Find the incorrectness of the process.

Clash Royale CLAN TAG#URR8PPP

up vote
0
down vote

favorite

Assume $$Axi=lambda_1xi $$
then $$Axixi^T=lambda_1xixi^T $$
and $$|Axixi^T|=|lambda_1xixi^T| $$
which we will have
$$|A|cdot |xixi^T|=lambda_1^ncdot|xixi^T|$$
then $$|A|=lambda_1^n$$
But $$|A|=prod_i=1^nlambda_i$$

What's the problem in this process?

linear-algebra proof-verification

share|cite|improve this question

edited Jul 17 at 9:18

Jneven

514218

asked Jul 17 at 5:19

Jaqen Chou

1646

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Writing $|Axixi^T| = |A||xixi^T|$ is incorrect. Here I am interpreting $|A|$ to mean $supx$.
  – caffeinemachine
  Jul 17 at 5:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @caffeinemachine In that case we wouldn't have $|A|=prod_i=1^nlambda_i$. It's most likely meant as determinant.
  – Arthur
  Jul 17 at 5:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Also, perhaps putting brackets would help avoid confusion. For instance, $Axixi^T$ is $(Axi)xi^T$.
  – caffeinemachine
  Jul 17 at 5:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Arthur I see. Thanks. Then $det((Axi)xi^T) = det(A)det(xixi^T) = det(A)xixi^T$ is also not correct. :)
  – caffeinemachine
  Jul 17 at 5:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @caffeinemachine Is also not claimed in the post. At least the last equality. The first equality you wrote is correct though.
  – Arthur
  Jul 17 at 5:33
  
  

  
  
  
  

 | 
show 1 more comment

up vote
0
down vote

favorite

Assume $$Axi=lambda_1xi $$
then $$Axixi^T=lambda_1xixi^T $$
and $$|Axixi^T|=|lambda_1xixi^T| $$
which we will have
$$|A|cdot |xixi^T|=lambda_1^ncdot|xixi^T|$$
then $$|A|=lambda_1^n$$
But $$|A|=prod_i=1^nlambda_i$$

What's the problem in this process?

linear-algebra proof-verification

share|cite|improve this question

edited Jul 17 at 9:18

Jneven

514218

asked Jul 17 at 5:19

Jaqen Chou

1646

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Writing $|Axixi^T| = |A||xixi^T|$ is incorrect. Here I am interpreting $|A|$ to mean $supx$.
  – caffeinemachine
  Jul 17 at 5:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @caffeinemachine In that case we wouldn't have $|A|=prod_i=1^nlambda_i$. It's most likely meant as determinant.
  – Arthur
  Jul 17 at 5:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Also, perhaps putting brackets would help avoid confusion. For instance, $Axixi^T$ is $(Axi)xi^T$.
  – caffeinemachine
  Jul 17 at 5:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Arthur I see. Thanks. Then $det((Axi)xi^T) = det(A)det(xixi^T) = det(A)xixi^T$ is also not correct. :)
  – caffeinemachine
  Jul 17 at 5:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @caffeinemachine Is also not claimed in the post. At least the last equality. The first equality you wrote is correct though.
  – Arthur
  Jul 17 at 5:33
  
  

  
  
  
  

 | 
show 1 more comment

up vote
0
down vote

favorite

up vote
0
down vote

favorite

Assume $$Axi=lambda_1xi $$
then $$Axixi^T=lambda_1xixi^T $$
and $$|Axixi^T|=|lambda_1xixi^T| $$
which we will have
$$|A|cdot |xixi^T|=lambda_1^ncdot|xixi^T|$$
then $$|A|=lambda_1^n$$
But $$|A|=prod_i=1^nlambda_i$$

What's the problem in this process?

linear-algebra proof-verification

share|cite|improve this question

edited Jul 17 at 9:18

Jneven

514218

asked Jul 17 at 5:19

Jaqen Chou

1646

Assume $$Axi=lambda_1xi $$
then $$Axixi^T=lambda_1xixi^T $$
and $$|Axixi^T|=|lambda_1xixi^T| $$
which we will have
$$|A|cdot |xixi^T|=lambda_1^ncdot|xixi^T|$$
then $$|A|=lambda_1^n$$
But $$|A|=prod_i=1^nlambda_i$$

What's the problem in this process?

linear-algebra proof-verification

share|cite|improve this question

edited Jul 17 at 9:18

Jneven

514218

asked Jul 17 at 5:19

Jaqen Chou

1646

share|cite|improve this question

share|cite|improve this question

edited Jul 17 at 9:18

Jneven

514218

edited Jul 17 at 9:18

Jneven

514218

edited Jul 17 at 9:18

Jneven

514218

514218

asked Jul 17 at 5:19

Jaqen Chou

1646

asked Jul 17 at 5:19

Jaqen Chou

1646

asked Jul 17 at 5:19

Jaqen Chou

1646

1646

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Writing $|Axixi^T| = |A||xixi^T|$ is incorrect. Here I am interpreting $|A|$ to mean $supx$.
  – caffeinemachine
  Jul 17 at 5:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @caffeinemachine In that case we wouldn't have $|A|=prod_i=1^nlambda_i$. It's most likely meant as determinant.
  – Arthur
  Jul 17 at 5:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Also, perhaps putting brackets would help avoid confusion. For instance, $Axixi^T$ is $(Axi)xi^T$.
  – caffeinemachine
  Jul 17 at 5:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Arthur I see. Thanks. Then $det((Axi)xi^T) = det(A)det(xixi^T) = det(A)xixi^T$ is also not correct. :)
  – caffeinemachine
  Jul 17 at 5:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @caffeinemachine Is also not claimed in the post. At least the last equality. The first equality you wrote is correct though.
  – Arthur
  Jul 17 at 5:33
  
  

  
  
  
  

 | 
show 1 more comment

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Writing $|Axixi^T| = |A||xixi^T|$ is incorrect. Here I am interpreting $|A|$ to mean $supx$.
  – caffeinemachine
  Jul 17 at 5:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @caffeinemachine In that case we wouldn't have $|A|=prod_i=1^nlambda_i$. It's most likely meant as determinant.
  – Arthur
  Jul 17 at 5:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Also, perhaps putting brackets would help avoid confusion. For instance, $Axixi^T$ is $(Axi)xi^T$.
  – caffeinemachine
  Jul 17 at 5:26
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Arthur I see. Thanks. Then $det((Axi)xi^T) = det(A)det(xixi^T) = det(A)xixi^T$ is also not correct. :)
  – caffeinemachine
  Jul 17 at 5:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @caffeinemachine Is also not claimed in the post. At least the last equality. The first equality you wrote is correct though.
  – Arthur
  Jul 17 at 5:33
  
  

  
  
  
  

1

1

Writing $|Axixi^T| = |A||xixi^T|$ is incorrect. Here I am interpreting $|A|$ to mean $supx$.
– caffeinemachine
Jul 17 at 5:24

Writing $|Axixi^T| = |A||xixi^T|$ is incorrect. Here I am interpreting $|A|$ to mean $supx$.
– caffeinemachine
Jul 17 at 5:24

@caffeinemachine In that case we wouldn't have $|A|=prod_i=1^nlambda_i$. It's most likely meant as determinant.
– Arthur
Jul 17 at 5:26

@caffeinemachine In that case we wouldn't have $|A|=prod_i=1^nlambda_i$. It's most likely meant as determinant.
– Arthur
Jul 17 at 5:26

Also, perhaps putting brackets would help avoid confusion. For instance, $Axixi^T$ is $(Axi)xi^T$.
– caffeinemachine
Jul 17 at 5:26

Also, perhaps putting brackets would help avoid confusion. For instance, $Axixi^T$ is $(Axi)xi^T$.
– caffeinemachine
Jul 17 at 5:26

@Arthur I see. Thanks. Then $det((Axi)xi^T) = det(A)det(xixi^T) = det(A)xixi^T$ is also not correct. :)
– caffeinemachine
Jul 17 at 5:29

@Arthur I see. Thanks. Then $det((Axi)xi^T) = det(A)det(xixi^T) = det(A)xixi^T$ is also not correct. :)
– caffeinemachine
Jul 17 at 5:29

@caffeinemachine Is also not claimed in the post. At least the last equality. The first equality you wrote is correct though.
– Arthur
Jul 17 at 5:33

@caffeinemachine Is also not claimed in the post. At least the last equality. The first equality you wrote is correct though.
– Arthur
Jul 17 at 5:33

 | 
show 1 more comment

1 Answer
1

active

oldest

votes

up vote
3
down vote



accepted

Hint: Take a good look at $|xixi^T|$. Think about, say, the rank of that matrix and what that means for the determinant.

share|cite|improve this answer

answered Jul 17 at 5:22

Arthur

98.9k793175

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Tks. Its rank is less than n.
  – Jaqen Chou
  Jul 17 at 5:28
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JaqenChou So, either $n=1$, in which case ..., or $nneq 1$, in which case ...
  – Arthur
  Jul 17 at 5:29
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
down vote



accepted

Hint: Take a good look at $|xixi^T|$. Think about, say, the rank of that matrix and what that means for the determinant.

share|cite|improve this answer

answered Jul 17 at 5:22

Arthur

98.9k793175

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Tks. Its rank is less than n.
  – Jaqen Chou
  Jul 17 at 5:28
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JaqenChou So, either $n=1$, in which case ..., or $nneq 1$, in which case ...
  – Arthur
  Jul 17 at 5:29
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

Hint: Take a good look at $|xixi^T|$. Think about, say, the rank of that matrix and what that means for the determinant.

share|cite|improve this answer

answered Jul 17 at 5:22

Arthur

98.9k793175

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Tks. Its rank is less than n.
  – Jaqen Chou
  Jul 17 at 5:28
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JaqenChou So, either $n=1$, in which case ..., or $nneq 1$, in which case ...
  – Arthur
  Jul 17 at 5:29
  

  
  
  
  

add a comment | 

up vote
3
down vote



accepted

up vote
3
down vote



accepted

Hint: Take a good look at $|xixi^T|$. Think about, say, the rank of that matrix and what that means for the determinant.

share|cite|improve this answer

answered Jul 17 at 5:22

Arthur

98.9k793175

Hint: Take a good look at $|xixi^T|$. Think about, say, the rank of that matrix and what that means for the determinant.

share|cite|improve this answer

answered Jul 17 at 5:22

Arthur

98.9k793175

share|cite|improve this answer

share|cite|improve this answer

answered Jul 17 at 5:22

Arthur

98.9k793175

answered Jul 17 at 5:22

Arthur

98.9k793175

answered Jul 17 at 5:22

Arthur

98.9k793175

98.9k793175

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Tks. Its rank is less than n.
  – Jaqen Chou
  Jul 17 at 5:28
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JaqenChou So, either $n=1$, in which case ..., or $nneq 1$, in which case ...
  – Arthur
  Jul 17 at 5:29
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Tks. Its rank is less than n.
  – Jaqen Chou
  Jul 17 at 5:28
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @JaqenChou So, either $n=1$, in which case ..., or $nneq 1$, in which case ...
  – Arthur
  Jul 17 at 5:29
  

  
  
  
  

Tks. Its rank is less than n.
– Jaqen Chou
Jul 17 at 5:28

Tks. Its rank is less than n.
– Jaqen Chou
Jul 17 at 5:28

@JaqenChou So, either $n=1$, in which case ..., or $nneq 1$, in which case ...
– Arthur
Jul 17 at 5:29

@JaqenChou So, either $n=1$, in which case ..., or $nneq 1$, in which case ...
– Arthur
Jul 17 at 5:29

add a comment | 

 

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