Unitary Operation On A Complex Matrix

[Bashyboy]

Hello everyone,

Let $A = (\alpha_{ij})$ be an $n \times n# complex matrix. Define $\hat$ acting on $A$ as producing the matrix $\hat{A} = (\alpha_{ij} I_n)$.

I don't understand what this is saying. Isn't $I_n$ the identity matrix, and therefore the product of it with any matrix results in the other matrix? If this is so, wouldn't $\hat{A} = (\alpha_{ij} I_n) = (\alpha_{ij}) = A$?

 

[FactChecker]

Without seeing more of the context, I would think that $\hat{A}$ is an n² by n² matrix where every element, $\alpha_{ij}$, of A is replaced by the n by n submatrix $\alpha_{ij}I_n$

 

[Fredrik]

Hello everyone,

Let $A = (\alpha_{ij})$ be an $n \times n$ complex matrix. Define $\hat{}$ acting on $A$ as producing the matrix $\hat{A} = (\alpha_{ij} I_n)$.

I don't understand what this is saying. Isn't $I_n$ the identity matrix, and therefore the product of it with any matrix results in the other matrix? If this is so, wouldn't $\hat{A} = (\alpha_{ij} I_n) = (\alpha_{ij}) = A$?

$\alpha_{ij}$ isn't a matrix.

The notation in the sentence you're asking about is very strange. If $(\alpha_{ij})$ denotes the matrix with $\alpha_{ij}$ on row i, column j, then $(\alpha_{ij}I_n)$ should denote the matrix with $\alpha_{ij}I_n$ on row i, column j. But this is a matrix whose every component is a matrix. Is it possible that this is what they meant? If yes, then I agree with FactChecker. If no, then maybe they meant $\hat A=(\alpha_{ij}(I_n)_{ij}) =(\alpha_{ij}\delta_{ij})$.