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Euler's formula states that if a finite, connected, planar graph is drawn in the plane without any edge intersections, and v is the number of vertices, e is the number of edges and f is the number of faces (regions bounded by edges, including the outer, infinitely large region), then

$$ v − e + f = 2 $$

-Wikipedia, Planar graph.

There are many proofs exist to the above result. One e-page that surmise many of them is well known, and could be found here - http://www.ics.uci.edu/~eppstein/junkyard/euler/all.html.

Yesterday I have been watching lecture 12 in linear algebra by Gilbert Strang, MIT. He introduced some application of the four fundamental subspaces, and one of them were Euler's formula, but I couldn't to follow his explanation this time.

Can someone sketch me a proof of Euler's result in linear algebra terms?

Thank you.

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    When in the video does he mention it? – 2017-01-25
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    He mention it in 39:20. – 2017-01-25
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    For those not wanting to watch: he derives Euler's formula by interpreting the kernel of the transpose of a graph's oriented incidence matrix as counting faces. – 2017-01-25

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Let $G$ be a connected and planar graph, where each edge is assigned an (arbitrary) orientation. Let $m = E$ denote the number of edges, and let $n = V$ denote the number of columns. The incidence matrix of $G$ is the $m \times n$ matrix $A$ for which $$ A_{ij} = \begin{cases} -1 & \text{edge $i$ leaves node $j$}\\ 1 & \text{edge $i$ enters node $j$}\\ 0 & \text{otherwise} \end{cases} $$ Note: The incidence matrix is usually defined to be the transpose of this matrix (as it is defined here for example), but we'll stick to Strang's convention.

With that, Strang tries to convince us of the following:

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

Claim: $\dim N (A^T)$ is the number of (internal) "faces" of the graph $G$. In the usual notation, $\dim N (A^T) = F-1$.

To prove the first claim, we see that the pivot columns upon row-reducing $A^T$ produce a spanning tree of $G$. Alternatively, it suffices to note that since $G$ is connected, $N(A)$ must be the space spanned by $(1,\dots,1)$.

The second result is not strictly "proven", but it is convincingly argued with Strang's example.

With that, the dimension formula yields $$ N(A^T) = m - r(A^T) \implies\\ F - 1 = m - (n - 1) \implies\\ F - 1 = E - (V - 1) \implies\\ V - E + F = 2 $$