MHB Definition of matrix transformation

[Carla1985]

Hi all,

I have the definition of a linear transformation in terms of a transformation matrix. So the mapping is a function $f:\mathbb{R}^m\rightarrow\mathbb{R}^n$, where $f(\textbf{x})=A\textbf{x}$ and $A$ is a $n\times m$ matrix.

I'm looking for a similar definition for a transformation that takes a matrix to another matrix. I.e a $2\times 2$ matrix to a $4\times 4$ one. I think the mapping will be of the form $sAs^T$, where $s$ is a matrix but I'm sure there's more to it than that.

Also when doing a linear transformation I can say that if I want to take the first element of $\textbf{x}$ to the third element of $f(\textbf{x})$ then there will be a $1$ in the first column and third row of $A$. Ideally I would like a similar explanation for the matrix transformation.

Can someone point me in the right direction please.

Thanks
Carla

 

[Evgeny.Makarov]

You can represent 2x2 matrices as vectors of length 4 and 4x4 matrices as vectors of length 16. Then a linear transformation from 2x2 matrices to 4x4 matrices is represented by a 16x4 matrix.

I think the mapping will be of the form $sAs^T$, where $s$ is a matrix but I'm sure there's more to it than that.

If $s$ is the argument, then $sAs^T$ is not a linear transformation.

 

[Opalg]

Hi all,

I have the definition of a linear transformation in terms of a transformation matrix. So the mapping is a function $f:\mathbb{R}^m\rightarrow\mathbb{R}^n$, where $f(\textbf{x})=A\textbf{x}$ and $A$ is a $n\times m$ matrix.

I'm looking for a similar definition for a transformation that takes a matrix to another matrix. I.e a $2\times 2$ matrix to a $4\times 4$ one. I think the mapping will be of the form $sAs^T$, where $s$ is a matrix but I'm sure there's more to it than that.

If $s$ is a fixed $4\times2$ matrix (with linearly independent columns) then the map $A\mapsto sAs^{\textsf{T}}$ takes the $2\times2$ matrix $A$ to a $4\times4$ matrix. If $A$ represents a linear transformation of $\mathbb{R}^2$ then $sAs^{\textsf{T}}$ represents the linear transformation of $\mathbb{R}^4$ which acts like $A$ on the 2-dimensional subspace spanned by the columns of $s$, and is zero on the orthogonal subspace.

 

[Carla1985]

You can represent 2x2 matrices as vectors of length 4 and 4x4 matrices as vectors of length 16. Then a linear transformation from 2x2 matrices to 4x4 matrices is represented by a 16x4 matrix.

If $s$ is the argument, then $sAs^T$ is not a linear transformation.

I hadn't thought of this route, I will definitely explore it to so if it makes things easier. Thank you

If $s$ is a fixed $4\times2$ matrix (with linearly independent columns) then the map $A\mapsto sAs^{\textsf{T}}$ takes the $2\times2$ matrix $A$ to a $4\times4$ matrix. If $A$ represents a linear transformation of $\mathbb{R}^2$ then $sAs^{\textsf{T}}$ represents the linear transformation of $\mathbb{R}^4$ which acts like $A$ on the 2-dimensional subspace spanned by the columns of $s$, and is zero on the orthogonal subspace.

This is precisely the way I have been attacking my problem thus far. The columns will be linearly independent as they consist of just zeros and ones. Do you know if there is a way of being able to tell where an element of the $2\times 2$ will end up in the $4\times 4$. I have been doing trial and error to figure it out but can't yet see the pattern.