(need explanation) Compute singular values of $ X^{(t)} $ via reduced gramians

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background information: I just started with the topic of singular values. Please be understanding.

groundwork:

So lets say we a matrix $ X^{(t)} $. $U_t$ is a basis for the columb spaces of $ X^{(t)} $. Hence theres a matrix $V_t$ such that $ X^{(t)} = U_tV_t^H $. We define the reduced gramian at $t$ as $ G_t = V_t^HV_t $.

From

$$X^{(t)}(X^{(t)})^H = U_tV_t^HV_tU_t^H = U_tG_tU_t^H$$

it follows that the singular values of $X^{(t)} $ are the square roots of the eigenvalues of the reduced Gramian $ G_t $, provided that $ U_t^HU_t = I $.

Is it possible to explain this in more detail?