Kronecker delta

Question

$D=C\cdot V$ ; 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.

Answer 1

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.