What is about the circular convolution in OFDM
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 $C$.