# Confusing notation in Block-Encoding

I'm reading the lecture notes from Ronald de Wolf and got confused at the part where it introduces Block encoding, specifically by this notation at page 76:

More generally we can define an $$a$$-qubit block-encoding of $$A$$, which is an $$(a + n)$$-qubit unitary $$U$$ with the property that $$(\langle 0^a|\otimes I)U(|0^a\rangle \otimes I)=A$$.

How do I interpret $$(\langle 0^a|\otimes I)U(|0^a\rangle \otimes I)$$? Like having a tensor product between a ket vector and a $$2^n\times 2^n$$ matrix?

I will try to answer this question for myself. We know that an $$a$$-qubit quantum state $$|\psi\rangle$$ lives in a $$2^a$$-dimensional Hilbert space $$\mathcal{H}$$. The operator $$\mathbb{1}$$ acting on $$|\psi\rangle$$ is a map $$\mathcal{H}\rightarrow\mathcal{H}$$, but also a vector in $$\mathcal{B}(\mathcal{H})$$, the vector space of bounded linear operators over $$\mathcal{H}$$. We can define elements in the tensor product space $$\mathcal{H}\otimes \mathcal{B}(\mathcal{H})$$. If $$\mathcal{H}$$ and $$\mathcal{B}(\mathcal{H})$$ are both finite dimensional, then $$\dim(\mathcal{H}\otimes \mathcal{B}(\mathcal{H})) = \dim(\mathcal{H})\times\dim(\mathcal{B}(\mathcal{H}))$$. In our case, $$\dim(\mathcal{H}) = 2^a$$ and $$\dim(\mathcal{B}(\mathcal{H}))=2^a\times2^a$$, so $$\dim(\mathcal{H}\otimes \mathcal{B}(\mathcal{H}))=2^a\times2^a\times2^a$$.
Generally, if $$A\in \mathbb{C}^{m\times n}$$ and $$B\in \mathbb{C}^{k\times l}$$ are two matrices, we can define $$A\otimes B$$ as $$A\otimes B =\begin{bmatrix}a_{11} B & \cdots & a_{1n}B \\ \vdots & & \vdots \\ a_{m1}B & \cdots & a_{mn}B\end{bmatrix}.$$ In the case of block-encoding (let's assume $$a=1$$ and $$A\in\mathbb{C}^{2\times 2}$$for simplicity), \begin{align} \langle 0 | &= \begin{bmatrix}1 & 0\end{bmatrix}, \\ |0\rangle &= \begin{bmatrix}1 \\ 0\end{bmatrix} \\ \mathbb{1} &= \begin{bmatrix}1 & 0\\0 & 1 \end{bmatrix} \end{align} hence \begin{align} \langle 0 | \otimes \mathbb{1} &= \begin{bmatrix}1&0&0&0\\0&1&0&0\end{bmatrix} = \begin{bmatrix}\mathbb{1} & 0\end{bmatrix},\\ |0\rangle \otimes \mathbb{1} &= \begin{bmatrix} 1 & 0 \\ 0&1\\0&0\\0&0\end{bmatrix} = \begin{bmatrix}\mathbb{1} \\ 0\end{bmatrix}. \end{align} Therefore, $$(\langle 0 | \otimes \mathbb{1}) U (| 0 \rangle \otimes \mathbb{1}) = \begin{bmatrix}\mathbb{1} & 0\end{bmatrix} \begin{bmatrix}A&B\\C&D\end{bmatrix}\begin{bmatrix}\mathbb{1} \\ 0\end{bmatrix} = \begin{bmatrix}\mathbb{1} & 0\end{bmatrix} \begin{bmatrix}A \\ C\end{bmatrix} = A.$$ The case when $$A\in \mathbb{C}^{2^n\times 2^n}$$ and $$a>1$$ is similar to the above.