Question - RNN math

How do we arrive at this answer ?

Hi @Prateek_Sinha

The dimensionality of h^{<t>} never changes - it’s always 4x1 at this case. So that means that h^{t-1} is also 4x1.
The dimensionality of x^{<t>} is said to be 10x1.
So h^{t-1} and x^{<t>} concatenated will be 14x1.

In order for the math (dot product) to work out, the “inner” dimension (in this case 14) has to match:

(4x14) @ (14x1) → (4x1)