LINEAR-ALGEBRA · Unit 12 · Video 2 · Interactive Practice

One Matrix Unlocks Kirchhoff's Law, Loop Currents, and Euler's Formula

IKey Formulas

Formula	Name	Description
$A_{ij} = -1$ (start), $+1$ (end), $0$ (else)	Incidence matrix rule	Each row has one $-1$ and one $+1$
$A^\top y = 0$	Kirchhoff's Current Law	Currents balance at every node
$\text{rank}(A) = n - 1$	Rank (connected graph)	One less than number of nodes
$n - m + \ell = 1$	Euler's formula	Nodes $-$ edges $+$ independent loops $= 1$

IIVisualization 1 — Graph ↔ Incidence Matrix

Each edge becomes one matrix row: $-1$ at its start node, $+1$ at its end, $0$ elsewhere.

Edge

💡 Adding the rows of $e_1, e_2, e_3$ gives zero — the closed loop $1 \to 2 \to 3 \to 1$ makes those incidence rows linearly dependent.

IIIVisualization 2 — Loop Currents Satisfy KCL

Any combination $y = a\,\text{loop}_1 + b\,\text{loop}_2$ balances the current at every node: $A^\top y = 0$ .

Loop 1 (a) 1.0 Loop 2 (b) 0.0

💡 The basis currents $(1,1,{-1},0,0)$ and $(0,0,1,{-1},1)$ span the null space of $A^\top$ , whose dimension equals the number of independent loops, $\ell = 2$ .

IVVisualization 3 — Why $n - m + \ell = 1$

Split the edges into a spanning tree ( $n-1$ ) plus $\ell$ loop edges, and the alternating count always lands on $1$ .

Nodes (n) 4 Loops (ℓ) 2

VQuiz Questions

Question 1

In the video's graph, edge 3 goes from node 1 to node 3. What is the entry in row 3 (edge 3), column 2 (node 2) of the incidence matrix $A$ ?

✅ Correct! Node 2 is not an endpoint of edge 3, so the entry is 0.

❌ Not quite. Only the start node gets −1 and the end node gets +1. All other entries are 0.

Show solution

Solution:

Edge 3 goes from node 1 to node 3. By the incidence matrix rule:

Column 1 (start node): $-1$
Column 3 (end node): $+1$
All other columns: $0$

Row 3 of $A$ : $[-1, \; 0, \; 1, \; 0]$

Node 2 is not an endpoint of edge 3, so the entry at (row 3, column 2) is 0.

Question 2

True or False: For a connected graph with $n$ nodes, the rank of its incidence matrix $A$ equals $n$ (the number of nodes).

✅ Correct! The rank is n − 1 because the columns of A always sum to zero — there is always one dependency.

❌ Not quite. Consider: what happens when you add all columns of A together?

Show solution

Solution:

False. The rank is $n - 1$ , not $n$ .

Every row of $A$ has entries that sum to zero (one $-1$ and one $+1$ ). This means the columns of $A$ always sum to the zero vector — a linear dependence. So the column space has dimension at most $n - 1$ .

For a connected graph, the rank is exactly $n - 1$ . The null space $N(A)$ has dimension 1, spanned by the constant vector $(1, 1, \ldots, 1)$ — this represents the "ground" reference potential.

Question 3

A connected graph has 6 nodes and 10 edges. How many independent loops does it have?

✅ Correct! ℓ = m − n + 1 = 10 − 6 + 1 = 5 independent loops.

❌ Not quite. Apply the formula: ℓ = m − n + 1.

Show solution

Solution:

For a connected graph with $n$ nodes and $m$ edges, the number of independent loops is:

$\ell = m - n + 1$

With $n = 6$ and $m = 10$ :

$\ell = 10 - 6 + 1 = 5$

Why? A spanning tree uses $n - 1 = 5$ edges. The remaining $10 - 5 = 5$ edges each create exactly one independent loop when added back to the tree. This count equals $\dim N(A^\top) = m - r = m - (n-1)$ .

Question 4

In the video's incidence matrix, Row 1 is $[-1, 1, 0, 0]$ (edge 1→2) and Row 2 is $[0, -1, 1, 0]$ (edge 2→3). Their sum equals Row 3: $[-1, 0, 1, 0]$ (edge 1→3). What does this linear dependence represent in the graph?

✅ Correct! The row dependence Row 1 + Row 2 = Row 3 directly corresponds to the closed triangle loop 1→2→3→1 in the graph.

❌ Not quite. Think about what it means when traversing edges 1 then 2 arrives at the same place as edge 3 alone.

Show solution

Solution:

Row 1 + Row 2 = Row 3 means the trip along edge 1 ( $1 \to 2$ ) followed by edge 2 ( $2 \to 3$ ) reaches the same destination as edge 3 ( $1 \to 3$ ). Reversing edge 3, we get a closed loop: $1 \to 2 \to 3 \to 1$ .

Every closed loop in a graph creates a linear dependence among the corresponding rows of the incidence matrix. This is the fundamental connection between graph topology and linear algebra.

In general, the number of independent row dependencies equals $m - (n-1)$ — exactly the number of independent loops.

Solved: 0 / 4

One Matrix Unlocks Kirchhoff's Law, Loop Currents, and Euler's Formula

IKey Formulas

IIVisualization 1 — Graph ↔ Incidence Matrix

IIIVisualization 2 — Loop Currents Satisfy KCL

IVVisualization 3 — Why n−m+ℓ=1n - m + \ell = 1n−m+ℓ=1

VQuiz Questions

IVVisualization 3 — Why $n - m + \ell = 1$