LINEAR-ALGEBRA · Unit 12 · Video 1 · Interactive Practice

From Nodes to Null Spaces: Why Only Differences Ever Matter

IKey Formulas

Formula	Name	Description
Row of $A$ : $-1$ at "from" node, $+1$ at "to" node, $0$ elsewhere	Incidence Matrix	One row per edge, one column per node
$(Ax)_k = x_{\text{to}} - x_{\text{from}}$ for edge $k$	Discrete Differentiation	$A$ computes differences across edges
$\ker(A) = \operatorname{span}\{(1,1,\ldots,1)\}$	Null Space	Uniform shift produces zero differences
$\operatorname{rank}(A) = n - 1$	Rank	For a connected graph with $n$ nodes

IIVisualization 1 — Building the Incidence Matrix

Each directed edge becomes one row of $A$ : $-1$ at its tail node, $+1$ at its head, zeros elsewhere.

IIIVisualization 2 — Reading the Jump Across Each Edge

Multiplying by $A$ returns the potential difference $x_{\text{to}} - x_{\text{from}}$ along every edge.

IVVisualization 3 — The Null Space

Adding the same constant to all four nodes leaves every edge difference untouched — the shift lives in $\ker(A)$ .

Uniform shift c 0.0

💡 This is why Celsius and Kelvin always agree on temperature differences: the constant $273$ offset between the scales is a null-space shift, invisible to every subtraction.

VVisualization 4 — Grounding the Reference

Grounding one node to zero removes the ambiguity: the null space collapses from a line to $\{0\}$ .

VIQuiz Questions

Question 1

Edge $e_4$ goes from node 1 to node 4. What is the corresponding row of the incidence matrix $A$ ?

✅ Correct! Place $-1$ at the 'from' node (column 1) and $+1$ at the 'to' node (column 4).

❌ Not quite. Remember: $-1$ goes at the 'from' column, $+1$ at the 'to' column, zeros elsewhere.

Show solution

Solution:

Edge $e_4$ goes from node 1 to node 4:

Place $-1$ in column 1 (the "from" node)
Place $+1$ in column 4 (the "to" node)
Columns 2 and 3 get zeros

$\text{Row } e_4 = [\,-1, \; 0, \; 0, \; +1\,]$

Question 2

Given node potentials $x = (3, \; 7, \; 5, \; 1)$ , what is $(Ax)_3$ — the third entry of the product $Ax$ ?

(Recall: edge $e_3$ goes from node 1 to node 3.)

✅ Correct! The difference across edge $e_3$ is $x_3 - x_1 = 5 - 3 = 2$ .

❌ Not quite. For edge $e_3$ (node 1 → node 3), compute $x_{\text{to}} - x_{\text{from}} = x_3 - x_1$ .

Show solution

Solution:

Row 3 of $A$ is $[-1, \; 0, \; 1, \; 0]$ (edge $e_3$ : node 1 → node 3).

$(Ax)_3 = (-1)(3) + (0)(7) + (1)(5) + (0)(1) = -3 + 5 = 2$

Equivalently: $x_{\text{to}} - x_{\text{from}} = x_3 - x_1 = 5 - 3 = 2$ .

Common mistakes:

Getting $-2$ : reversed the subtraction order ( $x_1 - x_3$ instead of $x_3 - x_1$ )
Getting $4$ : used the wrong edge ( $e_1$ : $x_2 - x_1 = 7 - 3 = 4$ )
Getting $8$ : added instead of subtracting ( $x_1 + x_3 = 3 + 5 = 8$ )

Question 3

True or False: For any connected directed graph with $n$ nodes, the null space of its incidence matrix has dimension exactly 1.

✅ Correct! The rank is $n-1$ , leaving exactly one null space dimension — spanned by the all-ones vector.

❌ Not quite. Since $\operatorname{rank}(A) = n - 1$ for a connected graph, the null space dimension is $n - (n-1) = 1$ .

Show solution

Answer: True

For a connected graph with $n$ nodes:

$\operatorname{rank}(A) = n - 1$

$\dim(\ker(A)) = n - \operatorname{rank}(A) = n - (n - 1) = 1$

The null space is spanned by $(1, 1, \ldots, 1)$ : adding the same constant to every node produces zero differences on all edges. This single degree of freedom is why we need to choose a reference (ground) to make potentials unique.

Note: This holds for connected graphs. A graph with $k$ connected components would have null space dimension $k$ .

Question 4

You add $c = 100$ to every node potential, replacing $x$ with $x + 100 \cdot (1,1,1,1)^T$ . What happens to the vector of edge differences $Ax$ ?

✅ Correct! The all-ones vector is in the null space, so $A$ annihilates any uniform shift — differences are completely unchanged.

❌ Not quite. Recall that $(1,1,1,1)^T$ is in the null space: $A \cdot (1,1,1,1)^T = \mathbf{0}$ . What does that imply for $A(x + c \cdot \mathbf{1})$ ?

Show solution

Answer: Every difference stays exactly the same.

$A(x + 100 \cdot \mathbf{1}) = Ax + 100 \cdot A\mathbf{1} = Ax + 100 \cdot \mathbf{0} = Ax$

Since $(1,1,1,1)^T$ is in the null space of $A$ , the matrix maps it to zero. The uniform shift is completely invisible to $A$ .

This is the same reason Celsius and Kelvin always agree on temperature differences — the 273-degree offset cancels in every subtraction:

$373\text{K} - 273\text{K} = 100 \quad = \quad 100^{\circ}\text{C} - 0^{\circ}\text{C}$

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