I am having trouble simplifying an equation that contains both pointwise (Hadamard) and standard matricial multiplication.
Given that $I_{n\times1}$ and $K_{n\times1}$ are real $n\times1$ matrices; $T_{n\times n}$ is a complex $n\times n$ DFT Matrix; and $C_{n\times1}$ is a real $n\times1$ matrix representing the convolution of both arrays we have:
$$C = \left ( IT \cdot KT \right )T$$
where $\cdot$ denotes pointwise product and $AB$ denotes standard matricial product.
How can I best simplify this equation?
It sounds reasonable to me that $C = \left ( I \cdot K \right )T^2$ but I do not know how to prove it.
Edit:
In fact, $C = \left ( I \cdot K \right )T^2$ cannot be true because the DFT Matrix is involutory, meaning $T^2=I$ (the Identity matrix) and that would mean that convolution in time domain is equal to the pointwise product in the time domain, which is false.
$ \def\o{{\tt1}} \def\LR#1{\left(#1\right)} \def\op#1{\operatorname{#1}} \def\diag#1{\op{diag}\LR{#1}} \def\Diag#1{\op{Diag}\LR{#1}} \def\qiq{\quad\implies\quad} $Let's use a convention where uppercase letters denote matrices and lowercase letters denote column vectors. Then rename the problem variables $$\eqalign{ \{I,K,C\} \qiq \{a,b,c\} \\ }$$ Define the following diagonal matrices $$\eqalign{ A &= \Diag{Ta},\quad B = \Diag{Tb},\quad C = \Diag{Tc} \\ }$$ Multiply the given equation by $T$ and use its involutory property $$\eqalign{ Tc &= T^2\LR{Ta\odot Tb} = \LR{Ta\odot Tb} \\ }$$ For diagonal matrices, an elementwise product is the same as a matrix product. So we can replace the vector equation by an equivalent matrix equation $$\eqalign{ C &= AB = BA \\ }$$ The latter equality holds since diagonal matrices commute.
The vectors can be easily recovered $$\eqalign{ a &= T\,\diag{A},\quad b = T\,\diag{B},\quad c = T\,\diag{C} \\ }$$