 matrices with numbers,
T ( p + q ) = ( p + q ) 2 = p 2 + 2 p q + q 2 p 2 + 2 p q (small numbers like  q  are "negligible" if we square them) = T ( p ) + 2 p q = T ( p ) + T ( p ) q ,
so long as q is sufficiently small. You may recognize that T ( p ) + T ( p ) q is really the first-order Taylor polynomial of T at p

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