If $|A_ij|leq 1$ for all entries of symmetric $A$, then the largest possible spectral norm of $A$ is smaller than $n$?

Clash Royale CLAN TAG#URR8PPP

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In other words, for any symmetric or antisymmetric matrix $AinmathbbR^ntimes n$ that has $|A_ij|leq 1$, can we conclude that
beginalign*
|A| leq |J|,
endalign*
where $|cdot|$ denotes spectral norm and $J$ represents all one matrix?

Since we already know that all one matrix $J$ has the smallest eigenvalue $sigma_n = 0$ and all other eigenvalues $sigma_1 = sigma_2 = ldots = sigma_n-1 = n$, it is to say whether such $A$ can have singular value greater than $n$.

linear-algebra eigenvalues-eigenvectors spectral-theory symmetric-matrices singularvalues

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asked Jul 23 at 21:14

ChristophorusX

1439

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up vote
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down vote

favorite

1

In other words, for any symmetric or antisymmetric matrix $AinmathbbR^ntimes n$ that has $|A_ij|leq 1$, can we conclude that
beginalign*
|A| leq |J|,
endalign*
where $|cdot|$ denotes spectral norm and $J$ represents all one matrix?

Since we already know that all one matrix $J$ has the smallest eigenvalue $sigma_n = 0$ and all other eigenvalues $sigma_1 = sigma_2 = ldots = sigma_n-1 = n$, it is to say whether such $A$ can have singular value greater than $n$.

linear-algebra eigenvalues-eigenvectors spectral-theory symmetric-matrices singularvalues

share|cite|improve this question

asked Jul 23 at 21:14

ChristophorusX

1439

add a comment | 

up vote
3
down vote

favorite

1

up vote
3
down vote

favorite

1

1

In other words, for any symmetric or antisymmetric matrix $AinmathbbR^ntimes n$ that has $|A_ij|leq 1$, can we conclude that
beginalign*
|A| leq |J|,
endalign*
where $|cdot|$ denotes spectral norm and $J$ represents all one matrix?

Since we already know that all one matrix $J$ has the smallest eigenvalue $sigma_n = 0$ and all other eigenvalues $sigma_1 = sigma_2 = ldots = sigma_n-1 = n$, it is to say whether such $A$ can have singular value greater than $n$.

linear-algebra eigenvalues-eigenvectors spectral-theory symmetric-matrices singularvalues

share|cite|improve this question

asked Jul 23 at 21:14

ChristophorusX

1439

In other words, for any symmetric or antisymmetric matrix $AinmathbbR^ntimes n$ that has $|A_ij|leq 1$, can we conclude that
beginalign*
|A| leq |J|,
endalign*
where $|cdot|$ denotes spectral norm and $J$ represents all one matrix?

Since we already know that all one matrix $J$ has the smallest eigenvalue $sigma_n = 0$ and all other eigenvalues $sigma_1 = sigma_2 = ldots = sigma_n-1 = n$, it is to say whether such $A$ can have singular value greater than $n$.

linear-algebra eigenvalues-eigenvectors spectral-theory symmetric-matrices singularvalues

share|cite|improve this question

asked Jul 23 at 21:14

ChristophorusX

1439

share|cite|improve this question

share|cite|improve this question

asked Jul 23 at 21:14

ChristophorusX

1439

asked Jul 23 at 21:14

ChristophorusX

1439

asked Jul 23 at 21:14

ChristophorusX

1439

1439

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

up vote
2
down vote



accepted

This is true for any matrix. In particular, we note that $|A| leq |A|_F$, where $|cdot|_F$ denotes the Frobenius norm. From there, we have
$$
|A| leq |A|_F = sqrtsum_i,j=1^nA_ij^2 leq sqrtsum_i,j=1^n 1 = n
$$

share|cite|improve this answer

answered Jul 23 at 21:22

Omnomnomnom

121k784170

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2
down vote

We have

$$|J| ge frac_2(1, ldots, 1) = frac(n, ldots, n)sqrtn = n$$

On the other hand, for $x = (x_1,ldots, x_n)$ with $|x|_2 = 1$ we have

$$|Ax|_2^2 = left|left(sum_j=1^n A_ijx_jright)_i=1^nright|_2^2 = sum_i=1^n left|sum_j=1^n A_ijx_jright|^2 stackrelCSBle sum_i=1^n underbraceleft(sum_j=1^n A_ij^2right)_le n underbraceleft(sum_j=1^n x_j^2right)_= 1 le n^2$$

so taking the supremum over the unit sphere gives $|A| le n$.

Therefore

$$|A| le n le |J|$$

It basically boils down to @Omnomnomnom's answer, but it might be useful to someone.

share|cite|improve this answer

answered Jul 23 at 22:11

mechanodroid

22.2k52041

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What is "CSB" here?
  – Omnomnomnom
  Jul 23 at 22:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is that "Cauchy Schwarz Bunyakovsky"?
  – Omnomnomnom
  Jul 23 at 22:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Omnomnomnom Yes it is.
  – mechanodroid
  Jul 23 at 22:15
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Neat. If you're interested in writing this using traces rather than summations, we have $$ |Ax|_2^2 = (x^TA^TAx) = operatornameTr(A^TAxx^T)^2 leq operatornameTr(A^TA)operatornameTr(xx^T) $$ where again, the idea is that we have applied CSB to the Hilbert-Schmidt (AKA Frobenius) inner product over $Bbb R^n times n$
  – Omnomnomnom
  Jul 23 at 22:19
  
  

  
  
  
  

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote



accepted

This is true for any matrix. In particular, we note that $|A| leq |A|_F$, where $|cdot|_F$ denotes the Frobenius norm. From there, we have
$$
|A| leq |A|_F = sqrtsum_i,j=1^nA_ij^2 leq sqrtsum_i,j=1^n 1 = n
$$

share|cite|improve this answer

answered Jul 23 at 21:22

Omnomnomnom

121k784170

add a comment | 

up vote
2
down vote



accepted

This is true for any matrix. In particular, we note that $|A| leq |A|_F$, where $|cdot|_F$ denotes the Frobenius norm. From there, we have
$$
|A| leq |A|_F = sqrtsum_i,j=1^nA_ij^2 leq sqrtsum_i,j=1^n 1 = n
$$

share|cite|improve this answer

answered Jul 23 at 21:22

Omnomnomnom

121k784170

add a comment | 

up vote
2
down vote



accepted

up vote
2
down vote



accepted

This is true for any matrix. In particular, we note that $|A| leq |A|_F$, where $|cdot|_F$ denotes the Frobenius norm. From there, we have
$$
|A| leq |A|_F = sqrtsum_i,j=1^nA_ij^2 leq sqrtsum_i,j=1^n 1 = n
$$

share|cite|improve this answer

answered Jul 23 at 21:22

Omnomnomnom

121k784170

This is true for any matrix. In particular, we note that $|A| leq |A|_F$, where $|cdot|_F$ denotes the Frobenius norm. From there, we have
$$
|A| leq |A|_F = sqrtsum_i,j=1^nA_ij^2 leq sqrtsum_i,j=1^n 1 = n
$$

share|cite|improve this answer

answered Jul 23 at 21:22

Omnomnomnom

121k784170

share|cite|improve this answer

share|cite|improve this answer

answered Jul 23 at 21:22

Omnomnomnom

121k784170

answered Jul 23 at 21:22

Omnomnomnom

121k784170

answered Jul 23 at 21:22

Omnomnomnom

121k784170

121k784170

add a comment | 

add a comment | 

up vote
2
down vote

We have

$$|J| ge frac_2(1, ldots, 1) = frac(n, ldots, n)sqrtn = n$$

On the other hand, for $x = (x_1,ldots, x_n)$ with $|x|_2 = 1$ we have

$$|Ax|_2^2 = left|left(sum_j=1^n A_ijx_jright)_i=1^nright|_2^2 = sum_i=1^n left|sum_j=1^n A_ijx_jright|^2 stackrelCSBle sum_i=1^n underbraceleft(sum_j=1^n A_ij^2right)_le n underbraceleft(sum_j=1^n x_j^2right)_= 1 le n^2$$

so taking the supremum over the unit sphere gives $|A| le n$.

Therefore

$$|A| le n le |J|$$

It basically boils down to @Omnomnomnom's answer, but it might be useful to someone.

share|cite|improve this answer

answered Jul 23 at 22:11

mechanodroid

22.2k52041

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What is "CSB" here?
  – Omnomnomnom
  Jul 23 at 22:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is that "Cauchy Schwarz Bunyakovsky"?
  – Omnomnomnom
  Jul 23 at 22:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Omnomnomnom Yes it is.
  – mechanodroid
  Jul 23 at 22:15
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Neat. If you're interested in writing this using traces rather than summations, we have $$ |Ax|_2^2 = (x^TA^TAx) = operatornameTr(A^TAxx^T)^2 leq operatornameTr(A^TA)operatornameTr(xx^T) $$ where again, the idea is that we have applied CSB to the Hilbert-Schmidt (AKA Frobenius) inner product over $Bbb R^n times n$
  – Omnomnomnom
  Jul 23 at 22:19
  
  

  
  
  
  

add a comment | 

up vote
2
down vote

We have

$$|J| ge frac_2(1, ldots, 1) = frac(n, ldots, n)sqrtn = n$$

On the other hand, for $x = (x_1,ldots, x_n)$ with $|x|_2 = 1$ we have

$$|Ax|_2^2 = left|left(sum_j=1^n A_ijx_jright)_i=1^nright|_2^2 = sum_i=1^n left|sum_j=1^n A_ijx_jright|^2 stackrelCSBle sum_i=1^n underbraceleft(sum_j=1^n A_ij^2right)_le n underbraceleft(sum_j=1^n x_j^2right)_= 1 le n^2$$

so taking the supremum over the unit sphere gives $|A| le n$.

Therefore

$$|A| le n le |J|$$

It basically boils down to @Omnomnomnom's answer, but it might be useful to someone.

share|cite|improve this answer

answered Jul 23 at 22:11

mechanodroid

22.2k52041

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What is "CSB" here?
  – Omnomnomnom
  Jul 23 at 22:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is that "Cauchy Schwarz Bunyakovsky"?
  – Omnomnomnom
  Jul 23 at 22:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Omnomnomnom Yes it is.
  – mechanodroid
  Jul 23 at 22:15
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Neat. If you're interested in writing this using traces rather than summations, we have $$ |Ax|_2^2 = (x^TA^TAx) = operatornameTr(A^TAxx^T)^2 leq operatornameTr(A^TA)operatornameTr(xx^T) $$ where again, the idea is that we have applied CSB to the Hilbert-Schmidt (AKA Frobenius) inner product over $Bbb R^n times n$
  – Omnomnomnom
  Jul 23 at 22:19
  
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

We have

$$|J| ge frac_2(1, ldots, 1) = frac(n, ldots, n)sqrtn = n$$

On the other hand, for $x = (x_1,ldots, x_n)$ with $|x|_2 = 1$ we have

$$|Ax|_2^2 = left|left(sum_j=1^n A_ijx_jright)_i=1^nright|_2^2 = sum_i=1^n left|sum_j=1^n A_ijx_jright|^2 stackrelCSBle sum_i=1^n underbraceleft(sum_j=1^n A_ij^2right)_le n underbraceleft(sum_j=1^n x_j^2right)_= 1 le n^2$$

so taking the supremum over the unit sphere gives $|A| le n$.

Therefore

$$|A| le n le |J|$$

It basically boils down to @Omnomnomnom's answer, but it might be useful to someone.

share|cite|improve this answer

answered Jul 23 at 22:11

mechanodroid

22.2k52041

We have

$$|J| ge frac_2(1, ldots, 1) = frac(n, ldots, n)sqrtn = n$$

On the other hand, for $x = (x_1,ldots, x_n)$ with $|x|_2 = 1$ we have

$$|Ax|_2^2 = left|left(sum_j=1^n A_ijx_jright)_i=1^nright|_2^2 = sum_i=1^n left|sum_j=1^n A_ijx_jright|^2 stackrelCSBle sum_i=1^n underbraceleft(sum_j=1^n A_ij^2right)_le n underbraceleft(sum_j=1^n x_j^2right)_= 1 le n^2$$

so taking the supremum over the unit sphere gives $|A| le n$.

Therefore

$$|A| le n le |J|$$

It basically boils down to @Omnomnomnom's answer, but it might be useful to someone.

share|cite|improve this answer

answered Jul 23 at 22:11

mechanodroid

22.2k52041

share|cite|improve this answer

share|cite|improve this answer

answered Jul 23 at 22:11

mechanodroid

22.2k52041

answered Jul 23 at 22:11

mechanodroid

22.2k52041

answered Jul 23 at 22:11

mechanodroid

22.2k52041

22.2k52041

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What is "CSB" here?
  – Omnomnomnom
  Jul 23 at 22:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is that "Cauchy Schwarz Bunyakovsky"?
  – Omnomnomnom
  Jul 23 at 22:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Omnomnomnom Yes it is.
  – mechanodroid
  Jul 23 at 22:15
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Neat. If you're interested in writing this using traces rather than summations, we have $$ |Ax|_2^2 = (x^TA^TAx) = operatornameTr(A^TAxx^T)^2 leq operatornameTr(A^TA)operatornameTr(xx^T) $$ where again, the idea is that we have applied CSB to the Hilbert-Schmidt (AKA Frobenius) inner product over $Bbb R^n times n$
  – Omnomnomnom
  Jul 23 at 22:19
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What is "CSB" here?
  – Omnomnomnom
  Jul 23 at 22:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Is that "Cauchy Schwarz Bunyakovsky"?
  – Omnomnomnom
  Jul 23 at 22:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Omnomnomnom Yes it is.
  – mechanodroid
  Jul 23 at 22:15
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Neat. If you're interested in writing this using traces rather than summations, we have $$ |Ax|_2^2 = (x^TA^TAx) = operatornameTr(A^TAxx^T)^2 leq operatornameTr(A^TA)operatornameTr(xx^T) $$ where again, the idea is that we have applied CSB to the Hilbert-Schmidt (AKA Frobenius) inner product over $Bbb R^n times n$
  – Omnomnomnom
  Jul 23 at 22:19
  
  

  
  
  
  

What is "CSB" here?
– Omnomnomnom
Jul 23 at 22:13

What is "CSB" here?
– Omnomnomnom
Jul 23 at 22:13

Is that "Cauchy Schwarz Bunyakovsky"?
– Omnomnomnom
Jul 23 at 22:15

Is that "Cauchy Schwarz Bunyakovsky"?
– Omnomnomnom
Jul 23 at 22:15

@Omnomnomnom Yes it is.
– mechanodroid
Jul 23 at 22:15

@Omnomnomnom Yes it is.
– mechanodroid
Jul 23 at 22:15

1

1

Neat. If you're interested in writing this using traces rather than summations, we have $$ |Ax|_2^2 = (x^TA^TAx) = operatornameTr(A^TAxx^T)^2 leq operatornameTr(A^TA)operatornameTr(xx^T) $$ where again, the idea is that we have applied CSB to the Hilbert-Schmidt (AKA Frobenius) inner product over $Bbb R^n times n$
– Omnomnomnom
Jul 23 at 22:19

Neat. If you're interested in writing this using traces rather than summations, we have $$ |Ax|_2^2 = (x^TA^TAx) = operatornameTr(A^TAxx^T)^2 leq operatornameTr(A^TA)operatornameTr(xx^T) $$ where again, the idea is that we have applied CSB to the Hilbert-Schmidt (AKA Frobenius) inner product over $Bbb R^n times n$
– Omnomnomnom
Jul 23 at 22:19

add a comment | 

 

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