# Nielsen & Chuang Exercise 2.3 - “Matrix representation for operator products” [closed]

Reproduced from Exercise 2.3 of Nielsen & Chuang's Quantum Computation and Quantum Information (10th Anniversary Edition):

Suppose $$A$$ is a linear operator from vector space $$V$$ to vector space $$W$$, and $$B$$ is a linear operator from vector space $$W$$ to vector space $$X$$. Let $$|v_i⟩$$, $$|w_j⟩$$, and $$|x_k⟩$$ be bases for the vector spaces $$V$$, $$W$$, and $$X$$, respectively. Show that the matrix representation for the linear transformation $$BA$$ is the matrix product of the matrix representations for $$B$$ and $$A$$, with respect to the appropriate bases.

Note: This question is part of a series attempting to provide worked solutions to the exercises provided in the above book.

Consider the linear maps $A: V\to W$ and $B: W\to X$. The composition $BA$ is a linear map from $V$ to $X$. Now, how can $\mathcal{M}(BA)$ be computed from $\mathcal{M}(B)$ and $\mathcal{M}(A)$? $\mathcal{M}(A)$ is the $n\times p$ matrix representation of the linear map $A$ w.r.t the basis $\{v_1,...,v_p\}$ and $\{w_1,...,w_n\}$. $\mathcal{M}(B)$ is the $m\times n$ matrix representation of the linear map $B$ w.r.t the basis $\{w_1,...,w_n\}$ and $\{x_1,...,x_m\}$.

Say, $\mathcal{M}(A) = \begin{bmatrix}a_{11} & ... & a_{1p}\\... & ... & ...\\a_{n1} & ... &a_{np}\end{bmatrix}$ & $\mathcal{M}(B) = \begin{bmatrix}b_{11} & ... & b_{1n}\\... & ... & ...\\b_{m1} & ... &b_{mn}\end{bmatrix}$.

Now, let's see the action of the linear map $BA$ on a certain basis vector of $V$, say $v_k$ (as it is sufficient to just know the action of any linear transformation on the basis vectors, in order to determine the transformation):

$$\therefore BA(v_k)=B(\sum_{r=1}^{n} a_{r,k}w_r)$$ $$= \sum_{r=1}^{n}a_{r,k}Bw_r$$ $$= \sum_{r=1}^{n}a_{r,k}\sum_{j=1}^{m}b_{j,r}x_j$$ $$= \sum_{j=1}^{m}(\sum_{r=1}^{n}b_{j,r}a_{r,k})x_j$$

Thus, $\mathcal{M}(BA)$ is the $m\times p$ matrix whose entry in row $j$, column $k$ equals $\sum_{r=1}^{n}b_{j,r}a_{r,k}$. This is exactly equal to the $j,k$-th element of the matrix we get after multiplying $\mathcal{M}(B)$ and $\mathcal{M}(A)$. Hence,

$$\mathcal{M}(BA)=\mathcal{M}(B)\mathcal{M}(A)$$

Note: I've omitted the ket notation given in the original question, for ease of typing. By default, $|v_i\rangle = v_i,|w_j\rangle = w_j$ and $|x_k\rangle = x_k$ for indices $i,j,k$.