Dimension of Matrix in MIMO with multi-path
The short answer is: if you transmit one block of $N$ symbols (no data before it for at least $L$ symbol times) over a frequency-selective channel of length $L$, then the channel matrix will be of dimension $NN_TN_R\times N$. 

The long answer: start with $N_T=N_R=1$. Then the $n^{\text{th}}$ received sample can be written as

$$y_n=\sum_{l=0}^Lh_lx_{n-l}+z_n$$

If you arrange these samples in a vector-matrix form you get

$$\mathbf{y}_{N\times 1}=\mathbf{H}_{N\times N}\mathbf{x}_{N\times 1} + \mathbf{z}_{N\times 1}$$

Now let $N_T=1$ while the number of receive antennas $N_R$, then the $n^{\text{th}}$ received sample over $m^{\text{th}}$ receive antenna is

$$y_n^{(m)}=\sum_{l=0}^Lh_l^{(m)}x_{n-l}+z^{(m)}_n$$

for $m=1,\,2,\,\ldots,\,N_R$. If you arrange these sample in a vector-Matrix form, you get

$$\mathbf{y}_{NN_R\times 1}=\mathbf{H}_{NN_R\times N}\mathbf{x}_{N\times 1} + \mathbf{z}_{NN_R\times 1}$$

Similarly, you can do the general case for $N_T$ transmit and $N_R$ receive antennas.

As you can see, to make this work on block-by-block basis, you need to padding $L$ zeros before each block to eliminate the inter-block interference. Alternatively, you can use cyclic-prefixed (CP)-OFDM, which is more efficient at the receiver side.