### Eigenvalues and expansion

Before trying to describe how one reasons about the eigenvalues of the adjacency matrix of a Cayley graph, I would like to describe the "easy direction" of the relationships between eigenvalues and expansion.

First, a one-minute summary of linear algebra. If $A$ is an $n\times n$ matrix, then a (column) non-zero vector $x$ is an

If $A$ is a symmetric real-valued matrix, then a series of miracles (or obvious things, depending on whether you have studied this material or not) happen:

From this fact, it is easy to conclude that starting from symmetric real matrix $A$ we can find a sequence of $n$ numbers

$\lambda_1\geq \ldots \geq \lambda_n$

and a sequence of $n$ unit vectors $v_1,\ldots,v_n$ such that the $v_i$ are orthogonal to each other and such that $A v_i = \lambda_i v_i$.

Now the claim is that these $n$ numbers and $n$ vectors completely describe $A$. Indeed, suppose that we are given a vector $x$ and that we want to compute $Ax$. First, we can write $x$ as a linear combination of the $v_i$:

$x= \alpha_1 v_1 + \cdots \alpha_n v_n$

and since the $v_i$ are an orthonormal basis, we know what the coefficients $\alpha_i$ are like: each $\alpha_i$ is just the inner product of $x$ and $v_i$.

Now, by linearity, we have

$Ax = \alpha_1 A v_1 + \cdots \alpha_n A v_n$

and, since the $v_i$ are eigenvectors,

$= \alpha_1 \lambda_1 v_1 + \cdots + \alpha_n \lambda_n v_n$

So, once we express vectors in the basis defined by the $v_i$, we see that multiplication by $A$ is just the effect of multiplying coordinate $i$ by the value $\lambda_i$.

But there is more. Recall that $\alpha_i = v_i^T x$ and so we can also write

$Ax = (\lambda_1 v_1^T v_1) x + \cdots + (\lambda_n v_n^T v_n) x$

and so, in particular, matrix $A$ is just

$A = \lambda_1 v_1^T v_1 + \cdots + \lambda_n v_n^T v_n $

a sum of $n$ rank-1 matrices, with the coefficients of the sum being the eigenvalues.

Let now $A$ be the adjacency matrix of a $d$-regular undirected graph (so $A$ is real-valued and symmetric), $\lambda_1\geq \cdots \geq \lambda_n$ be its eigenvalues, and $v_1,\ldots,v_n$ be a corresponding set of orthonormal eigenvectors.

It is not hard to see (no, really, it's two lines, see the bottom of page 2 of these notes) that all eigenvalues are between $d$ and $-d$. Also, $d$ is always an eigenvalue, as shown by the eigenvector $(1,\ldots,1)$. This means that $\lambda_1 = d$ and we can take

$v_1 = \left( \frac 1 {\sqrt n},\cdots, \frac 1 {\sqrt n} \right)$

and we have

$A = (d v_1^T v_1) x + \cdots + (\lambda_n v_n^T v_n) x$

Suppose now that all the other $\lambda_i$, $i=2,\ldots,n$ are much smaller than $d$ in absolute value. Then, intuitively, the sum is going to be dominated by the first term, and $A$ behaves sort of like the matrix $(d v_i^T v_i)$, that is, the matrix that has the value $d/n$ in each entry.

You may be skeptical that such an approximation makes sense, but see where this leads. If our graph is $G=(V,E)$, and $(S,V-S)$ is a cut of the set of vertices into two subsets, let $1_S$ be the $n$-dimensional vector that is 1 in coordinates corresponding to $S$ and 0 in the other coordinates, and define $1_{V-S}$ similarly. Then the number of edges crossing the cut is precisely

$1_S^T A 1_{V-S}$

and if we replace $A$ by the matrix that has $d/n$ in each coordinate we get the approximation $|S|\cdot|V-S|\cdot d /n$, which is the expected number of edges that would cross the cut if we had a random graph of degree $d$. This is pretty much the right result: if $|\lambda_2|,\ldots, |\lambda_n|$ are all much smaller than $d$, and if $|S|$ and $|V-S|$ are at least a constant times $n$, then the number of edges crossing the cut is indeed very close to $|S|\cdot |V-S| \cdot d/n$. This result is known as the

It is perhaps less intuitive that, provided that all other eigenvalues are at least

To summarize: the adjacency matrix of an undirected graph is symmetric, and so all eigenvalues are real. If the graph is $d$-regular, then $d$ is the largest eigenvalue, and all others are between $d$ and $-d$. If all the others are between $d-\epsilon$ and $-d$, then the graph has edge-expansion $\epsilon/2$ (or better).

First, a one-minute summary of linear algebra. If $A$ is an $n\times n$ matrix, then a (column) non-zero vector $x$ is an

*eigenvector*of $A$ provided that, for some scalar $\lambda$, we have $Ax= \lambda x$. If so, then $\lambda$ is an*eigenvalue*of $A$. A couple of immediate observations:- The equation $Ax=\lambda x$ can be rewritten as $(A-\lambda I)x = 0$, so if $\lambda$ is an eigenvalue of $A$ then the columns of the matrix $A-\lambda I$ are not linearly independent, and so the determinant of $A-\lambda I$ is zero. But we can write $det(A-\lambda I)$ as a degree-$n$ polynomial in the unknown $\lambda$, and a degree-$n$ polynomial cannot have more than $n$ roots.

So, an $n\times n$ matrix has at most $n$ eigenvalues. - If $\lambda$ is an eigenvalue of $A$, then the set of vectors $x$ such that $Ax=\lambda x$ is a linear space. If $\lambda_1$ and $\lambda_2$ are different eigenvalues, then their respective spaces of eigenvectors are independent (they only intersect at $0$).

If $A$ is a symmetric real-valued matrix, then a series of miracles (or obvious things, depending on whether you have studied this material or not) happen:

- All the roots of the polynomial $det(A-\lambda I)$ are real; so, counting multiplicities, $A$ has $n$ real eigenvalues.
- If $\lambda$ is an eigenvalue of $A$, and it is a root of multiplicity $k$ of $det(A-\lambda I)$, then the space of vectors $x$ such that $Ax=\lambda x$ has dimension $k$.
- If $\lambda_1 \neq \lambda_2$ are two different eigenvalues, then their respective spaces of eigenvectors are orthogonal

From this fact, it is easy to conclude that starting from symmetric real matrix $A$ we can find a sequence of $n$ numbers

$\lambda_1\geq \ldots \geq \lambda_n$

and a sequence of $n$ unit vectors $v_1,\ldots,v_n$ such that the $v_i$ are orthogonal to each other and such that $A v_i = \lambda_i v_i$.

Now the claim is that these $n$ numbers and $n$ vectors completely describe $A$. Indeed, suppose that we are given a vector $x$ and that we want to compute $Ax$. First, we can write $x$ as a linear combination of the $v_i$:

$x= \alpha_1 v_1 + \cdots \alpha_n v_n$

and since the $v_i$ are an orthonormal basis, we know what the coefficients $\alpha_i$ are like: each $\alpha_i$ is just the inner product of $x$ and $v_i$.

Now, by linearity, we have

$Ax = \alpha_1 A v_1 + \cdots \alpha_n A v_n$

and, since the $v_i$ are eigenvectors,

$= \alpha_1 \lambda_1 v_1 + \cdots + \alpha_n \lambda_n v_n$

So, once we express vectors in the basis defined by the $v_i$, we see that multiplication by $A$ is just the effect of multiplying coordinate $i$ by the value $\lambda_i$.

But there is more. Recall that $\alpha_i = v_i^T x$ and so we can also write

$Ax = (\lambda_1 v_1^T v_1) x + \cdots + (\lambda_n v_n^T v_n) x$

and so, in particular, matrix $A$ is just

$A = \lambda_1 v_1^T v_1 + \cdots + \lambda_n v_n^T v_n $

a sum of $n$ rank-1 matrices, with the coefficients of the sum being the eigenvalues.

Let now $A$ be the adjacency matrix of a $d$-regular undirected graph (so $A$ is real-valued and symmetric), $\lambda_1\geq \cdots \geq \lambda_n$ be its eigenvalues, and $v_1,\ldots,v_n$ be a corresponding set of orthonormal eigenvectors.

It is not hard to see (no, really, it's two lines, see the bottom of page 2 of these notes) that all eigenvalues are between $d$ and $-d$. Also, $d$ is always an eigenvalue, as shown by the eigenvector $(1,\ldots,1)$. This means that $\lambda_1 = d$ and we can take

$v_1 = \left( \frac 1 {\sqrt n},\cdots, \frac 1 {\sqrt n} \right)$

and we have

$A = (d v_1^T v_1) x + \cdots + (\lambda_n v_n^T v_n) x$

Suppose now that all the other $\lambda_i$, $i=2,\ldots,n$ are much smaller than $d$ in absolute value. Then, intuitively, the sum is going to be dominated by the first term, and $A$ behaves sort of like the matrix $(d v_i^T v_i)$, that is, the matrix that has the value $d/n$ in each entry.

You may be skeptical that such an approximation makes sense, but see where this leads. If our graph is $G=(V,E)$, and $(S,V-S)$ is a cut of the set of vertices into two subsets, let $1_S$ be the $n$-dimensional vector that is 1 in coordinates corresponding to $S$ and 0 in the other coordinates, and define $1_{V-S}$ similarly. Then the number of edges crossing the cut is precisely

$1_S^T A 1_{V-S}$

and if we replace $A$ by the matrix that has $d/n$ in each coordinate we get the approximation $|S|\cdot|V-S|\cdot d /n$, which is the expected number of edges that would cross the cut if we had a random graph of degree $d$. This is pretty much the right result: if $|\lambda_2|,\ldots, |\lambda_n|$ are all much smaller than $d$, and if $|S|$ and $|V-S|$ are at least a constant times $n$, then the number of edges crossing the cut is indeed very close to $|S|\cdot |V-S| \cdot d/n$. This result is known as the

*Expander Mixing Lemma*, and the proof is just a matter of doing the "obvious" calculations. (Write $1_S$ and $1_{V-S}$ in the eigenvector basis, distribute the product, remember that the $v_i$ are orthogonal, use Cauchy-Schwarz ...)It is perhaps less intuitive that, provided that all other eigenvalues are at least

*a little bit*smaller than $d$, then a reasonably large number of edges crosses the cut. In particular, if $|S| \leq n/2$ and if $\lambda_2 \leq d-\epsilon$, then at least $\epsilon |S|/2$ edges cross the cut. That is, if all eigenvalues except the largest one are at most $d-\epsilon$ then the edge expansion of the graph is at least $\epsilon/2$. (It takes about five lines to prove this, see the beginning of the proof of Thm 1 in these notes.)To summarize: the adjacency matrix of an undirected graph is symmetric, and so all eigenvalues are real. If the graph is $d$-regular, then $d$ is the largest eigenvalue, and all others are between $d$ and $-d$. If all the others are between $d-\epsilon$ and $-d$, then the graph has edge-expansion $\epsilon/2$ (or better).

## 3 Comments:

12/13/2006 10:50:00 AM

One minor correction,

Luca:"..., and $A$ behaves sort of like the matrix $(d v_i^T v_i)$, that is, the matrix ..."

$v_i$ should be $v_1$.

Great job, Luca. We are awaiting the rest of this series of posts.

12/13/2006 11:44:00 AM

I greatly enjoy your technical posts - they're both intuitive and informative. Looking forward to more of them...

3/12/2007 08:32:00 PM

Wonderful post! Thanks a ton.

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