LINEAR-ALGEBRA · Unit 12 · Video 3 · Interactive Practice

Three Laws, One Equation: The Hidden Symmetry Governing Every Network

IKey Formulas

Step	Formula	Name	Description
1. Geometry	$e = Ax$	Potential differences	$A$ = incidence matrix; each row has $-1$ (start) and $+1$ (end)
2. Physics	$y = Ce$	Ohm's law	$C$ = diagonal matrix of conductances; $y_k = c_k \, e_k$
3. Conservation	$A^\top y = f$	Kirchhoff's current law	Net current at each node = external source
Combined	$A^\top C A \, x = f$	Equilibrium equation	Substitute steps 1 & 2 into step 3

Symmetry property: $(A^\top C A)^\top = A^\top C^\top (A^\top)^\top = A^\top C A$ , since $C$ is diagonal (so $C^\top = C$ ) and $(A^\top)^\top = A$ .

IIVisualization 1 — Building the Incidence Matrix

Each directed edge becomes one row of $A$ : $-1$ at its start node, $+1$ at its end node.

💡 $A$ is pure topology — the physics (conductances) enters only later, through $C$ .

IIIVisualization 2 — The Three-Step Pipeline

Node potentials flow through $e = Ax \to y = Ce \to A^\top y = f$ , and the node forces always sum to zero.

Conductance c 1.0

💡 Challenge: set all four potentials equal — why does every drop $e$ , current $y$ , and force $f$ then vanish?

IVVisualization 3 — The Symmetry of $A^\top C A$

Whatever conductances you choose, $K = A^\top C A$ equals its own transpose.

c₁ 1.0 c₂ 2.0 c₃ 1.5 c₄ 1.0 c₅ 3.0

💡 Symmetry is structural: $(A^\top C A)^\top = A^\top C^\top (A^\top)^\top = A^\top C A$ , because $C^\top = C$ for any diagonal $C$ . That is exactly what guarantees real eigenvalues and a Cholesky factorization.

VQuiz Questions

Question 1

A directed edge runs from node 3 to node 1 in a network with 4 nodes. What does the corresponding row of the incidence matrix $A$ look like?

✅ Correct! Start node 3 gets $-1$ , end node 1 gets $+1$ .

❌ Not quite. Remember: $-1$ at the start node, $+1$ at the end node. The edge goes from 3 to 1.

Show solution

Solution:

The rule is: place $-1$ at the start node column and $+1$ at the end node column.

Start node = node 3 → column 3 gets $-1$
End node = node 1 → column 1 gets $+1$

$\text{Row} = [\underbrace{+1}_{\text{n}_1}, \; \underbrace{0}_{\text{n}_2}, \; \underbrace{-1}_{\text{n}_3}, \; \underbrace{0}_{\text{n}_4}]$

Question 2

In the equilibrium equation $A^\top C A \, x = f$ , what does the diagonal matrix $C$ encode?

✅ Correct! $C$ holds the material property (conductance, stiffness, etc.) for each edge, and it's diagonal because edges are independent.

❌ Not quite. Think about which of the three steps — geometry, physics, or conservation — introduces material properties.

Show solution

Solution:

In $A^\top C A \, x = f$ , each symbol plays a distinct role:

Symbol	Encodes
$A$	Connectivity — which nodes each edge connects (geometry)
$C$	Material properties — conductance, stiffness, or pipe conductance on each edge (physics)
$f$	External forcing — current, force, or flow injected at each node
$x$	Unknowns — potentials, displacements, or pressures at each node

$C$ is diagonal because each edge's material property is independent of the others. It enters in Step 2 (physics): $y = Ce$ .

Question 3

Consider the 4-node network from the video. Edge 3 runs from node 1 to node 3, so its row in $A$ is $[-1, \; 0, \; +1, \; 0]$ .

If the node potentials are $x_1 = 4$ , $x_2 = 2$ , $x_3 = -1$ , $x_4 = 0$ , what is the voltage drop $e_3$ across edge 3?

✅ Correct! $e_3 = x_3 - x_1 = -1 - 4 = -5$ . The potential drops along this edge.

❌ Not quite. Use the formula: $e_3 = (+1) \cdot x_3 + (-1) \cdot x_1 = x_3 - x_1$ . Be careful with the sign!

Show solution

Solution:

Edge 3 goes from node 1 to node 3. Its row in $A$ is $[-1, \; 0, \; +1, \; 0]$ .

The voltage drop is the dot product of this row with $x$ :

$e_3 = (-1)(x_1) + (0)(x_2) + (+1)(x_3) + (0)(x_4) = x_3 - x_1$

$e_3 = (-1) - 4 = -5$

The formula is always end node minus start node: $e_k = x_{\text{end}} - x_{\text{start}}$ .

The negative sign means potential decreases from node 1 to node 3.

Question 4

True or False: The matrix $A^\top C A$ is symmetric only when all edge conductances are equal (i.e., only when $C = cI$ for some scalar $c$ ).

✅ Correct! Symmetry holds for any diagonal $C$ , not just scalar multiples of $I$ . The proof only requires that $C^\top = C$ , which is true for every diagonal matrix.

❌ Not quite. Look at the symmetry proof — which property of $C$ does it actually use? Is that property limited to $C = cI$ ?

Show solution

Solution: False.

$A^\top C A$ is symmetric for any diagonal matrix $C$ with positive entries — the conductances do not need to be equal.

Proof:

$(A^\top C A)^\top = A^\top \cdot C^\top \cdot (A^\top)^\top$

Now simplify each piece:

$(A^\top)^\top = A$ — transposing twice returns the original
$C^\top = C$ — any diagonal matrix is symmetric (not just $cI$ )

Substituting back:

$= A^\top \cdot C \cdot A = A^\top C A \quad\checkmark$

The symmetry is a structural property guaranteed by the algebraic form $A^\top C A$ , not by any special relationship between the values in $C$ .

You can verify this in Visualization 3 above — try any combination of conductances!

Solved: 0 / 4

Three Laws, One Equation: The Hidden Symmetry Governing Every Network

IKey Formulas

IIVisualization 1 — Building the Incidence Matrix

IIIVisualization 2 — The Three-Step Pipeline

IVVisualization 3 — The Symmetry of A⊤CAA^\top C AA⊤CA

VQuiz Questions

IVVisualization 3 — The Symmetry of $A^\top C A$