Unitary Operation On A Complex Matrix

  • Thread starter Bashyboy
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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

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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##
 
Last edited:

Fredrik

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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})##.
 
Last edited:

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