$D=C\cdot V$ ; C and V are both matrices and C is a square by square matrix $C_{ij}=1$ if i=j and $C_{ij}=0$ for $i\neq j$ (Kronecker delta). $\mathcal{F}^{-1} D = \mathcal{F}^{-1} C$ $ * \mathcal{F}^{-1}V $; Pleaase could some one help? what is the $\mathcal{F}^{-1}C$? a constant or a delta function ? where $\mathcal{F}^{-1}$ is the inverse Fourier transform.
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1The Kronecker delta matrix is the unit or identity matrix, so $CA=A$ for all matrices $A$. Please give more details if there are. What is the center of a matrix? The diagonal? – Lutz Lehmann Feb 21 '14 at 07:48
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My apologies, C is a square by square matrix, $C_{ij}=1$ if i=j and $C_{ij}=0$ if $i\neq j$ – armando Feb 21 '14 at 08:01
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Yes, this is the identity matrix, the multiplicative unit of matrix algebra. What kind of product are you using in $D=C\cdot V$? The normal matrix product results in $D=V$, the Kronecker (tensor) product will give a different result. Or is there something with pointwise products and convolution, since you are using $*$ in the Fourier picture? – Lutz Lehmann Feb 21 '14 at 08:50
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Thanks, yeah is the normal matrices product, $\mathcal{F}^{-1} D=\mathcal{F}^{-1}C * \mathcal{F}^{-1} V$ from the convolution theorem. For the simple case, let us say c_{ij}=1 for $i=j=\frac{n}{2}$ and zero else where. n$\times n$ the size of C – armando Feb 21 '14 at 12:10
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Thanks, yeah is the normal matrices product, $\mathcal{F}^{-1} D=\mathcal{F}^{-1}C * \mathcal{F}^{-1} V$ from the convolution theorem. For the simple case, let us say c_{ij}=1 for $i=j=\frac{n}{2}$ and zero else where. n$\times n$ the size of C with n even – armando Feb 21 '14 at 12:17
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Since $\mathbf{C} = \mathbf{I}$, where $\mathbf{I}$ is the identity matrix, then $\mathcal{F}^{-1} \mathbf{C}$ will be the inverse DFT matrix. You don't say whether the the FT is 2-D, along the rows, or along the columns, so nothing else can really be said about it.
Also, I do not think that the convolution property is generally applicable to matrix multiplication.
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