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Expander Graphs and Their Applications (II)
Yijia ChenShanghai Jiaotong University
Last Lecture
Graph Reachability Problem on Good Graphs
For our purpose, let c, d ∈ N, a graph G = (V , E) is (c, d)-good if each v ∈ Vhas degree at most d and for all u, v ∈ V , if there is a path from u to v , then
dist(u, v) ≤ c · log |V |.
TheoremLet c, d ∈ N. For (c, d)-good graphs G = (V , E), the reachability problem canbe solved in space
O(c · log |V | · log d) = O(log |V |),
i.e., in LogSpace.
Graph Reachability Problem on Good Graphs
For our purpose, let c, d ∈ N, a graph G = (V , E) is (c, d)-good if each v ∈ Vhas degree at most d and for all u, v ∈ V , if there is a path from u to v , then
dist(u, v) ≤ c · log |V |.
TheoremLet c, d ∈ N. For (c, d)-good graphs G = (V , E), the reachability problem canbe solved in space
O(c · log |V | · log d) = O(log |V |),
i.e., in LogSpace.
Graph Reachability Problem on Good Graphs
For our purpose, let c, d ∈ N, a graph G = (V , E) is (c, d)-good if each v ∈ Vhas degree at most d and for all u, v ∈ V , if there is a path from u to v , then
dist(u, v) ≤ c · log |V |.
TheoremLet c, d ∈ N. For (c, d)-good graphs G = (V , E), the reachability problem canbe solved in space
O(c · log |V | · log d) = O(log |V |),
i.e., in LogSpace.
Graph Expansion
For every S , T ⊆ V we let
E(S , T ) :={~e | ~e is a direction of some e ∈ E with tail in S and head in T
}.
Let S ⊆ V . Then the edge boundary of S is
∂S := E(S , S).
The (edge) expansion ratio of G is
h(G) := min{S||S|≤|V |/2}
|∂S ||S | .
DefinitionLet d ∈ N. A sequence of d-regular graphs {Gi}i∈N of size increasing with i isa family of expander graphs if there exists ε > 0 such that h(Gi ) ≥ ε for all i .
Graph Expansion
For every S , T ⊆ V we let
E(S , T ) :={~e | ~e is a direction of some e ∈ E with tail in S and head in T
}.
Let S ⊆ V . Then the edge boundary of S is
∂S := E(S , S).
The (edge) expansion ratio of G is
h(G) := min{S||S|≤|V |/2}
|∂S ||S | .
DefinitionLet d ∈ N. A sequence of d-regular graphs {Gi}i∈N of size increasing with i isa family of expander graphs if there exists ε > 0 such that h(Gi ) ≥ ε for all i .
Graph Expansion
For every S , T ⊆ V we let
E(S , T ) :={~e | ~e is a direction of some e ∈ E with tail in S and head in T
}.
Let S ⊆ V . Then the edge boundary of S is
∂S := E(S , S).
The (edge) expansion ratio of G is
h(G) := min{S||S|≤|V |/2}
|∂S ||S | .
DefinitionLet d ∈ N. A sequence of d-regular graphs {Gi}i∈N of size increasing with i isa family of expander graphs if there exists ε > 0 such that h(Gi ) ≥ ε for all i .
Graph Expansion
For every S , T ⊆ V we let
E(S , T ) :={~e | ~e is a direction of some e ∈ E with tail in S and head in T
}.
Let S ⊆ V . Then the edge boundary of S is
∂S := E(S , S).
The (edge) expansion ratio of G is
h(G) := min{S||S|≤|V |/2}
|∂S ||S | .
DefinitionLet d ∈ N. A sequence of d-regular graphs {Gi}i∈N of size increasing with i isa family of expander graphs if there exists ε > 0 such that h(Gi ) ≥ ε for all i .
Graph Expansion
For every S , T ⊆ V we let
E(S , T ) :={~e | ~e is a direction of some e ∈ E with tail in S and head in T
}.
Let S ⊆ V . Then the edge boundary of S is
∂S := E(S , S).
The (edge) expansion ratio of G is
h(G) := min{S||S|≤|V |/2}
|∂S ||S | .
DefinitionLet d ∈ N. A sequence of d-regular graphs {Gi}i∈N of size increasing with i isa family of expander graphs if there exists ε > 0 such that h(Gi ) ≥ ε for all i .
Graphs as Matrices
The adjacency matrix of an n-vertex graph G , denoted A = A(G), is an n × nmatrix (ai,j)i,j∈[n] where
ai,j = the number of edges in G between the i-th and the j-th vertices.
Note, A is a real and symmetric matrix.
TheoremIf A is real and symmetric, then
I A has exactly n (not necessarily distinct) eigenvalues, i.e., λ1 ≥ λ2 ≥ . . .≥ λn.
I There exists a set of n eigenvectors v1, v2, . . . , vn ∈ Rn, one for eacheigenvalue (i.e., Avi = λivi ), that are orthonormal, i.e., all vi s are of length1 and orthogonal with each other.
Note, v1, . . . , vn form a basis for Rn.
Graphs as Matrices
The adjacency matrix of an n-vertex graph G , denoted A = A(G), is an n × nmatrix (ai,j)i,j∈[n] where
ai,j = the number of edges in G between the i-th and the j-th vertices.
Note, A is a real and symmetric matrix.
TheoremIf A is real and symmetric, then
I A has exactly n (not necessarily distinct) eigenvalues, i.e., λ1 ≥ λ2 ≥ . . .≥ λn.
I There exists a set of n eigenvectors v1, v2, . . . , vn ∈ Rn, one for eacheigenvalue (i.e., Avi = λivi ), that are orthonormal, i.e., all vi s are of length1 and orthogonal with each other.
Note, v1, . . . , vn form a basis for Rn.
Graphs as Matrices
The adjacency matrix of an n-vertex graph G , denoted A = A(G), is an n × nmatrix (ai,j)i,j∈[n] where
ai,j = the number of edges in G between the i-th and the j-th vertices.
Note, A is a real and symmetric matrix.
TheoremIf A is real and symmetric, then
I A has exactly n (not necessarily distinct) eigenvalues, i.e., λ1 ≥ λ2 ≥ . . .≥ λn.
I There exists a set of n eigenvectors v1, v2, . . . , vn ∈ Rn, one for eacheigenvalue (i.e., Avi = λivi ), that are orthonormal, i.e., all vi s are of length1 and orthogonal with each other.
Note, v1, . . . , vn form a basis for Rn.
Graphs as Matrices
The adjacency matrix of an n-vertex graph G , denoted A = A(G), is an n × nmatrix (ai,j)i,j∈[n] where
ai,j = the number of edges in G between the i-th and the j-th vertices.
Note, A is a real and symmetric matrix.
TheoremIf A is real and symmetric, then
I A has exactly n (not necessarily distinct) eigenvalues, i.e., λ1 ≥ λ2 ≥ . . .≥ λn.
I There exists a set of n eigenvectors v1, v2, . . . , vn ∈ Rn, one for eacheigenvalue (i.e., Avi = λivi ), that are orthonormal, i.e., all vi s are of length1 and orthogonal with each other.
Note, v1, . . . , vn form a basis for Rn.
Graph Spectrum
Let G be a graph and A = A(G). The spectrum of G is the eigenvalues of A,i.e., λ1 ≥ λ2 ≥ . . . ≥ λn.
Let G be a d-regular graph. Then λ1 = d and the corresponding eigenvectorcan be chosen as
v1 =1√n
=
(1√n
, . . . ,1√n
).
TheoremLet G be a d-regular graph with spectrum λ1 ≥ λ2 ≥ . . . ≥ λn. Then
d − λ2
2≤ h(G) ≤
√2d(d − λ2).
Note d − λ2 = λ1 − λ2 is the spectral gap of G.
Graph Spectrum
Let G be a graph and A = A(G).
The spectrum of G is the eigenvalues of A,i.e., λ1 ≥ λ2 ≥ . . . ≥ λn.
Let G be a d-regular graph. Then λ1 = d and the corresponding eigenvectorcan be chosen as
v1 =1√n
=
(1√n
, . . . ,1√n
).
TheoremLet G be a d-regular graph with spectrum λ1 ≥ λ2 ≥ . . . ≥ λn. Then
d − λ2
2≤ h(G) ≤
√2d(d − λ2).
Note d − λ2 = λ1 − λ2 is the spectral gap of G.
Graph Spectrum
Let G be a graph and A = A(G). The spectrum of G is the eigenvalues of A,i.e., λ1 ≥ λ2 ≥ . . . ≥ λn.
Let G be a d-regular graph. Then λ1 = d and the corresponding eigenvectorcan be chosen as
v1 =1√n
=
(1√n
, . . . ,1√n
).
TheoremLet G be a d-regular graph with spectrum λ1 ≥ λ2 ≥ . . . ≥ λn. Then
d − λ2
2≤ h(G) ≤
√2d(d − λ2).
Note d − λ2 = λ1 − λ2 is the spectral gap of G.
Graph Spectrum
Let G be a graph and A = A(G). The spectrum of G is the eigenvalues of A,i.e., λ1 ≥ λ2 ≥ . . . ≥ λn.
Let G be a d-regular graph. Then λ1 = d and the corresponding eigenvectorcan be chosen as
v1 =1√n
=
(1√n
, . . . ,1√n
).
TheoremLet G be a d-regular graph with spectrum λ1 ≥ λ2 ≥ . . . ≥ λn. Then
d − λ2
2≤ h(G) ≤
√2d(d − λ2).
Note d − λ2 = λ1 − λ2 is the spectral gap of G.
Graph Spectrum
Let G be a graph and A = A(G). The spectrum of G is the eigenvalues of A,i.e., λ1 ≥ λ2 ≥ . . . ≥ λn.
Let G be a d-regular graph. Then λ1 = d and the corresponding eigenvectorcan be chosen as
v1 =1√n
=
(1√n
, . . . ,1√n
).
TheoremLet G be a d-regular graph with spectrum λ1 ≥ λ2 ≥ . . . ≥ λn. Then
d − λ2
2≤ h(G) ≤
√2d(d − λ2).
Note d − λ2 = λ1 − λ2 is the spectral gap of G.
Expander Mixing Lemma
We let λ := λ(G) := max{|λ2|, |λn|}.
LemmaLet G = (V , E) be a d-regular graph with n-vertices. Then for all S , T ⊆ V ,∣∣∣∣∣∣E(S , T )
∣∣− d |S ||T |n
∣∣∣∣ ≤ λ√|S ||T |.
Expander Mixing Lemma
We let λ := λ(G) := max{|λ2|, |λn|}.
LemmaLet G = (V , E) be a d-regular graph with n-vertices. Then for all S , T ⊆ V ,∣∣∣∣∣∣E(S , T )
∣∣− d |S ||T |n
∣∣∣∣ ≤ λ√|S ||T |.
Expander Mixing Lemma
We let λ := λ(G) := max{|λ2|, |λn|}.
LemmaLet G = (V , E) be a d-regular graph with n-vertices. Then for all S , T ⊆ V ,∣∣∣∣∣∣E(S , T )
∣∣− d |S ||T |n
∣∣∣∣ ≤ λ√|S ||T |.
d-Regular Graphs as Matrices
d-Regular Graphs as Matrices
Lemmaλ1 = d and the corresponding eigenvector is v1 = 1/
√n.
Proof.Let λ be any eigenvalue of A = A(G) = (ai,j)i,j∈[n] with correspondingeigenvector v = (e1, . . . , en). We choose an i ∈ [n] such that for every j ∈ [n]
|ei | ≥ |ej |.
By Av = λv , ∑k∈[n]
ai,kek = λei ,
Then,
d |ei | = |ei |∑k∈[n]
|ai,k | ≥∑k∈[n]
|ai,k ||ek | ≥
∣∣∣∣∣∣∑k∈[n]
ai,kek
∣∣∣∣∣∣ = |λ||ei |.
d-Regular Graphs as Matrices
Lemmaλ1 = d and the corresponding eigenvector is v1 = 1/
√n.
Proof.Let λ be any eigenvalue of A = A(G) = (ai,j)i,j∈[n] with correspondingeigenvector v = (e1, . . . , en). We choose an i ∈ [n] such that for every j ∈ [n]
|ei | ≥ |ej |.
By Av = λv , ∑k∈[n]
ai,kek = λei ,
Then,
d |ei | = |ei |∑k∈[n]
|ai,k | ≥∑k∈[n]
|ai,k ||ek | ≥
∣∣∣∣∣∣∑k∈[n]
ai,kek
∣∣∣∣∣∣ = |λ||ei |.
d-Regular Graphs as Matrices
Lemmaλ1 = d and the corresponding eigenvector is v1 = 1/
√n.
Proof.Let λ be any eigenvalue of A = A(G) = (ai,j)i,j∈[n] with correspondingeigenvector v = (e1, . . . , en). We choose an i ∈ [n] such that for every j ∈ [n]
|ei | ≥ |ej |.
By Av = λv , ∑k∈[n]
ai,kek = λei ,
Then,
d |ei | = |ei |∑k∈[n]
|ai,k | ≥∑k∈[n]
|ai,k ||ek | ≥
∣∣∣∣∣∣∑k∈[n]
ai,kek
∣∣∣∣∣∣ = |λ||ei |.
d-Regular Graphs as Matrices (cont’d)
LemmaThe graph is connected if and only if λ1 > λ2.
Proof.(⇐) Easy.
(⇒) Let v = (e1, . . . , en) be an eigenvector for d . We choose an i ∈ [n] suchthat for every j ∈ [n]
|ei | ≥ |ej |.
Without loss of generality, we assume ei > 0. Then,
dei = ei
∑k∈[n]
ai,k = ei
∑{i,k}∈E
ai,k ≥∑
{i,k}∈E
ai,kek =∑k∈[n]
ai,kek = dei ,
where {i , k} ∈ E means that there is at least one edge between the i-th andthe k-th vertices.
Therefore, if {i , k} ∈ E , then ek = ei . Then we repeat the above argument onek .
d-Regular Graphs as Matrices (cont’d)
LemmaThe graph is connected if and only if λ1 > λ2.
Proof.(⇐) Easy.
(⇒) Let v = (e1, . . . , en) be an eigenvector for d . We choose an i ∈ [n] suchthat for every j ∈ [n]
|ei | ≥ |ej |.
Without loss of generality, we assume ei > 0. Then,
dei = ei
∑k∈[n]
ai,k = ei
∑{i,k}∈E
ai,k ≥∑
{i,k}∈E
ai,kek =∑k∈[n]
ai,kek = dei ,
where {i , k} ∈ E means that there is at least one edge between the i-th andthe k-th vertices.
Therefore, if {i , k} ∈ E , then ek = ei . Then we repeat the above argument onek .
d-Regular Graphs as Matrices (cont’d)
LemmaThe graph is connected if and only if λ1 > λ2.
Proof.(⇐) Easy.
(⇒) Let v = (e1, . . . , en) be an eigenvector for d . We choose an i ∈ [n] suchthat for every j ∈ [n]
|ei | ≥ |ej |.
Without loss of generality, we assume ei > 0. Then,
dei = ei
∑k∈[n]
ai,k = ei
∑{i,k}∈E
ai,k ≥∑
{i,k}∈E
ai,kek =∑k∈[n]
ai,kek = dei ,
where {i , k} ∈ E means that there is at least one edge between the i-th andthe k-th vertices.
Therefore, if {i , k} ∈ E , then ek = ei . Then we repeat the above argument onek .
d-Regular Graphs as Matrices (cont’d)
LemmaThe graph is connected if and only if λ1 > λ2.
Proof.(⇐) Easy.
(⇒) Let v = (e1, . . . , en) be an eigenvector for d .
We choose an i ∈ [n] suchthat for every j ∈ [n]
|ei | ≥ |ej |.
Without loss of generality, we assume ei > 0. Then,
dei = ei
∑k∈[n]
ai,k = ei
∑{i,k}∈E
ai,k ≥∑
{i,k}∈E
ai,kek =∑k∈[n]
ai,kek = dei ,
where {i , k} ∈ E means that there is at least one edge between the i-th andthe k-th vertices.
Therefore, if {i , k} ∈ E , then ek = ei . Then we repeat the above argument onek .
d-Regular Graphs as Matrices (cont’d)
LemmaThe graph is connected if and only if λ1 > λ2.
Proof.(⇐) Easy.
(⇒) Let v = (e1, . . . , en) be an eigenvector for d . We choose an i ∈ [n] suchthat for every j ∈ [n]
|ei | ≥ |ej |.
Without loss of generality, we assume ei > 0.
Then,
dei = ei
∑k∈[n]
ai,k = ei
∑{i,k}∈E
ai,k ≥∑
{i,k}∈E
ai,kek =∑k∈[n]
ai,kek = dei ,
where {i , k} ∈ E means that there is at least one edge between the i-th andthe k-th vertices.
Therefore, if {i , k} ∈ E , then ek = ei . Then we repeat the above argument onek .
d-Regular Graphs as Matrices (cont’d)
LemmaThe graph is connected if and only if λ1 > λ2.
Proof.(⇐) Easy.
(⇒) Let v = (e1, . . . , en) be an eigenvector for d . We choose an i ∈ [n] suchthat for every j ∈ [n]
|ei | ≥ |ej |.
Without loss of generality, we assume ei > 0. Then,
dei = ei
∑k∈[n]
ai,k = ei
∑{i,k}∈E
ai,k ≥∑
{i,k}∈E
ai,kek =∑k∈[n]
ai,kek = dei ,
where {i , k} ∈ E means that there is at least one edge between the i-th andthe k-th vertices.
Therefore, if {i , k} ∈ E , then ek = ei . Then we repeat the above argument onek .
d-Regular Graphs as Matrices (cont’d)
LemmaThe graph is connected if and only if λ1 > λ2.
Proof.(⇐) Easy.
(⇒) Let v = (e1, . . . , en) be an eigenvector for d . We choose an i ∈ [n] suchthat for every j ∈ [n]
|ei | ≥ |ej |.
Without loss of generality, we assume ei > 0. Then,
dei = ei
∑k∈[n]
ai,k = ei
∑{i,k}∈E
ai,k ≥∑
{i,k}∈E
ai,kek =∑k∈[n]
ai,kek = dei ,
where {i , k} ∈ E means that there is at least one edge between the i-th andthe k-th vertices.
Therefore, if {i , k} ∈ E , then ek = ei .
Then we repeat the above argument onek .
d-Regular Graphs as Matrices (cont’d)
LemmaThe graph is connected if and only if λ1 > λ2.
Proof.(⇐) Easy.
(⇒) Let v = (e1, . . . , en) be an eigenvector for d . We choose an i ∈ [n] suchthat for every j ∈ [n]
|ei | ≥ |ej |.
Without loss of generality, we assume ei > 0. Then,
dei = ei
∑k∈[n]
ai,k = ei
∑{i,k}∈E
ai,k ≥∑
{i,k}∈E
ai,kek =∑k∈[n]
ai,kek = dei ,
where {i , k} ∈ E means that there is at least one edge between the i-th andthe k-th vertices.
Therefore, if {i , k} ∈ E , then ek = ei . Then we repeat the above argument onek .
d-Regular Graphs as Matrices (cont’d)
LemmaThe graph is bipartite if and only if λ1 = −λn.
Proof.(⇒) Easy.
(⇐) Let v = (e1, . . . , en) be an eigenvector for −d . We choose an i ∈ [n] suchthat for every j ∈ [n]
|ei | ≥ |ej |.
Then,
dei = ei
∑k∈[n]
ai,k = ei
∑{i,k}∈E
ai,k
On the other hand,
dei = −(−dei ) = −∑k∈[n]
ai,kek =∑
{i,k}∈E
ai,k(−ek).
Therefore, if {i , k} ∈ E , then ek = −ei . Then we repeat the above argumenton ek .
d-Regular Graphs as Matrices (cont’d)
LemmaThe graph is bipartite if and only if λ1 = −λn.
Proof.(⇒) Easy.
(⇐) Let v = (e1, . . . , en) be an eigenvector for −d . We choose an i ∈ [n] suchthat for every j ∈ [n]
|ei | ≥ |ej |.
Then,
dei = ei
∑k∈[n]
ai,k = ei
∑{i,k}∈E
ai,k
On the other hand,
dei = −(−dei ) = −∑k∈[n]
ai,kek =∑
{i,k}∈E
ai,k(−ek).
Therefore, if {i , k} ∈ E , then ek = −ei . Then we repeat the above argumenton ek .
d-Regular Graphs as Matrices (cont’d)
LemmaThe graph is bipartite if and only if λ1 = −λn.
Proof.(⇒) Easy.
(⇐) Let v = (e1, . . . , en) be an eigenvector for −d . We choose an i ∈ [n] suchthat for every j ∈ [n]
|ei | ≥ |ej |.
Then,
dei = ei
∑k∈[n]
ai,k = ei
∑{i,k}∈E
ai,k
On the other hand,
dei = −(−dei ) = −∑k∈[n]
ai,kek =∑
{i,k}∈E
ai,k(−ek).
Therefore, if {i , k} ∈ E , then ek = −ei . Then we repeat the above argumenton ek .
d-Regular Graphs as Matrices (cont’d)
LemmaThe graph is bipartite if and only if λ1 = −λn.
Proof.(⇒) Easy.
(⇐) Let v = (e1, . . . , en) be an eigenvector for −d .
We choose an i ∈ [n] suchthat for every j ∈ [n]
|ei | ≥ |ej |.
Then,
dei = ei
∑k∈[n]
ai,k = ei
∑{i,k}∈E
ai,k
On the other hand,
dei = −(−dei ) = −∑k∈[n]
ai,kek =∑
{i,k}∈E
ai,k(−ek).
Therefore, if {i , k} ∈ E , then ek = −ei . Then we repeat the above argumenton ek .
d-Regular Graphs as Matrices (cont’d)
LemmaThe graph is bipartite if and only if λ1 = −λn.
Proof.(⇒) Easy.
(⇐) Let v = (e1, . . . , en) be an eigenvector for −d . We choose an i ∈ [n] suchthat for every j ∈ [n]
|ei | ≥ |ej |.
Then,
dei = ei
∑k∈[n]
ai,k = ei
∑{i,k}∈E
ai,k
On the other hand,
dei = −(−dei ) = −∑k∈[n]
ai,kek =∑
{i,k}∈E
ai,k(−ek).
Therefore, if {i , k} ∈ E , then ek = −ei . Then we repeat the above argumenton ek .
d-Regular Graphs as Matrices (cont’d)
LemmaThe graph is bipartite if and only if λ1 = −λn.
Proof.(⇒) Easy.
(⇐) Let v = (e1, . . . , en) be an eigenvector for −d . We choose an i ∈ [n] suchthat for every j ∈ [n]
|ei | ≥ |ej |.
Then,
dei = ei
∑k∈[n]
ai,k = ei
∑{i,k}∈E
ai,k
On the other hand,
dei = −(−dei ) = −∑k∈[n]
ai,kek =∑
{i,k}∈E
ai,k(−ek).
Therefore, if {i , k} ∈ E , then ek = −ei . Then we repeat the above argumenton ek .
d-Regular Graphs as Matrices (cont’d)
LemmaThe graph is bipartite if and only if λ1 = −λn.
Proof.(⇒) Easy.
(⇐) Let v = (e1, . . . , en) be an eigenvector for −d . We choose an i ∈ [n] suchthat for every j ∈ [n]
|ei | ≥ |ej |.
Then,
dei = ei
∑k∈[n]
ai,k = ei
∑{i,k}∈E
ai,k
On the other hand,
dei = −(−dei ) = −∑k∈[n]
ai,kek =∑
{i,k}∈E
ai,k(−ek).
Therefore, if {i , k} ∈ E , then ek = −ei . Then we repeat the above argumenton ek .
d-Regular Graphs as Matrices (cont’d)
LemmaThe graph is bipartite if and only if λ1 = −λn.
Proof.(⇒) Easy.
(⇐) Let v = (e1, . . . , en) be an eigenvector for −d . We choose an i ∈ [n] suchthat for every j ∈ [n]
|ei | ≥ |ej |.
Then,
dei = ei
∑k∈[n]
ai,k = ei
∑{i,k}∈E
ai,k
On the other hand,
dei = −(−dei ) = −∑k∈[n]
ai,kek =∑
{i,k}∈E
ai,k(−ek).
Therefore, if {i , k} ∈ E , then ek = −ei .
Then we repeat the above argumenton ek .
d-Regular Graphs as Matrices (cont’d)
LemmaThe graph is bipartite if and only if λ1 = −λn.
Proof.(⇒) Easy.
(⇐) Let v = (e1, . . . , en) be an eigenvector for −d . We choose an i ∈ [n] suchthat for every j ∈ [n]
|ei | ≥ |ej |.
Then,
dei = ei
∑k∈[n]
ai,k = ei
∑{i,k}∈E
ai,k
On the other hand,
dei = −(−dei ) = −∑k∈[n]
ai,kek =∑
{i,k}∈E
ai,k(−ek).
Therefore, if {i , k} ∈ E , then ek = −ei . Then we repeat the above argumenton ek .
Expander Graphs are Almost Good
Recall: A graph G = (V , E) is (c, d)-good if each v ∈ V has degree at most dand for all u, v ∈ V , if there is a path from u to v , then
dist(u, v) ≤ c · log |V |.
DefinitionA d-regular graph G on n vertices is called an (n, d)-graph. Moreover, it is an(n, d , α)-graph if λ(G) ≤ αd .
TheoremLet α < 1/2. Then all (n, d , α) graphs are (c, d)-good for some appropriatec ∈ N.
Expander Graphs are Almost Good
Recall: A graph G = (V , E) is (c, d)-good if each v ∈ V has degree at most dand for all u, v ∈ V , if there is a path from u to v , then
dist(u, v) ≤ c · log |V |.
DefinitionA d-regular graph G on n vertices is called an (n, d)-graph. Moreover, it is an(n, d , α)-graph if λ(G) ≤ αd .
TheoremLet α < 1/2. Then all (n, d , α) graphs are (c, d)-good for some appropriatec ∈ N.
Expander Graphs are Almost Good
Recall: A graph G = (V , E) is (c, d)-good if each v ∈ V has degree at most dand for all u, v ∈ V , if there is a path from u to v , then
dist(u, v) ≤ c · log |V |.
DefinitionA d-regular graph G on n vertices is called an (n, d)-graph.
Moreover, it is an(n, d , α)-graph if λ(G) ≤ αd .
TheoremLet α < 1/2. Then all (n, d , α) graphs are (c, d)-good for some appropriatec ∈ N.
Expander Graphs are Almost Good
Recall: A graph G = (V , E) is (c, d)-good if each v ∈ V has degree at most dand for all u, v ∈ V , if there is a path from u to v , then
dist(u, v) ≤ c · log |V |.
DefinitionA d-regular graph G on n vertices is called an (n, d)-graph. Moreover, it is an(n, d , α)-graph if λ(G) ≤ αd .
TheoremLet α < 1/2. Then all (n, d , α) graphs are (c, d)-good for some appropriatec ∈ N.
Expander Graphs are Almost Good
Recall: A graph G = (V , E) is (c, d)-good if each v ∈ V has degree at most dand for all u, v ∈ V , if there is a path from u to v , then
dist(u, v) ≤ c · log |V |.
DefinitionA d-regular graph G on n vertices is called an (n, d)-graph. Moreover, it is an(n, d , α)-graph if λ(G) ≤ αd .
TheoremLet α < 1/2. Then all (n, d , α) graphs are (c, d)-good for some appropriatec ∈ N.
Proof
Fix an (n, d , α)- graph G = (V , E). For every v ∈ V and r ∈ N let
B(x , r) :={y ∈ V | dist(x , y) ≤ r
}.
Claim. If |B(x , r)| ≤ n/2, then for some fixed ε > 0∣∣B(x , r + 1)∣∣ ≥ (1 + ε)
∣∣B(x , r)∣∣.
By the Expander Mixing Lemma, for every S ⊆ V
|E(S , S)||S | ≤ d
(|S |n
+ α
).
Then by |E(S , S)|+ |E(S , S)| = d |S |,
|E(S , S)||S | ≥ d
((1− α)− |S |
n
).
S has at least |E(S , S)|/d neighbors outside S . The claim follows by takingε := 1/2− α. a
Then for some r = O(log n) and every x ∈ V we have |B(x , r)| > n/2. �
Proof
Fix an (n, d , α)- graph G = (V , E).
For every v ∈ V and r ∈ N let
B(x , r) :={y ∈ V | dist(x , y) ≤ r
}.
Claim. If |B(x , r)| ≤ n/2, then for some fixed ε > 0∣∣B(x , r + 1)∣∣ ≥ (1 + ε)
∣∣B(x , r)∣∣.
By the Expander Mixing Lemma, for every S ⊆ V
|E(S , S)||S | ≤ d
(|S |n
+ α
).
Then by |E(S , S)|+ |E(S , S)| = d |S |,
|E(S , S)||S | ≥ d
((1− α)− |S |
n
).
S has at least |E(S , S)|/d neighbors outside S . The claim follows by takingε := 1/2− α. a
Then for some r = O(log n) and every x ∈ V we have |B(x , r)| > n/2. �
Proof
Fix an (n, d , α)- graph G = (V , E). For every v ∈ V and r ∈ N let
B(x , r) :={y ∈ V | dist(x , y) ≤ r
}.
Claim. If |B(x , r)| ≤ n/2, then for some fixed ε > 0∣∣B(x , r + 1)∣∣ ≥ (1 + ε)
∣∣B(x , r)∣∣.
By the Expander Mixing Lemma, for every S ⊆ V
|E(S , S)||S | ≤ d
(|S |n
+ α
).
Then by |E(S , S)|+ |E(S , S)| = d |S |,
|E(S , S)||S | ≥ d
((1− α)− |S |
n
).
S has at least |E(S , S)|/d neighbors outside S . The claim follows by takingε := 1/2− α. a
Then for some r = O(log n) and every x ∈ V we have |B(x , r)| > n/2. �
Proof
Fix an (n, d , α)- graph G = (V , E). For every v ∈ V and r ∈ N let
B(x , r) :={y ∈ V | dist(x , y) ≤ r
}.
Claim. If |B(x , r)| ≤ n/2, then for some fixed ε > 0∣∣B(x , r + 1)∣∣ ≥ (1 + ε)
∣∣B(x , r)∣∣.
By the Expander Mixing Lemma, for every S ⊆ V
|E(S , S)||S | ≤ d
(|S |n
+ α
).
Then by |E(S , S)|+ |E(S , S)| = d |S |,
|E(S , S)||S | ≥ d
((1− α)− |S |
n
).
S has at least |E(S , S)|/d neighbors outside S . The claim follows by takingε := 1/2− α. a
Then for some r = O(log n) and every x ∈ V we have |B(x , r)| > n/2. �
Proof
Fix an (n, d , α)- graph G = (V , E). For every v ∈ V and r ∈ N let
B(x , r) :={y ∈ V | dist(x , y) ≤ r
}.
Claim. If |B(x , r)| ≤ n/2, then for some fixed ε > 0∣∣B(x , r + 1)∣∣ ≥ (1 + ε)
∣∣B(x , r)∣∣.
By the Expander Mixing Lemma, for every S ⊆ V
|E(S , S)||S | ≤ d
(|S |n
+ α
).
Then by |E(S , S)|+ |E(S , S)| = d |S |,
|E(S , S)||S | ≥ d
((1− α)− |S |
n
).
S has at least |E(S , S)|/d neighbors outside S . The claim follows by takingε := 1/2− α. a
Then for some r = O(log n) and every x ∈ V we have |B(x , r)| > n/2. �
Proof
Fix an (n, d , α)- graph G = (V , E). For every v ∈ V and r ∈ N let
B(x , r) :={y ∈ V | dist(x , y) ≤ r
}.
Claim. If |B(x , r)| ≤ n/2, then for some fixed ε > 0∣∣B(x , r + 1)∣∣ ≥ (1 + ε)
∣∣B(x , r)∣∣.
By the Expander Mixing Lemma, for every S ⊆ V
|E(S , S)||S | ≤ d
(|S |n
+ α
).
Then by |E(S , S)|+ |E(S , S)| = d |S |,
|E(S , S)||S | ≥ d
((1− α)− |S |
n
).
S has at least |E(S , S)|/d neighbors outside S . The claim follows by takingε := 1/2− α. a
Then for some r = O(log n) and every x ∈ V we have |B(x , r)| > n/2. �
Proof
Fix an (n, d , α)- graph G = (V , E). For every v ∈ V and r ∈ N let
B(x , r) :={y ∈ V | dist(x , y) ≤ r
}.
Claim. If |B(x , r)| ≤ n/2, then for some fixed ε > 0∣∣B(x , r + 1)∣∣ ≥ (1 + ε)
∣∣B(x , r)∣∣.
By the Expander Mixing Lemma, for every S ⊆ V
|E(S , S)||S | ≤ d
(|S |n
+ α
).
Then by |E(S , S)|+ |E(S , S)| = d |S |,
|E(S , S)||S | ≥ d
((1− α)− |S |
n
).
S has at least |E(S , S)|/d neighbors outside S .
The claim follows by takingε := 1/2− α. a
Then for some r = O(log n) and every x ∈ V we have |B(x , r)| > n/2. �
Proof
Fix an (n, d , α)- graph G = (V , E). For every v ∈ V and r ∈ N let
B(x , r) :={y ∈ V | dist(x , y) ≤ r
}.
Claim. If |B(x , r)| ≤ n/2, then for some fixed ε > 0∣∣B(x , r + 1)∣∣ ≥ (1 + ε)
∣∣B(x , r)∣∣.
By the Expander Mixing Lemma, for every S ⊆ V
|E(S , S)||S | ≤ d
(|S |n
+ α
).
Then by |E(S , S)|+ |E(S , S)| = d |S |,
|E(S , S)||S | ≥ d
((1− α)− |S |
n
).
S has at least |E(S , S)|/d neighbors outside S . The claim follows by takingε := 1/2− α. a
Then for some r = O(log n) and every x ∈ V we have |B(x , r)| > n/2. �
Proof
Fix an (n, d , α)- graph G = (V , E). For every v ∈ V and r ∈ N let
B(x , r) :={y ∈ V | dist(x , y) ≤ r
}.
Claim. If |B(x , r)| ≤ n/2, then for some fixed ε > 0∣∣B(x , r + 1)∣∣ ≥ (1 + ε)
∣∣B(x , r)∣∣.
By the Expander Mixing Lemma, for every S ⊆ V
|E(S , S)||S | ≤ d
(|S |n
+ α
).
Then by |E(S , S)|+ |E(S , S)| = d |S |,
|E(S , S)||S | ≥ d
((1− α)− |S |
n
).
S has at least |E(S , S)|/d neighbors outside S . The claim follows by takingε := 1/2− α. a
Then for some r = O(log n) and every x ∈ V we have |B(x , r)| > n/2. �
Random Walks on Expander Graphs
Random Walks
A vector p = (p1, . . . , pn) ∈ Rn is a probability (distribution) vector if all pi ≥ 0and
∑i∈[n] pi = 1.
The probability vector for the uniform distribution is
u :=1
n=
⟨1
n, . . . ,
1
n
⟩.
DefinitionA random walk on a graph G = (V , E) is a discrete-time stochastic process(X0, X1, . . .) taking values in V . The vertex X0 is sampled from some initialdistribution on V , and Xi+1 is chosen uniformly at random from the neighborsof Xi .
Random Walks
A vector p = (p1, . . . , pn) ∈ Rn is a probability (distribution) vector if all pi ≥ 0and
∑i∈[n] pi = 1. The probability vector for the uniform distribution is
u :=1
n=
⟨1
n, . . . ,
1
n
⟩.
DefinitionA random walk on a graph G = (V , E) is a discrete-time stochastic process(X0, X1, . . .) taking values in V . The vertex X0 is sampled from some initialdistribution on V , and Xi+1 is chosen uniformly at random from the neighborsof Xi .
Random Walks
A vector p = (p1, . . . , pn) ∈ Rn is a probability (distribution) vector if all pi ≥ 0and
∑i∈[n] pi = 1. The probability vector for the uniform distribution is
u :=1
n=
⟨1
n, . . . ,
1
n
⟩.
DefinitionA random walk on a graph G = (V , E) is a discrete-time stochastic process(X0, X1, . . .) taking values in V .
The vertex X0 is sampled from some initialdistribution on V , and Xi+1 is chosen uniformly at random from the neighborsof Xi .
Random Walks
A vector p = (p1, . . . , pn) ∈ Rn is a probability (distribution) vector if all pi ≥ 0and
∑i∈[n] pi = 1. The probability vector for the uniform distribution is
u :=1
n=
⟨1
n, . . . ,
1
n
⟩.
DefinitionA random walk on a graph G = (V , E) is a discrete-time stochastic process(X0, X1, . . .) taking values in V . The vertex X0 is sampled from some initialdistribution on V ,
and Xi+1 is chosen uniformly at random from the neighborsof Xi .
Random Walks
A vector p = (p1, . . . , pn) ∈ Rn is a probability (distribution) vector if all pi ≥ 0and
∑i∈[n] pi = 1. The probability vector for the uniform distribution is
u :=1
n=
⟨1
n, . . . ,
1
n
⟩.
DefinitionA random walk on a graph G = (V , E) is a discrete-time stochastic process(X0, X1, . . .) taking values in V . The vertex X0 is sampled from some initialdistribution on V , and Xi+1 is chosen uniformly at random from the neighborsof Xi .
Normalized Adjacency Matrices
Let G be a d-regular graph with adjacency matrix A = A(G). Then itsnormalized adjacency matrix is
A :=1
dA.
I The random walk on G = (V , E) is a Markov Chain with state set V andtransition matrix A.
I A s real, symmetric, and doubly stochastic; i.e. every column and every rowsums up to 1.
I If d = λ1 ≥ . . . ≥ λn are the eigenvalues of A, then 1 = λ1 ≥ . . . ≥ λn
with λi = λi/d for i ∈ [n] are the eigenvalues of A with the sameeigenvectors as of A.
I Sampling a vertex x from some probability distribution p on V and thenmoving to a random neighbor of x is equivalent to sampling a vertex fromthe distribution Ap.
Normalized Adjacency Matrices
Let G be a d-regular graph with adjacency matrix A = A(G).
Then itsnormalized adjacency matrix is
A :=1
dA.
I The random walk on G = (V , E) is a Markov Chain with state set V andtransition matrix A.
I A s real, symmetric, and doubly stochastic; i.e. every column and every rowsums up to 1.
I If d = λ1 ≥ . . . ≥ λn are the eigenvalues of A, then 1 = λ1 ≥ . . . ≥ λn
with λi = λi/d for i ∈ [n] are the eigenvalues of A with the sameeigenvectors as of A.
I Sampling a vertex x from some probability distribution p on V and thenmoving to a random neighbor of x is equivalent to sampling a vertex fromthe distribution Ap.
Normalized Adjacency Matrices
Let G be a d-regular graph with adjacency matrix A = A(G). Then itsnormalized adjacency matrix is
A :=1
dA.
I The random walk on G = (V , E) is a Markov Chain with state set V andtransition matrix A.
I A s real, symmetric, and doubly stochastic; i.e. every column and every rowsums up to 1.
I If d = λ1 ≥ . . . ≥ λn are the eigenvalues of A, then 1 = λ1 ≥ . . . ≥ λn
with λi = λi/d for i ∈ [n] are the eigenvalues of A with the sameeigenvectors as of A.
I Sampling a vertex x from some probability distribution p on V and thenmoving to a random neighbor of x is equivalent to sampling a vertex fromthe distribution Ap.
Normalized Adjacency Matrices
Let G be a d-regular graph with adjacency matrix A = A(G). Then itsnormalized adjacency matrix is
A :=1
dA.
I The random walk on G = (V , E) is a Markov Chain with state set V andtransition matrix A.
I A s real, symmetric, and doubly stochastic; i.e. every column and every rowsums up to 1.
I If d = λ1 ≥ . . . ≥ λn are the eigenvalues of A, then 1 = λ1 ≥ . . . ≥ λn
with λi = λi/d for i ∈ [n] are the eigenvalues of A with the sameeigenvectors as of A.
I Sampling a vertex x from some probability distribution p on V and thenmoving to a random neighbor of x is equivalent to sampling a vertex fromthe distribution Ap.
Normalized Adjacency Matrices
Let G be a d-regular graph with adjacency matrix A = A(G). Then itsnormalized adjacency matrix is
A :=1
dA.
I The random walk on G = (V , E) is a Markov Chain with state set V andtransition matrix A.
I A s real, symmetric, and doubly stochastic; i.e. every column and every rowsums up to 1.
I If d = λ1 ≥ . . . ≥ λn are the eigenvalues of A, then 1 = λ1 ≥ . . . ≥ λn
with λi = λi/d for i ∈ [n] are the eigenvalues of A with the sameeigenvectors as of A.
I Sampling a vertex x from some probability distribution p on V and thenmoving to a random neighbor of x is equivalent to sampling a vertex fromthe distribution Ap.
Normalized Adjacency Matrices
Let G be a d-regular graph with adjacency matrix A = A(G). Then itsnormalized adjacency matrix is
A :=1
dA.
I The random walk on G = (V , E) is a Markov Chain with state set V andtransition matrix A.
I A s real, symmetric, and doubly stochastic; i.e. every column and every rowsums up to 1.
I If d = λ1 ≥ . . . ≥ λn are the eigenvalues of A, then 1 = λ1 ≥ . . . ≥ λn
with λi = λi/d for i ∈ [n] are the eigenvalues of A with the sameeigenvectors as of A.
I Sampling a vertex x from some probability distribution p on V and thenmoving to a random neighbor of x is equivalent to sampling a vertex fromthe distribution Ap.
Normalized Adjacency Matrices
Let G be a d-regular graph with adjacency matrix A = A(G). Then itsnormalized adjacency matrix is
A :=1
dA.
I The random walk on G = (V , E) is a Markov Chain with state set V andtransition matrix A.
I A s real, symmetric, and doubly stochastic; i.e. every column and every rowsums up to 1.
I If d = λ1 ≥ . . . ≥ λn are the eigenvalues of A, then 1 = λ1 ≥ . . . ≥ λn
with λi = λi/d for i ∈ [n] are the eigenvalues of A with the sameeigenvectors as of A.
I Sampling a vertex x from some probability distribution p on V and thenmoving to a random neighbor of x is equivalent to sampling a vertex fromthe distribution Ap.
Normalized Adjacency Matrices (cont’d)
I The matrix At
is the transition matrix of the Markov Chain defined by
random walks of length t, i.e., (At)i,j is the probability a random walk
starting at i is at j after t steps.
I The stationary distribution of the random walk on G is the uniformdistribution, namely, uA =Au = u.
Normalized Adjacency Matrices (cont’d)
I The matrix At
is the transition matrix of the Markov Chain defined by
random walks of length t,
i.e., (At)i,j is the probability a random walk
starting at i is at j after t steps.
I The stationary distribution of the random walk on G is the uniformdistribution, namely, uA =Au = u.
Normalized Adjacency Matrices (cont’d)
I The matrix At
is the transition matrix of the Markov Chain defined by
random walks of length t, i.e., (At)i,j is the probability a random walk
starting at i is at j after t steps.
I The stationary distribution of the random walk on G is the uniformdistribution, namely, uA =Au = u.
Normalized Adjacency Matrices (cont’d)
I The matrix At
is the transition matrix of the Markov Chain defined by
random walks of length t, i.e., (At)i,j is the probability a random walk
starting at i is at j after t steps.
I The stationary distribution of the random walk on G is the uniformdistribution, namely, uA =Au = u.
The norms of vectors
Let x = (x1, . . . , xn) ∈ Rn. Then:
- ‖x‖1 :=∑
i∈[n] |xi |.
- ‖x‖2 :=√∑
i∈[n] x2i .
- ‖x‖∞ := maxi∈[n] |xi |.
The norms of vectors
Let x = (x1, . . . , xn) ∈ Rn. Then:
- ‖x‖1 :=∑
i∈[n] |xi |.
- ‖x‖2 :=√∑
i∈[n] x2i .
- ‖x‖∞ := maxi∈[n] |xi |.
The norms of vectors
Let x = (x1, . . . , xn) ∈ Rn. Then:
- ‖x‖1 :=∑
i∈[n] |xi |.
- ‖x‖2 :=√∑
i∈[n] x2i .
- ‖x‖∞ := maxi∈[n] |xi |.
The norms of vectors
Let x = (x1, . . . , xn) ∈ Rn. Then:
- ‖x‖1 :=∑
i∈[n] |xi |.
- ‖x‖2 :=√∑
i∈[n] x2i .
- ‖x‖∞ := maxi∈[n] |xi |.
The norms of vectors
Let x = (x1, . . . , xn) ∈ Rn. Then:
- ‖x‖1 :=∑
i∈[n] |xi |.
- ‖x‖2 :=√∑
i∈[n] x2i .
- ‖x‖∞ := maxi∈[n] |xi |.
Rapid Mixing of Walks
TheoremLet G be an (n, d , α)-graph with normalized adjacency matrix A. Then for anydistribution vector p and any positive integer t:
‖Atp − u‖1 ≤
√n αt
Corollary
(n, d , α)-graphs are (c, d)-good with
c = − 3
2 log α+ ε.
and ε > 0.
Rapid Mixing of Walks
TheoremLet G be an (n, d , α)-graph with normalized adjacency matrix A. Then for anydistribution vector p and any positive integer t:
‖Atp − u‖1 ≤
√n αt
Corollary
(n, d , α)-graphs are (c, d)-good with
c = − 3
2 log α+ ε.
and ε > 0.
Rapid Mixing of Walks
TheoremLet G be an (n, d , α)-graph with normalized adjacency matrix A. Then for anydistribution vector p and any positive integer t:
‖Atp − u‖1 ≤
√n αt
Corollary
(n, d , α)-graphs are (c, d)-good with
c = − 3
2 log α+ ε.
and ε > 0.
Rapid Mixing of Walks (cont’d)
TheoremLet G be an (n, d , α)-graph with normalized adjacency matrix A. Then for anydistribution vector p and any positive integer t:
‖Atp − u‖2 ≤ ‖p − u‖2α
t ≤ αt .
LemmaFor every probability vector p, ‖Ap − u‖2 ≤ ‖p − u‖2α ≤ α.
Rapid Mixing of Walks (cont’d)
TheoremLet G be an (n, d , α)-graph with normalized adjacency matrix A. Then for anydistribution vector p and any positive integer t:
‖Atp − u‖2 ≤ ‖p − u‖2α
t ≤ αt .
LemmaFor every probability vector p, ‖Ap − u‖2 ≤ ‖p − u‖2α ≤ α.
Rapid Mixing of Walks (cont’d)
TheoremLet G be an (n, d , α)-graph with normalized adjacency matrix A. Then for anydistribution vector p and any positive integer t:
‖Atp − u‖2 ≤ ‖p − u‖2α
t ≤ αt .
LemmaFor every probability vector p, ‖Ap − u‖2 ≤ ‖p − u‖2α ≤ α.
Proof of the Lemma
Obviously, 〈p − u, u〉 = 0, that is, p − u and u are orthogonal. It is easy toshow ‖p − u‖2 ≤ 1. Consequently, for some appropriate (αi )2≤i≤n we have
p − u =∑
2≤i≤n
βivi .
‖Ap − u‖2 = ‖Ap − Au‖2 = ‖A(p − u)‖2
=
∥∥∥∥∥∥∑
2≤i≤n
λiβivi
∥∥∥∥∥∥2
=
√√√√⟨ ∑2≤i≤n
λiβivi ,∑
2≤i≤n
λiβivi
⟩
=
√ ∑2≤i≤n
λ2i β
2i 〈vi , vi 〉 ≤ α
√ ∑2≤i≤n
β2i 〈vi , vi 〉
= α
∥∥∥∥∥∥∑
2≤i≤n
βivi
∥∥∥∥∥∥2
= α‖p − u‖2 ≤ α.
�
Proof of the Lemma
Obviously, 〈p − u, u〉 = 0, that is, p − u and u are orthogonal.
It is easy toshow ‖p − u‖2 ≤ 1. Consequently, for some appropriate (αi )2≤i≤n we have
p − u =∑
2≤i≤n
βivi .
‖Ap − u‖2 = ‖Ap − Au‖2 = ‖A(p − u)‖2
=
∥∥∥∥∥∥∑
2≤i≤n
λiβivi
∥∥∥∥∥∥2
=
√√√√⟨ ∑2≤i≤n
λiβivi ,∑
2≤i≤n
λiβivi
⟩
=
√ ∑2≤i≤n
λ2i β
2i 〈vi , vi 〉 ≤ α
√ ∑2≤i≤n
β2i 〈vi , vi 〉
= α
∥∥∥∥∥∥∑
2≤i≤n
βivi
∥∥∥∥∥∥2
= α‖p − u‖2 ≤ α.
�
Proof of the Lemma
Obviously, 〈p − u, u〉 = 0, that is, p − u and u are orthogonal. It is easy toshow ‖p − u‖2 ≤ 1.
Consequently, for some appropriate (αi )2≤i≤n we have
p − u =∑
2≤i≤n
βivi .
‖Ap − u‖2 = ‖Ap − Au‖2 = ‖A(p − u)‖2
=
∥∥∥∥∥∥∑
2≤i≤n
λiβivi
∥∥∥∥∥∥2
=
√√√√⟨ ∑2≤i≤n
λiβivi ,∑
2≤i≤n
λiβivi
⟩
=
√ ∑2≤i≤n
λ2i β
2i 〈vi , vi 〉 ≤ α
√ ∑2≤i≤n
β2i 〈vi , vi 〉
= α
∥∥∥∥∥∥∑
2≤i≤n
βivi
∥∥∥∥∥∥2
= α‖p − u‖2 ≤ α.
�
Proof of the Lemma
Obviously, 〈p − u, u〉 = 0, that is, p − u and u are orthogonal. It is easy toshow ‖p − u‖2 ≤ 1. Consequently, for some appropriate (αi )2≤i≤n we have
p − u =∑
2≤i≤n
βivi .
‖Ap − u‖2 = ‖Ap − Au‖2 = ‖A(p − u)‖2
=
∥∥∥∥∥∥∑
2≤i≤n
λiβivi
∥∥∥∥∥∥2
=
√√√√⟨ ∑2≤i≤n
λiβivi ,∑
2≤i≤n
λiβivi
⟩
=
√ ∑2≤i≤n
λ2i β
2i 〈vi , vi 〉 ≤ α
√ ∑2≤i≤n
β2i 〈vi , vi 〉
= α
∥∥∥∥∥∥∑
2≤i≤n
βivi
∥∥∥∥∥∥2
= α‖p − u‖2 ≤ α.
�
Proof of the Lemma
Obviously, 〈p − u, u〉 = 0, that is, p − u and u are orthogonal. It is easy toshow ‖p − u‖2 ≤ 1. Consequently, for some appropriate (αi )2≤i≤n we have
p − u =∑
2≤i≤n
βivi .
‖Ap − u‖2 = ‖Ap − Au‖2 = ‖A(p − u)‖2
=
∥∥∥∥∥∥∑
2≤i≤n
λiβivi
∥∥∥∥∥∥2
=
√√√√⟨ ∑2≤i≤n
λiβivi ,∑
2≤i≤n
λiβivi
⟩
=
√ ∑2≤i≤n
λ2i β
2i 〈vi , vi 〉 ≤ α
√ ∑2≤i≤n
β2i 〈vi , vi 〉
= α
∥∥∥∥∥∥∑
2≤i≤n
βivi
∥∥∥∥∥∥2
= α‖p − u‖2 ≤ α.
�