Reproduced from Nielsen & Chuang's Quantum Computation and Quantum Information (10th Anniversary Edition) in page 64:

We've seen that matrices can be regarded as linear operators. [...] Suppose $A: V \rightarrow W$ is a linear operator between vector spaces $V$ and $W$. Suppose $|v_{1}\rangle,...,|v_{m}\rangle$ is a basis for $V$ and $|w_{1}\rangle,...,|w_{m}\rangle$ is a basis for $W$. Then for each j in the range 1,...,m, there exist complex numbers $A_{1j}$ through $A_{nj}$ such that $$A|v_{j}\rangle = \sum_{i} A_{ij} |w_{i}\rangle \tag{2.12} $$

Looking for information about this equation, I was not able to find any demonstration. I saw in other websites and books that they use this equation, but not where it comes from.

I'm presenting this question because I made a mistake with equation (2.12) in this other question.

But I'm still not understanding why, once I've got members of the matrix of the operator, I can't use it directly with vector components. And I believe the problem comes from equation (2.12).

Thanks for all information or advice you can give to me.

  • 1
    $\begingroup$ If you're still confused about the linear algebra I recommend finding a copy and starting with Strangs highly accessible and well explained textbook, amazon.ca/Introduction-Linear-Algebra-Gilbert-Strang/dp/… $\endgroup$
    – Sam Palmer
    Jul 12 '20 at 0:12
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    $\begingroup$ I do not understand how this differs from the other question, and why the answers there do not satisfy you $\endgroup$
    – glS
    Jul 12 '20 at 22:37
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    $\begingroup$ @vcorle yes. That's not something you prove, that's the definition of matrix-vector multiplication $\endgroup$
    – glS
    Jul 13 '20 at 15:27

I'll try to explain the whole process here, I hope this will clear the equation for you!

So, what we are trying to do here is to find a matrix that represents the linear operator $A : V \rightarrow W$.
$V$ and $W$ are both vector spaces so we know we can find one basis for each set, here $|v_1\rangle, ..., |v_m\rangle$ and $|w_1\rangle, ..., |w_n\rangle$. Now, we want to study how $A$ acts on the set $V$, so we will look at how it affects the basis of the set (a result of linear algebra actually tells us that looking at the results of one basis suffices to completely describe the operator). Let's note $|y_i\rangle = A(|v_i\rangle)$. So since $A : V \rightarrow W$, we have $|y_i\rangle \in W$, so the direct result we have is that we can write this vector as a linear combination of the $|w_1\rangle, ..., |w_n\rangle$ since this is a basis of $W$. This is the definition of a basis, that every vector of the set can be written as a linear combination of the vectors of the basis.

This is the information $(2.12)$ is representing, and it is with this information that we will build the matrix associated to the linear operator. The equation is telling us that $A(|v_i\rangle)$ is written as a linear combination of the basis $|w_1\rangle, ..., |w_n\rangle$.

Please let me know if you need clarification on something :)

  • $\begingroup$ Actually yes it is, the property claiming that in a vector space, given a basis of the space, any element of the space can be described as a linear combination of elements of the basis is really well-known and doesn't need explicit and complicated proofs when one wants to use it since it's common. Here for this equation, this is exactly the property they use to write this, nothing more, the equation is coming from this property. If you are not familiar with it then I strongly suggest you look at books about bases of linear algebra, and you will have a general proof of this property :) $\endgroup$
    – Lena
    Jul 13 '20 at 15:23
  • $\begingroup$ One thing I'm confused about is the choice of indices. I realize we can call the indices whatever we want, but matrix dimensions are conventionally referred to as m by n, meaning the matrix has m rows and n columns. Has that convention been reversed in this discussion? If so (i.e. if A is n by m in this discussion), I understand it. If not (i.e. if A really is m by n, as usual), then I am totally confused. In summary: Should I think of A as an n by m matrix here, or is A somehow supposed to be m by n? $\endgroup$
    – Mike Bell
    Mar 5 at 16:26

Suppose that you have an basis vector $|v_i\rangle $ in 3-dimensions and you act A on it then in other 3-dimensions space you want to find an operator B that act on basis $|w_j\rangle $ and because two space are in the rotation position of each other you need some B that act on different basis of it. That we name it $A_{ij}$

  • $\begingroup$ Hi @armankashef ... As I said to Lena, I need a formal demonstration for eq. 2.12 $\endgroup$
    – user12503
    Jul 13 '20 at 14:43