Covariance matrix of data projected onto eigenvectors is diagonal. I am reading about PCA and found an exercise that says

Show that when a $N$-dim set of data points $X$ is projected onto the eigenvectors $V = [e_1 e_2...e_n]$ of its covariance matrix $C=XX^T$, the covariance matrix of the projected data $C_p = YY^T$ is diagonal and hence that, in the space of the eigenvector decomposition, the distribution of X is uncorrelated.

What I have so far is
$$Y = V^TX$$
Therefore
$$C_p = YY^T = V^TX(V^TX)^T = V^TXX^TV$$
but there I got stuck. Any advise on how to proceed, moreover, what does "The covariance matrix of the projected data is diagonal" mean?
 A: A diagonal matrix is one with zero everywhere and the diagonal entries can be zero or non zero.
Assuming the eigenvectors are normalized in magnitude $\|e_i\|=1$
$$
\begin{align}
\because \text{  } & V = [e_1 e_2 \dots e_n] \text{  and  } Ce_i =\lambda_ie_i \\\\
\therefore \text{  } &CV = V 
\begin{bmatrix} 
\lambda_1 & 0 & \cdots & 0\\
0 & \lambda_2 & \cdots & 0\\
\vdots & \vdots & \ddots & \vdots\\
0 & 0 & \cdots & \lambda_n
\end{bmatrix}\\\\
\because \text{  } & C_p = V^T (XX^T) V = V^TCV\\\\
\therefore \text{  } & C_p=V^TV\begin{bmatrix} 
\lambda_1 & 0 & \cdots & 0\\
0 & \lambda_2 & \cdots & 0\\
\vdots & \vdots & \ddots & \vdots\\
0 & 0 & \cdots & \lambda_n
\end{bmatrix}\\
\because \text{  } & V^TV = \mathbf{I}\\
&\text{ as eigenvecotrs of a symmetric matrix are orthogonal}\\\\
\therefore \text{  } &C_p= \begin{bmatrix} 
\lambda_1 & 0 & \cdots & 0\\
0 & \lambda_2 & \cdots & 0\\
\vdots & \vdots & \ddots & \vdots\\
0 & 0 & \cdots & \lambda_n
\end{bmatrix}\\
&\text{A diagonal matrix with variance in each eigenvector direction}
\end{align}
$$
