Adjoint, Transpose and Scalar Product

Adjoint, transpose and scalar product

Suppose that a scalar product is defined in the tangent space. If

Q_{0}

is a symmetric, definite positive matrix, it defines a scalar product as:

< p_{0} | q_{0} >_{0} \equiv p_{0}^{T} Q_{0} q_{0}

p_{0}

q_{0}

are generic (tangent) vectors (at time 0) and

p_{0}^{T}

is a row vector, the trasnpose of the column vector

p_{0}

. Remark that if the state vector components (then the tangent vector components) have physical dimensions, the components of

Q_{0}

must have inverse square dimensions make the sum of squares possible (the scalar product is then dimensionless).

At time 0 and at time

t

the scalar product may be different. So, a different matrix

Q_{t}

< y_{t} | z_{t} >_{t} \equiv y_{t}^{T} Q_{t} z_{t}

y_{t}

z_{t}

are generic tangent vector at time

t

M

be a linear operator, like the tangent linear model, applied to tangent vectors at time 0 and giving tangent vectors at time

t

. By definition its adjoint operator is

M^{†}

< p_{t} | M q_{0} >_{t} \equiv < M^{†} p_{t} | q_{0} >_{0}

Q

matrices, this is written as:

p_{t}^{T} Q_{t} M q_{0} = {(M^{†} p_{t})}^{T} Q_{0} q_{0}

This is the same as:

p_{t}^{T} Q_{t} M q_{0} = p_{t}^{T} {(M^{†})}^{T} Q_{0} q_{0}

Since this relation is valid for generic vectors

p_{t}

q_{0}

, it really is a relation between matrices:

Q_{t} M = {(M^{†})}^{T} Q_{0}

Now take the transpose:

M^{T} Q_{t} = Q_{0} M^{†}

The expression of the adjoint is obtained:

M^{†} = Q_{0}^{- 1} M^{T} Q_{t}

From this expression one sees that:

the adjoint intrinsicly depends on the scalar products at initial and final time;

the adjoint is related to the transpose;

the adjoint coincides with the transpose only when both $Q$ s coincide with the identity matrix: $Q_{t} = Q_{0} = I$ (in this case both scalar products are Euclidean, or L2); remark that this is possible only if all the state variables have the same physical dimensions.

Remark that the tangent linear operator is applied to tangent vectors at time

0

and gives tangent vectors at time

t

. Tangent vectors are state variations, or differentials, and are approximated by finite state differences, so they are indicated with

δ

δ x_{t} = M δ x_{0}

The transpose is applied to derivatives at time

t

and gives derivatives at time

0

, in this sense it goes backward in time:

\frac{\partial J}{\partial x_{0}} = M^{T} \frac{\partial J}{\partial x_{t}}

J = J (x_{t})

is a generic function of the state at time

t

The adjoint is defined on tangent vectors, not on derivatives (such as the tangent linear model), and goes backward in time (such as the transpose). A transformation between tangent vectors and “derivatives”, which at least accounts for physical dimensions, is then provided by the

Q

s in the above expression relating adjoint and transpose.

When (since?) what is really needed it the transpose (to compute derivatives with respect to initial conditions), instead of a scalar products one may define a duality form. Tangent vectors and derivatives belong to spaces that are dual to each other and, at time

0

(\frac{\partial J}{\partial x_{0}} | δ x_{0}) \equiv {(\frac{\partial J}{\partial x_{0}})}^{T} δ x_{0} = δ J

t

(\frac{\partial J}{\partial x_{t}} | δ x_{t}) \equiv {(\frac{\partial J}{\partial x_{t}})}^{T} δ x_{t} = δ J

These expressions are the same, both coincide with the first variation of

J

. In the duality form, the transpose behaves like the adjoint does in a scalar product:

(\frac{\partial J}{\partial x_{t}} | M δ x_{0}) \equiv (M^{T} \frac{\partial J}{\partial x_{t}} | δ x_{0})

That is to say:

{(\frac{\partial J}{\partial x_{t}})}^{T} M δ x_{0} = {(M^{T} \frac{\partial J}{\partial x_{t}})}^{T} δ x_{0}

The duality form (which here appears as a simple product of one row by one column) does not depend on the definition of scalar products (the

Q

s) and it does not depend on time.

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Francesco Uboldi 2014,2015,2016,2017