Schur Complements, Shorted Operators, and Energy Minimization

Basic definitions

Suppose that a matrix or operator is written in block form as

Assume that $a_{22}$ is invertible. The Schur complement of $a_{22}$ in $M$ is

Here $S$ is the operator that remains after the second group of variables has been eliminated. This same elimination appears in several different forms. It appears in block Gaussian elimination, in block matrix inversion, in the reduction of mechanical or electrical networks, in minimization of quadratic energies, and in the construction of transfer functions in systems theory.

Gaussian elimination

Consider the system of equations

Assume $a_{22}$ is invertible. Then the second equation gives

Substituting this into the first equation, we get

So the reduced equation is

This is the basic meaning of the Schur complement. The variable $y$ is eliminated, but its effect has not disappeared. Its influence is through the correction $-a_{12}a^{-1}_{22}a_{21}$ in the effective operator $S$ acting on $x$.

Block factorizations

The same calculation is in the factorization

This is a block $UDL$ factorization.

To see the relation with elimination, apply the right-hand factor $L$ to a vector

The corrected second variable is therefore

In the variables $x$ and $z$, the middle factor $D$ acts diagonally

The coupling has been moved into the triangular changes of variables. The diagonal part contains the effective operator $S$ acting on the first variable and the original operator $a_{22}$ acting on the corrected second variable.

Equivalently, one may perform a block row operation

Applying the same row operation to the right-hand side gives

The first row is exactly the reduced equation obtained above.

Determinants and invertibility

In finite dimensions, the factorization

immediately gives

Since $a_{22}$ is assumed invertible, it follows that

Block inverse

Suppose that both $a_{22}$ and $S=a_{11}-a_{12}a^{-1}_{22}a_{21}$ are invertible. Starting from

inverting the three factors gives

In particular, the upper-left block of $M^{-1}$ is

So the Schur complement is the effective coefficient in the reduced equation

The inverse Schur complement is the corresponding solution operator

Thus $S$ and $S^{-1}$ describe the same reduced system in opposite directions.

Which variable is the input?

A possible confusion is that the words input and output are used in different ways.

Suppose that

In a mechanical interpretation, $x$ may be a displacement, $F$ an applied force, and $K$ a stiffness matrix.

As a linear map, $K$ acts by

From this point of view, $x$ is the argument of the map and $F$ is its value.

In an experiment, however, one may prescribe the force $F$ and measure the displacement $x$. The experimentally relevant response map is then

The same issue occurs in many settings. A Dirichlet-to-Neumann map takes prescribed boundary values to boundary fluxes, while its inverse, when it exists, takes boundary fluxes to boundary values. A conductance operator takes voltages to currents, while a resistance operator goes in the opposite direction.

The words input and output depend on how the system is being used. They are not determined solely by which side of an equation a variable occupies.

Boundary and internal variables

A standard application of the Schur complement is the elimination of hidden internal variables.

Consider

Think of $x$ as a visible boundary variable, $y$ as an internal variable, and $f$ as an external force applied only at the boundary.

The internal equation is

Hence

The internal variable is not prescribed independently. It adjusts in response to the boundary value $x$.

Substituting into the boundary equation gives

The Schur complement therefore gives the reduced boundary law

It describes how much boundary force is required to maintain the boundary configuration $x$, after the internal variables have been allowed to equilibrate.

If the boundary force is prescribed instead, then one solves

In that use of the system, the response map is $S^{-1}$.

These terms come from physics and engineering analogies, but they have a very clean mathematical meaning.

Imagine a weight hanging from two springs connected in a row:

1. The top spring is attached to your hand at the very top (the boundary $x$).
2. The bottom spring connects to the weight at the bottom.
3. The junction where the two springs tie together is tucked inside the scale (the internal variable $y$).

1. What does “Internal” mean?

An internal variable ($y$) is a part of the system that you cannot directly grab, control, or see from the outside.

• The Boundary ($x$): This is your hand. You can physically pull the top of the spring to a specific position $x$. You have direct control over it.
• The Internal variable ($y$): This is the middle junction inside the scale casing. You cannot reach in and pull it. It just floats freely inside.

Mathematically, when you look at the right-hand side of your equation:

Notice that the force on the right side for $y$ is exactly $0$. This tells you that no external agent is pulling or pushing on $y$. It is left entirely alone to do whatever the system forces it to do.

2. What does “Equilibrate” mean?

To equilibrate means to settle into a steady, balanced state where all the forces acting on the internal part cancel out to zero.

Because the internal junction $y$ is free to move and has no external force acting on it ($0$), it will naturally shift until the tension from the top spring perfectly balances the tension from the bottom spring. It finds its comfort zone.

That balancing act is exactly what the second equation describes:

When we solve this for $y$:

We are saying: “If you move your hand to position $x$, the inside junction $y$ will automatically slide to this exact spot to keep the internal forces balanced.”

Bringing it together

It means you pull the boundary $x$, you let the inside mechanism $y$ wiggle and settle until it stops moving, and then you measure how much total force ($f$) your hand has to exert to hold everything still. The Schur complement $S$ is the single math shortcut that tells you that final force without making you calculate where the inside junction $y$ settled.

The energy interpretation

There is another way to see the same construction.

Consider the quadratic energy/form

Expanding $E\left(x,y\right)$ using

gives

Obviously, $M\geq0$ (positive definite) iff $S\geq0$ and $a_{22}\geq0$.

Assuming $M\geq0$, then $E\left(x,y\right)$ is minimized when

Consequently,

Thus the Schur complement is the effective energy seen by $x$ after the variable $y$ has been allowed to relax. It explains why Schur complements arise naturally in mechanics, network theory, finite element methods, boundary-value problems, and variational arguments.

Dynamic systems

In systems theory, one often begins with differential equations

Here the notation assigns specific roles to the variables.

The vector $x\left(t\right)$ is the internal state. It describes information inside the system that evolves with time.

The vector $u\left(t\right)$ is the external input. It is the signal applied to the system.

The vector $y\left(t\right)$ is the measured output.

The matrix $A$ governs the internal evolution. The matrix $B$ describes how the input enters the state equation. The matrix $C$ describes how the internal state contributes to the observed output. The matrix $D$ describes a direct contribution from the input to the output.

The goal is to eliminate the hidden state $x\left(t\right)$ and obtain a direct relation between $u\left(t\right)$ and $y\left(t\right)$.

Take the Laplace transform. Write

The transform of the derivative is

Therefore the state equation becomes

Whenever $\left(sI-A\right)$ is invertible,

The output equation becomes

The first term is the response caused by the external input. The second term is the response caused by the initial state.

The transfer function $G\left(s\right)$ describes the zero-initial-state response, since when $x\left(0\right)=0$,

For matrix-valued systems, $G\left(s\right)$ is usually called the transfer matrix.

Why the transfer function is a Schur complement

The transformed equations can be written as one block system

For the zero-initial-state problem,

The first block equation is

Substituting into the second block equation gives

Now compute the Schur complement of the top-left block $\left(sI-A\right)$:

Thus the transfer function is the Schur complement obtained by eliminating the internal state.

There is no need to invert the Schur complement in this formulation. After $X\left(s\right)$ has been eliminated, the remaining equation is already

The retained variable $U\left(s\right)$ is the designated input, and the right-hand side variable $Y\left(s\right)$ is the designated output.

This is different from the boundary stiffness equation

If the applied force $f$ is regarded as the input and the displacement $x$ as the output, then the response map is

In the state-space arrangement, however, the block signs and the placement of $U\left(s\right)$ and $Y\left(s\right)$ are chosen so that the Schur complement itself maps input to output.

There is therefore no general engineering rule saying that a Schur complement must be inverted to obtain an input-output map. Whether the Schur complement or its inverse is the relevant map depends on the variables and on the direction in which the reduced relation is being used.

What if the block is not invertible?

The standard formula

relies on $a_{22}$ being invertible, but in real-world applications, this requirement often fails. For instance, a physical stiffness matrix might have a kernel corresponding to rigid body motions, and a graph Laplacian inherently has constant vectors in its kernel. In infinite-dimensional settings, an operator might simply fail to have a closed range.

When this happens, there are a few ways to move forward. We can restrict the problem to a subspace where $a_{22}$ is invertible, apply a generalized inverse, enforce compatibility conditions, or bypass the matrix algebra entirely by defining the reduced object variationally.

Specifically, in the setting of a positive quadratic form, the energy expression

remains perfectly meaningful even when the direct algebraic inverse breaks down. Because of this, the minimization perspective is often viewed as much deeper and more fundamental than the explicit matrix formula.

Shorted operators

The energy interpretation leads naturally to the notion of a shorted operator. The Schur complement and the shorted operator describe the same elimination, but from slightly different points of view.

Let

and suppose that

We want to eliminate the second space $\mathcal{H}_{2}$ and retain the effective part of $M$ acting on $\mathcal{H}_{1}$.

If $D$ is invertible, then the usual Schur complement is

The corresponding operator on the original space $\mathcal{H}_{1}\oplus\mathcal{H}_{2}$ is

This is the shorted operator of $M$ to the subspace

The terminology comes from electrical network theory. One imagines short-circuiting part of a network and asking what effective positive operator remains on the part that is retained. For our purposes, the important point is that the shorted operator has an order-theoretic definition that does not require $D$ to be invertible.

The shorted operator of $M$ to $\mathcal{S}$ is the largest positive operator $X$ such that

and

Since $X$ is supported on the first summand, it has the form

for some positive operator $T$ on $\mathcal{H}_{1}$.

The inequality $X\leq M$ means that

for every $x\in\mathcal{H}_{1}$ and $y\in\mathcal{H}_{2}$. That is,

for every $y\in\mathcal{H}_{2}$. Taking the infimum over $y$ gives

The largest possible operator $T$ is characterized by

Thus the shorted operator is exactly the operator obtained by minimizing out the second variable.

When $D$ is invertible, completing the square gives

The second term is nonnegative and is minimized when

It follows that

Therefore

So the Schur complement is the upper-left block of the shorted operator when the block being eliminated is invertible. The shorted operator is the more general object because its definition still makes sense when $D$ is singular.

This also explains why the variational point of view is more general than the formula involving $D^{-1}$. Even when the infimum is not attained by a particular $y$, the reduced quadratic form may still define a bounded positive operator on $\mathcal{H}_{1}$.

In finite dimensions, one can often write the reduced operator as

where $D^{\dagger}$ is the Moore-Penrose inverse. The positivity of $M$ provides the necessary compatibility between $B$ and the range of $D$. In infinite dimensions, this formula requires more care, especially when the range of $D$ is not closed. The variational definition avoids these difficulties.

It is also useful to distinguish shorting from compression. Let $P$ be the orthogonal projection onto $\mathcal{H}_{1}$. The compression of $M$ to $\mathcal{H}_{1}$ is simply

This ignores the second variable. By contrast, shorting allows the second variable to adjust before the effective operator on $\mathcal{H}_{1}$ is computed. In the invertible case, it gives

Thus compression and shorting answer different questions. Compression asks what the upper-left block of the original operator is. Shorting asks what positive operator remains on the first space after the second variable has been eliminated.

In this sense, the shorted operator is not a separate construction from the Schur complement. It is the positive, order-theoretic, and variational version of the same elimination procedure.

What does it mean to short a port?

The word “shorted” comes from electrical network theory. It means short-circuited, not shortened. The main issue is this: in a two-port network, $i_{1},i_{2},v_{1},v_{2}$ are not “input” and “output” in the usual one-input, one-output sense. They are just the four port variables.

A two-port network is a black box with two pairs of terminals: Port 1 is one pair of terminals, and Port 2 is another pair of terminals. Each port has a voltage across it and a current through it. So the variables are:

• $v_{1}$ is the voltage at Port 1;
• $i_{1}$ is the current at Port 1;
• $v_{2}$ is the voltage at Port 2;
• $i_{2}$ is the current at Port 2.

Thus, in a two-port setup, both ports have a current and a voltage. That does not mean both currents are “inputs” in the usual causal sense. When one writes

this is just a constitutive law for the network. It says that if currents $i_{1}$ and $i_{2}$ are present at the two ports, then the resulting port voltages are $v_{1}$ and $v_{2}$. This is like a multi-port version of $v=Ri$ for a resistor. Here, the network behaves like a generalized impedance operator.

The standard convention in network theory is that, at each port, the current is defined as positive when it enters the box. So both $i_{1}$ and $i_{2}$ are simply port currents, one for each port. They are not both externally prescribed inputs unless both ports are driven. In fact, in the shorting discussion, both are not prescribed.

To short Port 2 means to connect the two terminals of Port 2 by an ideal wire. That forces

But it does not force $i_{2}=0$. Current may flow through the short, which is the whole point. So after shorting Port 2, the condition $v_{2}=0$ is imposed, and $i_{2}$ becomes whatever it must be to satisfy that condition. Thus $i_{2}$ is not an independent input; it is an induced current.

In the shorted-operator situation, the physical experiment consists of the following steps:

• At Port 1, apply a test current $i_{1}$;
• Port 2 is shorted, so $v_{2}=0$;
• Solve for the induced current $i_{2}$;
• Observe the resulting voltage $v_{1}$.

So the only externally applied quantity is $i_{1}$. The variable $i_{2}$ is determined completely by the constraint $v_{2}=0$. From the second row block equation,

the short-circuit condition gives

Assuming that $A_{22}$ is invertible,

Then the first row block equation,

becomes

So the shorted operator is the effective impedance seen at Port 1 when Port 2 is short-circuited.

To simplify the mental model, one can ignore “input-output systems” for a moment and think instead of a resistive device with two accessible terminal pairs. Port 1 is where the device is probed, while Port 2 is an extra terminal pair chosen to be bound together. The core question becomes: after Port 2 is shorted, what effective resistance or impedance is seen from Port 1? That effective operator is the shorted operator. This perspective is much closer to classical network reduction than to the usual engineering idea of a transfer function.

This formulation also makes the distinction between compression and shorting completely transparent. If Port 2 is left open, no current flows through it:

Then

So the compression $A_{11}$ describes the network when the second port is open. If Port 2 is short-circuited, then $v_{2}=0$, the current $i_{2}$ is allowed to adjust, and the effective impedance is $A_{11}-A_{12}A^{-1}_{22}A_{21}$. Thus,

Compression ignores the second port by imposing $i_{2}=0$. Shorting lets current flow through the second port and imposes $v_{2}=0$. So $A_{11}$ means “just look at the first block,” whereas $A_{11}-A_{12}A^{-1}_{22}A_{21}$ means “let the second port respond under the short-circuit condition.”

This directly explains why, for a positive impedance operator,

Leaving the unused port open prevents current from moving through that part of the network. Shorting the unused port creates an additional parallel path through which current can divert, so the total effective resistance or impedance seen at the remaining port can only decrease.

Caption: A two-port impedance representation and the effective impedance obtained by short-circuiting Port 2.

The square-root projection formula

Let $R\geq0$ be a bounded positive operator on a Hilbert space $\mathcal{H}$, and let $\mathcal{S}\subseteq\mathcal{H}$ be a closed subspace to be retained. The shorted operator of $R$ to $\mathcal{S}$ is the largest positive operator below $R$ whose range lies in $\mathcal{S}$.

A frequently used formula is

but what is $P$? It is not usually the orthogonal projection onto $\mathcal{S}$. Instead, define

Then $\mathcal{M}$ is a closed subspace (as the pre-image of $\mathcal{S}$ under $R^{1/2}$), and $P$ is the orthogonal projection onto $\mathcal{M}$:

Thus the precise formula is

Equivalently,

where the orthogonal complement automatically ignores whether $R^{1/2}\mathcal{S}^{\perp}$ is closed. Hence one can also write

The equivalence relies entirely on two facts: $\mathcal{S}$ is a closed subspace, and $R^{1/2}$ is a self-adjoint (symmetric) operator. Start with the first definition. A vector $h$ belongs to $\mathcal{M}$ if and only if

Because $\mathcal{S}$ is closed, saying a vector is in $\mathcal{S}$ is exactly the same as saying it is orthogonal to everything in the orthogonal complement $\mathcal{S}^{\perp}$. So, $R^{1/2}h\in\mathcal{S}$ means:

Because $R^{1/2}$ is self-adjoint, the operator can be shifted to the other side of the inner product:

This equation states that $h$ must be orthogonal to every single vector of the form $R^{1/2}y$, where $y$ comes from $\mathcal{S}^{\perp}$. In other words, $h$ is orthogonal to the entire set $R^{1/2}\mathcal{S}^{\perp}$.

The collection of all vectors orthogonal to a set is simply the orthogonal complement of that set. Therefore, the condition is exactly equivalent to:

So the set of vectors that $R^{1/2}$ maps into $\mathcal{S}$ is exactly the orthogonal complement of the image of $\mathcal{S}^{\perp}$ under $R^{1/2}$.

The cleanest explanation for why this formula appears comes from the energy interpretation already established. Suppose that $x\in\mathcal{S}$. The goal is to retain $x$ and allow a vector

to adjust. The reduced energy is

Since $R\geq0$,

Therefore

In taking the infimum, the vectors $R^{1/2}y$, with $y\in\mathcal{S}^{\perp}$, remove the component of $R^{1/2}x$ that lies in

If $R^{1/2}\mathcal{S}^{\perp}$ is not closed, this component may only be approximated arbitrarily well rather than cancelled by a single choice of $y$. What remains is its orthogonal projection onto

Thus

Since $P_{\mathcal{M}}$ is an orthogonal projection,

This is exactly why

There is a useful way to read what the three factors in the expression

are doing. The rightmost $R^{1/2}$ moves the vector into the geometry of the $R$-energy, since

The projection $P_{\mathcal{M}}$ removes the part that can be absorbed or approximated by changing the eliminated variable. The leftmost $R^{1/2}$ brings the resulting quadratic form back to an operator on the original Hilbert space. So this is not just an arbitrary positive sandwich. It is an ordinary orthogonal projection, but performed after passing to the square-root coordinates determined by $R$.

It is also important to clarify why $P$ is not simply $P_{\mathcal{S}}$. It would be tempting to write

That is generally not the shorted operator. The reason is that $P_{\mathcal{S}}$ projects onto $\mathcal{S}$ before accounting for the geometry introduced by $R^{1/2}$. The correct projection is onto

This guarantees that

for every $h\in\mathcal{H}$. So the projection depends on both $R$ and the retained subspace $\mathcal{S}$. If $R$ is invertible, then

and the formula becomes

This is a frequently cited finite-dimensional version.

To see how this becomes the Schur complement, take

and write

The shorted operator $R_{\mathcal{S}}$ is supported purely on the first summand, so its block structure must be

The link between the projection formula and the matrix blocks is the quadratic form. In this block setting, vectors in $\mathcal{S}$ have the form $\begin{bmatrix}x\\ 0 \end{bmatrix}$, and vectors in $\mathcal{S}^{\perp}$ have the form $\begin{bmatrix}0\\ y \end{bmatrix}$. The earlier energy derivation established that the quadratic form of the square-root projection operator is precisely the infimum over the orthogonal complement:

If $D$ is invertible, completing the square on the right-hand side explicitly evaluates this infimum as

Because both operators act solely on the $\mathcal{H}_{1}$ block and yield identical values for their quadratic forms on every vector $x$, the operators themselves must be identical on that block. Therefore, $T=A-BD^{-1}B^{*}$, which enforces the matrix identity:

So

is the intrinsic Hilbert-space formula, while

is the nonzero upper-left block of the exact same shorted operator when the eliminated block is invertible.

Radon-Nikodym derivatives

This is exactly Douglas' lemma, but one has to distinguish the Douglas contraction from the Radon-Nikodym derivative. Let

Apply Douglas' lemma to $A^{1/2}$ and $R^{1/2}$. Since

there is a contraction $C$ such that

Consequently,

Therefore the Radon-Nikodym derivative is

So Douglas gives the square root of the derivative:

not generally $C$ itself. On the support space

the positive contraction $C^{*}C$ is uniquely determined by

For the shorted operator. Now take

It has already been established that

where, on the support of $R$,

Comparing with the Douglas representation gives

Hence

Thus the reduced Douglas contraction $C$ is a partial isometry whose initial projection is

It need not itself equal $P_{\mathcal{M}}$. That is the precise reconciliation.

The invertible case. When $R$ is invertible, the Douglas contraction can be written explicitly:

Then

Therefore

For $A=R_{\mathcal{S}}$,

So in this case,

Again, $C$ itself is generally not the projection; its modulus is:

Hence the Douglas contraction has polar decomposition

for a suitable partial isometry $U$.

Another way to see Douglas directly. Starting with

set

Then

whereas

The inequality

is precisely the form of Douglas' lemma for $X^{*}X\leq Y^{*}Y$, and the factorization is visibly

Here the Douglas contraction is literally $P_{\mathcal{M}}$, because the factor $X=P_{\mathcal{M}}R^{1/2}$ was chosen, rather than the canonical positive factor $R^{1/2}_{\mathcal{S}}$.

This explains the apparent discrepancy: factoring the positive square root

gives a partial isometry-type contraction $C$, with

factoring the nonpositive square-root factor

gives the contraction $P_{\mathcal{M}}$ directly. So everything is consistent. The exact statement is

where $C$ is the reduced Douglas solution of

Dirichlet-to-Neumann operators and Schur complements

The connection is very direct:

A Dirichlet-to-Neumann operator is often a Schur complement obtained by eliminating the interior variables and retaining the boundary variables.

The cleanest place to see this is a finite resistor network or a discretized elliptic equation. Suppose the variables are divided into boundary values $u_{B}$ and interior values $u_{I}$. Write the Laplacian or stiffness matrix in block form:

The vector

records the potential at the boundary and interior nodes. Applying $L$ gives the corresponding currents or forces:

Here:

• $u_{B}$ is the prescribed boundary potential;
• $u_{I}$ is the unknown interior potential;
• $f_{B}$ is the resulting boundary current;
• $f_{I}$ is the current injected at the interior nodes.

For the Dirichlet problem, $u_{B}$ is prescribed and the interior is required to be source-free:

Thus the interior equation is

Assuming $L_{II}$ is invertible,

Now substitute this into the boundary equation:

This yields

Therefore the Dirichlet-to-Neumann operator is

That is exactly the Schur complement of $L_{II}$ in $L$. Depending on whether boundary current is taken to flow into or out of the network, one may instead obtain the negative of this operator. The Schur-complement mechanism is the same.

For a connected resistor network with a nonempty boundary set, $L_{II}$ is invertible. More generally, it is enough that every connected component of the network meet the boundary.

Why it is called Dirichlet-to-Neumann. The word “Dirichlet” refers to prescribing the value of the function on the boundary:

The word “Neumann” refers to the boundary flux or normal derivative:

So the Dirichlet-to-Neumann map takes

In the finite network model, this is exactly

The boundary potentials are the Dirichlet data, and the boundary currents are the Neumann data.

The PDE version. Consider, for example, the elliptic equation

with prescribed boundary value

Assume that $\Omega$ is a bounded Lipschitz domain, that $a$ is bounded, symmetric, and uniformly elliptic, and that the corresponding Dirichlet problem is uniquely solvable. For example, one may assume that $q\geq0$.

Let $u_{g}$ be the solution. The Dirichlet-to-Neumann operator is

up to the sign convention used for the outward normal. In the standard weak formulation, it acts as

Conceptually, the function is again divided into:

• boundary data $g$;
• an interior function determined by solving the PDE.

The interior function is eliminated by solving the interior equation. What remains is an operator acting only on the boundary data. That reduced boundary operator is the infinite-dimensional analogue of the Schur complement.

The formula may not literally be written as

unless one has chosen a suitable block decomposition and function spaces, but the mechanism is the same.

The energy interpretation. This also fits perfectly with the section on quadratic energy. Let the energy be

Under the assumptions above, for a prescribed boundary value $g$, the solution $u_{g}$ minimizes the energy among all functions having boundary trace $g$:

The reduced boundary energy is then

So the Dirichlet-to-Neumann operator is the operator associated with the energy that remains after the interior variables have been minimized out. That is exactly the variational meaning of a Schur complement or shorted operator.

The resistor-network picture. The resistor-network version is especially concrete.

• Prescribe voltages on the boundary nodes.
• Let the interior voltages adjust so that Kirchhoff's current law holds at every interior node.
• Measure the currents entering the network through the boundary nodes.

The map

is the Dirichlet-to-Neumann map. If the full network Laplacian is

then

This same operation is also called Kron reduction in network theory.

Compression is not the Dirichlet-to-Neumann map. If $L$ is simply compressed to the boundary space, the result is

But that ignores the fact that the interior voltages respond to the boundary values. The actual boundary operator is

So, just as before:

while

The Dirichlet-to-Neumann map is the second one.

Relation to the shorted operator. For a positive stiffness or Laplacian operator, the reduced boundary energy is

In the finite network or discretized setting, the operator associated with this reduced quadratic form is precisely the shorted operator to the boundary coordinate space. When $L_{II}$ is invertible, its nonzero boundary block is

In the continuum PDE setting, the boundary trace space is not literally a coordinate subspace of the interior energy space. One must first choose a suitable lifting of the boundary data or use a quotient-form description. The Dirichlet-to-Neumann map is nevertheless the corresponding boundary reduction of the energy.

So in the positive setting, the three descriptions are the same construction viewed from different directions:

The main qualification is that one should not say that every Dirichlet-to-Neumann operator is literally a finite block-matrix Schur complement. In PDEs, it is generally an infinite-dimensional boundary operator. But it is the same elimination mechanism, and under discretization or an appropriate operator decomposition, it becomes an actual Schur complement.