# In block encoding, how to determine the dimension of the elements from $\alpha(⟨0|^{\otimes a}\otimes I)U(|0〉^{\otimes a}\otimes|\psi〉)=A|\psi〉$?

The block-encoding framework shows the following statement in general, as discussed in the paper https://arxiv.org/abs/1804.01973.

A block-encoding of a matrix $$A \in \mathbb{C}^{N \times N}$$ is a unitary $$U$$ such that the top left block of $$U$$ is equal to $$A / \alpha$$ for some normalizing constant $$\alpha \geq\|A\|$$: $$U=\left(\begin{array}{cc} A / \alpha & \\ \cdot & \cdot \end{array}\right)$$ In other words, for some $$a$$, for any state $$|\psi\rangle$$ of appropriate dimension, $$\alpha\left(\left\langle\left. 0\right|^{\otimes a} \otimes I\right) U\left(|0\rangle^{\otimes a} \otimes\right.\right |\psi\rangle)=A|\psi\rangle.$$

My question is to find out the dimension of the elements from $$\alpha\left(\left\langle\left. 0\right|^{\otimes a} \otimes I\right) U\left(|0\rangle^{\otimes a} \otimes\right.\right.|\psi\rangle)=A|\psi\rangle.$$ I am not sure where $$a$$ came from in $$|0\rangle^{\otimes a}$$, and not sure what the dimension of $$I$$ is in this expression.

Any help could be highly appreciatedted!

$$U$$ can be written as: $$U=\begin{pmatrix}\frac{1}{\alpha}A&B\\C&D\end{pmatrix}$$ while $$|0\rangle^{\otimes a}\otimes|\psi\rangle$$ can be written as: $$|0\rangle^{\otimes a}\otimes|\psi\rangle=\begin{pmatrix}|\psi\rangle\\0\end{pmatrix}$$ We thus have: $$U\left(|0\rangle^{\otimes a}\otimes|\psi\rangle\right)=\begin{pmatrix}\frac{1}{\alpha}A&B\\C&D\end{pmatrix}\begin{pmatrix}|\psi\rangle\\0\end{pmatrix}=\begin{pmatrix}\frac{1}{\alpha}A|\psi\rangle\\C|\psi\rangle\end{pmatrix}$$ and then: $$\alpha\left(\langle0|^{\otimes a}\otimes I\right)U\left(|0\rangle^{\otimes a}\otimes|\psi\rangle\right)=\alpha\begin{pmatrix}I & 0\end{pmatrix}\begin{pmatrix}\frac{1}{\alpha}A|\psi\rangle\\C|\psi\rangle\end{pmatrix}=A|\psi\rangle$$
Is is now easy to see that $$I$$ must have the same size as $$A$$, which is the $$N\times N$$ identity matrix, while $$a$$ is equal to $$\frac{n}{N}$$, where $$n$$ is the dimension of $$U$$. You can also see in the last equation that the $$\alpha$$ at the beginning of the product cancels out with the $$\frac{1}{\alpha}$$ factor applied to $$A$$ in $$U$$.