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cnada
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I will give you a few elements for the demonstration on real vectors which you can extend to complex.

Let {$e_i$} be athe standard basis for the space where $U (n*n)$ is defined . Let {$e_j$} be athe standard basis for the space where $V (m*m)$ is defined.

First, it is a property that the basis {$e_i \otimes e_j$} is a basis for the n*m-matrices space.

$ U \otimes V $ is a linear mapping on the space and we have that : $$ U \otimes V (e_i \otimes e_j) = (U e_i) \otimes(V e_j) (1) $$

A remark to be given is that in linear algebra, when $W$ is linear and $W e_i$ is known, W is uniquely determined. As {$e_i \otimes e_j$} is a basis for the linear mapping $ U \otimes V $, it will be unique for the definition (1).

In particular,
$$ U \otimes V (x \otimes y) = (U x) \otimes(V y) $$

Indeed : $$ U \otimes V (x \otimes y) = U \otimes V (\sum_i x_i e_i \otimes \sum_j y_j e_j) $$ $$ = U \otimes V (\sum_{i,j} x_i y_j (e_i \otimes e_j)) $$ $$ = \sum_{i,j} x_i y_j U \otimes V (e_i \otimes e_j) $$ $$ = \sum_{i,j} x_i y_j (U e_i) \otimes(V e_j) $$ $$ = \sum_{i,j} x_i y_j (U e_i) (V e_j)^T $$ $$ = \sum_{i} x_i (U e_i) \sum_{j} y_j (V e_j)^T $$ $$ = U (\sum_{i} x_i e_i) (V (\sum_{j} y_je_j))^T $$ $$ = U x (V y)^T $$ $$ = U x \otimes V y $$

You can look at that PDF if it makes it more clear.

I will give you a few elements for the demonstration on real vectors which you can extend to complex.

Let {$e_i$} be a basis for the space where $U (n*n)$ is defined . Let {$e_j$} be a basis for the space where $V (m*m)$ is defined.

First, it is a property that the basis {$e_i \otimes e_j$} is a basis for the n*m-matrices space.

$ U \otimes V $ is a linear mapping on the space and we have that : $$ U \otimes V (e_i \otimes e_j) = (U e_i) \otimes(V e_j) (1) $$

A remark to be given is that in linear algebra, when $W$ is linear and $W e_i$ is known, W is uniquely determined. As {$e_i \otimes e_j$} is a basis for the linear mapping $ U \otimes V $, it will be unique for the definition (1).

In particular,
$$ U \otimes V (x \otimes y) = (U x) \otimes(V y) $$

Indeed : $$ U \otimes V (x \otimes y) = U \otimes V (\sum_i x_i e_i \otimes \sum_j y_j e_j) $$ $$ = U \otimes V (\sum_{i,j} x_i y_j (e_i \otimes e_j)) $$ $$ = \sum_{i,j} x_i y_j U \otimes V (e_i \otimes e_j) $$ $$ = \sum_{i,j} x_i y_j (U e_i) \otimes(V e_j) $$ $$ = \sum_{i,j} x_i y_j (U e_i) (V e_j)^T $$ $$ = \sum_{i} x_i (U e_i) \sum_{j} y_j (V e_j)^T $$ $$ = U (\sum_{i} x_i e_i) (V (\sum_{j} y_je_j))^T $$ $$ = U x (V y)^T $$ $$ = U x \otimes V y $$

You can look at that PDF if it makes it more clear.

I will give you a few elements for the demonstration on real vectors which you can extend to complex.

Let {$e_i$} be the standard basis for the space where $U (n*n)$ is defined . Let {$e_j$} be the standard basis for the space where $V (m*m)$ is defined.

First, it is a property that the basis {$e_i \otimes e_j$} is a basis for the n*m-matrices space.

$ U \otimes V $ is a linear mapping on the space and we have that : $$ U \otimes V (e_i \otimes e_j) = (U e_i) \otimes(V e_j) (1) $$

A remark to be given is that in linear algebra, when $W$ is linear and $W e_i$ is known, W is uniquely determined. As {$e_i \otimes e_j$} is a basis for the linear mapping $ U \otimes V $, it will be unique for the definition (1).

In particular,
$$ U \otimes V (x \otimes y) = (U x) \otimes(V y) $$

Indeed : $$ U \otimes V (x \otimes y) = U \otimes V (\sum_i x_i e_i \otimes \sum_j y_j e_j) $$ $$ = U \otimes V (\sum_{i,j} x_i y_j (e_i \otimes e_j)) $$ $$ = \sum_{i,j} x_i y_j U \otimes V (e_i \otimes e_j) $$ $$ = \sum_{i,j} x_i y_j (U e_i) \otimes(V e_j) $$ $$ = \sum_{i,j} x_i y_j (U e_i) (V e_j)^T $$ $$ = \sum_{i} x_i (U e_i) \sum_{j} y_j (V e_j)^T $$ $$ = U (\sum_{i} x_i e_i) (V (\sum_{j} y_je_j))^T $$ $$ = U x (V y)^T $$ $$ = U x \otimes V y $$

You can look at that PDF if it makes it more clear.

Source Link
cnada
  • 4.8k
  • 1
  • 9
  • 22

I will give you a few elements for the demonstration on real vectors which you can extend to complex.

Let {$e_i$} be a basis for the space where $U (n*n)$ is defined . Let {$e_j$} be a basis for the space where $V (m*m)$ is defined.

First, it is a property that the basis {$e_i \otimes e_j$} is a basis for the n*m-matrices space.

$ U \otimes V $ is a linear mapping on the space and we have that : $$ U \otimes V (e_i \otimes e_j) = (U e_i) \otimes(V e_j) (1) $$

A remark to be given is that in linear algebra, when $W$ is linear and $W e_i$ is known, W is uniquely determined. As {$e_i \otimes e_j$} is a basis for the linear mapping $ U \otimes V $, it will be unique for the definition (1).

In particular,
$$ U \otimes V (x \otimes y) = (U x) \otimes(V y) $$

Indeed : $$ U \otimes V (x \otimes y) = U \otimes V (\sum_i x_i e_i \otimes \sum_j y_j e_j) $$ $$ = U \otimes V (\sum_{i,j} x_i y_j (e_i \otimes e_j)) $$ $$ = \sum_{i,j} x_i y_j U \otimes V (e_i \otimes e_j) $$ $$ = \sum_{i,j} x_i y_j (U e_i) \otimes(V e_j) $$ $$ = \sum_{i,j} x_i y_j (U e_i) (V e_j)^T $$ $$ = \sum_{i} x_i (U e_i) \sum_{j} y_j (V e_j)^T $$ $$ = U (\sum_{i} x_i e_i) (V (\sum_{j} y_je_j))^T $$ $$ = U x (V y)^T $$ $$ = U x \otimes V y $$

You can look at that PDF if it makes it more clear.