What is about the circular convolution in OFDM In an OFDM system, serial-to-parallel conversion for data is done, then the DFT is performed and then adding the cyclic prefix (CP).

My question is related to that step of adding a CP. As I know, adding the CP is 
My question, if $H$ is the vector of $y$ in frequency domain, "$H = Fy$", so Can we diagonalize $H$ by left-right multiplying by proper DFT?  Which mean $A = F \, H \, F^H$ and $A$ is a diagonal matrix of $N \times N$ ??   Does that right in OFDM? And what's about after removing the CP when performing DFT operating in receiver side, will this propriety of circular convolution will be kept? 

thank you
Assuming you have $N$ symbols to transmit encoded in block $k$,
\begin{equation}
 s(k)
 =
 \begin{bmatrix}
  s_1(k) \\
  \vdots \\
  s_N(k)
 \end{bmatrix}
\end{equation}
Performing $N-$FFT at the transmitter, you now have
\begin{equation}
 x(k) = F^Hs(k)
\end{equation}
It is here that you insert a cyclic prefix of size $D &gt; L - 1$, where $L$ is the number of taps in your FIR channel, i.e. at the receiver you have
\begin{equation}
 r(k) 
 =H_{ISI}\begin{bmatrix}
  \overbrace{x_{N-D+1}(k)} \\
  \vdots \\
  \underbrace{x_N(k)}_{\text{CP}} \\
  x_1(k) \\
  \vdots \\
  x_N(k)
 \end{bmatrix}
 +
 H_{IBI}
\begin{bmatrix}
  x_{N-D+1}(k-1) \\
  \vdots \\
  x_N(k-1) \\
  x_1(k-1) \\
  \vdots \\
  x_N(k-1)
 \end{bmatrix} 
\end{equation}
where $\lbrace c_k \rbrace_{k=0}^{L-1}$ are the channel taps. Note that the channel matrix is decomposed into two parts, ISI (intersymbol interference) and IBI (interblock interference)
\begin{equation}
 H = H_{ISI} + H_{IBI}
\end{equation}
i.e.
\begin{align}
 H_{ISI}
 &amp;=
 \begin{bmatrix}
  c_0 &amp; 0 &amp; \ldots &amp; 0\\
  \vdots &amp; \ddots &amp; \ldots &amp; \vdots \\
  c_{L-1} &amp; \ddots &amp;  &amp; \vdots \\
  \vdots &amp; \ddots &amp;  &amp; \vdots \\
  0 &amp; \ldots &amp; c_{L-1} &amp; c_0
 \end{bmatrix} \\
 H_{IBI}
 &amp;=
 \begin{bmatrix}
  0 &amp; \ldots &amp; 0 &amp; c_{L-1} &amp; \ldots &amp; \ldots &amp; c_1 \\
  \vdots &amp; \ddots &amp; &amp;  &amp;  \ddots &amp;  &amp; \vdots \\
  \vdots &amp; \ddots &amp; &amp;  &amp;  \ddots &amp;  &amp; c_{L-1}\\
  0 &amp; \ldots &amp; &amp;  &amp;  \ldots &amp;  &amp; 0\\
  \vdots &amp; \ddots &amp; &amp;  &amp;  \ddots &amp;  &amp; \vdots \\
  0 &amp; \ldots &amp; &amp;  &amp;  \ldots &amp;  &amp; 0\\
 \end{bmatrix} 
\end{align}
If we remove the first $D$ samples of $r(k)$, we have 
\begin{equation}
\tilde{r}(k)
=
 \begin{bmatrix}
  r_{D+1}(k) \\
  \vdots \\
  r_{N+D}(k)
 \end{bmatrix}
 =
 \begin{bmatrix}
  0 &amp; \ldots &amp; 0 &amp; c_{L-1} \ldots &amp; c_0 &amp; 0 &amp; \ldots &amp; 0\\
  \vdots &amp;  &amp;  &amp;  \ddots &amp;  &amp; \ddots &amp;  &amp; \vdots \\
   0 &amp; \ldots &amp;  &amp;  \ldots &amp; 0 &amp; c_{L-1} &amp; \ldots &amp; c_0\\
 \end{bmatrix}
\begin{bmatrix}
  x_{N-D+1}(k) \\
  \vdots \\
  x_N(k) \\
  x_1(k) \\
  \vdots \\
  x_N(k)
 \end{bmatrix}
\end{equation}
After a little bit of manipulation, you can see that
\begin{equation}
 \tilde{r}(k)
 =
 C F s(k)
\end{equation}
where $C$ is a circular matrix that contains 
\begin{equation}
 \begin{bmatrix}
  c_0 &amp; 0 &amp; \ldots &amp; c_{L-1} &amp; \ldots c_1
 \end{bmatrix}
\end{equation}
in its first row. Now, applying a DFT, we get
\begin{equation}
 y(k)
 =
 F
 \tilde{r}(k)
 =
 F
 C F^H s(k)
\end{equation}
Since $C$ is circular, then $F
 C F^H$ is diagonalizable because any circular matrix is diagonal in a Fourier basis, hence you get
    \begin{equation}
  y(k) = D s(k)
 \end{equation}
    where $D$ is diagonal containing the eigenvalues of $C$.