The encoder waits for the decoder

Form $(\mat I-\mat D\mat E)\mat S$ once and use it in both gradients.

Here $\mat I-\mat D\mat E=\begin{bmatrix}0&0\\0&1\end{bmatrix}$, so

Multiplying on the right by $\mat E^{\mathsf T}=(1,0)^{\mathsf T}$ and on the left by $\mat D^{\mathsf T}=(1,\,0)$:

The decoder is told to raise its second weight: the residual, which lives in the second coordinate, is correlated with the code. The encoder is told nothing. Its gradient passes through $\mat D^{\mathsf T}$, and the decoder does not yet write to the coordinate where the error is.

After the step, $\mat D=(1,\,0.288)^{\mathsf T}$ and $\mat E$ is unchanged. The residual map is $\begin{bmatrix}0&0\\-0.288&1\end{bmatrix}$ and the loss is

down from $2.08$. On the next step the encoder’s gradient is no longer zero.

At $\mat E=\mat 0$ and $\mat D=\mat 0$, each gradient has a zero matrix as a factor, so both vanish. The loss is $\operatorname{tr}\mat S=5$ and gradient descent never leaves. An autoencoder needs a nonzero initialisation in both halves.