Undergrad Expressing the Matrix Transpose Function: Is There a Different Approach?

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The discussion centers on expressing the matrix transpose function and its relationship with power series. It highlights that while functions of a matrix can often be expressed using power series, the transpose function does not have a convergent power series expansion due to the non-commutative nature of matrices with their transposes. Transposition is characterized as a linear map, which leads to a different representation involving individual matrix elements rather than a power series. The conversation also touches on the complexity of defining matrix functions and the need to consider the underlying space of the matrix. Ultimately, the transpose function is framed as a linear transformation between two isomorphic vector spaces.
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How is the transpose function of a matrix expressed?
One way to express a function of a matrix A is by a power series (a Taylor expansion). It is not too difficult to show that two functions f(A) and g(A) with such a power series representation must commute, i.e. f(A)g(A) = g(A)f(A). But matrices typically do not commute with their own transpose, so presumably the transpose function does not have convergent a power series expansion? I had not previously appreciated that even simple matrix functions may not have a power series representation. Is there another way to express the matrix transpose function, or matrix functions in general?
 
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madness said:
Summary:: How is the transpose function of a matrix expressed?

One way to express a function of a matrix A is by a power series (a Taylor expansion). It is not too difficult to show that two functions f(A) and g(A) with such a power series representation must commute, i.e. f(A)g(A) = g(A)f(A). But matrices typically do not commute with their own transpose, so presumably the transpose function does not have convergent a power series expansion? I had not previously appreciated that even simple matrix functions may not have a power series representation. Is there another way to express the matrix transpose function, ...
Yes. Transposition is a linear map, so your power series should come to an end early: ##(f(a_{ij}))_{kl} = (f_{kl}(a_{ij}))=(a_{lk})##.
... or matrix functions in general?
No. Functions in general means almost complete arbitrariness. So how should a structure on everything work? The only meaningful way is by coordinates: ##f(a_{ij})=f_{kl}(a_{11},\ldots , a_{nm})##.
 
You have a matrix, but you talk about analysis. And a matrix from the analytical point of view is simply an ##n\cdot m## tuple of numbers or variables. You cannot expect a matrix to behave like a real or complex number. You have a linear function ##A\, : \,\mathbb{R}^n \longrightarrow \mathbb{R}^m##. If you want to consider the matrix itself as variable, then you have to determine the space the matrix is from, e.g. an algebraic group, and consider paths within this space, e.g. ##t \longmapsto t\cdot A##.

What is variable and what is constant?

Transposition is ##\tau\, : \,\mathbb{M}(n,m) \longrightarrow \mathbb{M}(m,n)##, i.e. a linear function between two isomorphic but not identical spaces of vectors of length ##n\times m##. In this case we have constants which represent ##\tau## and variables which represent the ##n\times m## input and ##m\times n## output variables. As transposition is linear, there is a matrix representation ##\tau \in \mathbb{M}(nm,nm)## with ##(nm)^2## many entries.
 
Thread 'How to define a vector field?'
Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...

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