# Fantastic Adjoints And How To Find Them

This is a very short note about the adjoints. I hope it would be as useful for those feeling a bit confused with this wonderful mathematical concept as it was for me. Although there is not a single word about sensitivity, this very short example can give you a sneek peek.

## So, what’s the matter?

I have a problem: I need to find a vector, $u$, satisfying the following linear algebra problem:

Here $A$ is some matrix (presumably very unpleasant and big) or a matrix operator, $F$ is called the external forcing, and both $A$ and $F$ are given. Naturally, I would invert the matrix and multiply the right hand side by it. The answer would be:

In practice, finding $u$ itself is rarely the goal. Some function of $u$ is almost always much more meaningful. This can be, for instance, a linear combination with another vector, $(l, u)$. Picking a vector of ones $l = \mathbf{1}$ will give us the average value of $u$.

We can take the direct operator and do the complex conjugation and matrix transpose. The new adjoint operator then works this way:

Now let’s introduce the dual problem, which depends on previously discussed vector $l$:

Using the definition of adjoint operator, I can show:

This is a wonderful result. It means, that $(l, u)$ can be found for any $F$ without recomputation of the very first matrix equation just by taking an inner product of $v$ with the $F$ we care at the moment.

The only cost of such computation is to calculate $v = \left(A^\dagger \right)^{-1} l$ once.

## Summing up

A problem of finding the inner product $(u, l)$ for $u$ being the solution of $A u = F$, and given $l$ can be cheaply solved for any forcing term $F$. First, we need to find the adjoint operator $A^\dagger$. Second, we solve the dual problem $A^\dagger v = l \$ for a vector $v$ once. Third, the target inner product equals to $(v, F)$.

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