3
votes
$\begingroup$

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.

$\endgroup$

closed as off-topic by Sanchayan Dutta Jan 7 at 15:12

  • This question does not appear to be about quantum computing or quantum information, within the scope defined in the help center.
If this question can be reworded to fit the rules in the help center, please edit the question.

locked by Sanchayan Dutta Jan 7 at 15:16

This question exists because it has historical significance, but it is not considered a good, on-topic question for this site, so please do not use it as evidence that you can ask similar questions here. This question and its answers are frozen and cannot be changed. More info: help center.

Read more about locked posts here.

4
votes
$\begingroup$

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 $i,j,k$.

$\endgroup$

This site is temporarily in read only mode and not accepting new answers.

Not the answer you're looking for? Browse other questions tagged .