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Eigenvalues of a $A^T A$

Show some eigenvalue properties for $A=xy^*$Some properties of a $2times 2$ matrix with repeated eigenvaluesProving that matrix is positive definiteEigenvalues of the sum of a stochastic matrix and a diagonal matrixShow non-symmetric matrix has non-orthogonal eigenvectorsRoots of Matrices and DiagonalizationEigenvalues of a block diagonal matrixEigenvalues of block matrixEigenvalues of Symmetric Pseudo-Toeplitz MatrixEigenvalues of same-row matricesEigenvalues of product of two matrices

4

$begingroup$

Given the matrix of order $1timesn$, $A=(a_1, a_2, ..., a_n)$ , where $a_i$ are real;
The question is to find all eigenvalues of $A^T A$.

I have proved that it is a non-invertible matrix, therefore $0$ is one of the values.
And the product matrix is an $nxn$ matrix with the diagonal elements being $a_1^2, a_2^2,...,a_n^2$.
I am struggling with finding the other eigenvalues, tried by calculating the det of $A - aI$, but didn't go anywhere.

linear-algebra matrices eigenvalues-eigenvectors

share|cite|improve this question

edited Mar 12 at 18:49

6005

37k751127

asked Mar 12 at 16:48

Kamal KhalailyKamal Khalaily

233

New contributor

Kamal Khalaily is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your might check out math.stackexchange.com/questions/3137885/…. I left an answer there which I believe addresses your question.
  $endgroup$
  – Robert Lewis
  Mar 12 at 16:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also, by "A(transpose)*" do you mean $(A^T)^* = A^dagger$? Cheers!
  $endgroup$
  – Robert Lewis
  Mar 12 at 16:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RobertLewis I think the * is probably supposed to indicate multiplication.
  $endgroup$
  – saulspatz
  Mar 12 at 17:00
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @RobertLewis By that I mean multiplying A transpose with A.
  $endgroup$
  – Kamal Khalaily
  Mar 12 at 17:01
  
  

  
  
  
  

add a comment | 

4

$begingroup$

Given the matrix of order $1timesn$, $A=(a_1, a_2, ..., a_n)$ , where $a_i$ are real;
The question is to find all eigenvalues of $A^T A$.

I have proved that it is a non-invertible matrix, therefore $0$ is one of the values.
And the product matrix is an $nxn$ matrix with the diagonal elements being $a_1^2, a_2^2,...,a_n^2$.
I am struggling with finding the other eigenvalues, tried by calculating the det of $A - aI$, but didn't go anywhere.

linear-algebra matrices eigenvalues-eigenvectors

share|cite|improve this question

edited Mar 12 at 18:49

6005

37k751127

asked Mar 12 at 16:48

Kamal KhalailyKamal Khalaily

233

New contributor

Kamal Khalaily is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your might check out math.stackexchange.com/questions/3137885/…. I left an answer there which I believe addresses your question.
  $endgroup$
  – Robert Lewis
  Mar 12 at 16:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also, by "A(transpose)*" do you mean $(A^T)^* = A^dagger$? Cheers!
  $endgroup$
  – Robert Lewis
  Mar 12 at 16:54
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RobertLewis I think the * is probably supposed to indicate multiplication.
  $endgroup$
  – saulspatz
  Mar 12 at 17:00
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @RobertLewis By that I mean multiplying A transpose with A.
  $endgroup$
  – Kamal Khalaily
  Mar 12 at 17:01
  
  

  
  
  
  

add a comment | 

$begingroup$

Given the matrix of order $1timesn$, $A=(a_1, a_2, ..., a_n)$ , where $a_i$ are real;
The question is to find all eigenvalues of $A^T A$.

I have proved that it is a non-invertible matrix, therefore $0$ is one of the values.
And the product matrix is an $nxn$ matrix with the diagonal elements being $a_1^2, a_2^2,...,a_n^2$.
I am struggling with finding the other eigenvalues, tried by calculating the det of $A - aI$, but didn't go anywhere.

linear-algebra matrices eigenvalues-eigenvectors

share|cite|improve this question

edited Mar 12 at 18:49

6005

37k751127

asked Mar 12 at 16:48

Kamal KhalailyKamal Khalaily

233

New contributor

Kamal Khalaily is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited Mar 12 at 18:49

6005

37k751127

asked Mar 12 at 16:48

Kamal KhalailyKamal Khalaily

233

New contributor

Kamal Khalaily is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited Mar 12 at 18:49

6005

37k751127

asked Mar 12 at 16:48

Kamal KhalailyKamal Khalaily

233

New contributor

Kamal Khalaily is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited Mar 12 at 18:49

6005

37k751127

edited Mar 12 at 18:49

6005

37k751127

asked Mar 12 at 16:48

Kamal KhalailyKamal Khalaily

233

New contributor

Kamal Khalaily is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Mar 12 at 16:48

Kamal KhalailyKamal Khalaily

233

3 Answers
3

active

oldest

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2

$begingroup$

$$A^tA = beginpmatrix
a_1a_1 & a_1a_2 & a_1a_3 & dots & a_1a_n\
a_2a_1 & a_2 a_2 & a_2a_3 & dots & a_2a_n \
vdots & vdots & vdots & vdots & vdots \
a_na_1 & a_na_2 & a_na_3 & dots & a_na_n
endpmatrix$$

All the columns can be obtained by mutiplying by a scalar the first column, so $rank(A^tA)=1$. So $0$ is an eigen value with multiplicity $n-1$. The sum of eigen values is equal to the trace of the matrix, thus you can easily find the last eigenvalue.

share|cite|improve this answer

answered Mar 12 at 17:49

JenniferJennifer

8,60521837

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much; this has been the most helpful!
  $endgroup$
  – Kamal Khalaily
  Mar 12 at 18:07
  

  
  
  
  

add a comment | 

2

$begingroup$

Hint: Suppose that $v$ is an $n times 1$ vector orthogonal to $A$ (or more precisely, $w$ is a $1 times n$ vector orthogonal to $A$, and $v = w^t$). What do you get when you compute $A^*Av$?

share|cite|improve this answer

answered Mar 12 at 17:48

John HughesJohn Hughes

64.6k24191

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for your time! I managed to solve the problem :)
  $endgroup$
  – Kamal Khalaily
  Mar 12 at 18:08
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Consider the definition of an eigenvalue / eigenvector pair,
$$
A^T A v = lambda v.
$$
Now extend $A$ to an orthogonal basis: so let $A_1, A_2, ldots, A_n$ be an orthogonal basis for $mathbbR^n$, where $A_1 = A^T$. Write $v$ in this basis:
$$
v = a_1 A^T + a_2 A_2 + a_3 A_3 + cdots + a_n A_n.
$$

Then,
$$
A^T Av = a_1 A^T A A^T + a_2 A^T A A_2 + a_3 A^T A A_3 + cdots + a_n A^T A A_n.
$$

Because the basis is orthogonal, $A A_k = 0$ for any $k ge 2$. Also $A A^T$ = $|A|^2$. So this just reduces to
$$
A^T A v = a_1 A^T |A|^2.
$$
The only way this can be equal to $lambda v$ is if $lambda = 0$ and the component $a_1 = 0$, or if $lambda = |A|^2$ and the components $a_2, a_3, ldots, a_n$ are all $0$.

So the only eigenvalues are $0$ and $|A|^2$, and we get the corresponding eigenvectors as well.

share|cite|improve this answer

answered Mar 12 at 18:44

60056005

37k751127

$endgroup$

add a comment | 

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2

$begingroup$

$$A^tA = beginpmatrix
a_1a_1 & a_1a_2 & a_1a_3 & dots & a_1a_n\
a_2a_1 & a_2 a_2 & a_2a_3 & dots & a_2a_n \
vdots & vdots & vdots & vdots & vdots \
a_na_1 & a_na_2 & a_na_3 & dots & a_na_n
endpmatrix$$

All the columns can be obtained by mutiplying by a scalar the first column, so $rank(A^tA)=1$. So $0$ is an eigen value with multiplicity $n-1$. The sum of eigen values is equal to the trace of the matrix, thus you can easily find the last eigenvalue.

share|cite|improve this answer

answered Mar 12 at 17:49

JenniferJennifer

8,60521837

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much; this has been the most helpful!
  $endgroup$
  – Kamal Khalaily
  Mar 12 at 18:07
  

  
  
  
  

add a comment | 

2

$begingroup$

$$A^tA = beginpmatrix
a_1a_1 & a_1a_2 & a_1a_3 & dots & a_1a_n\
a_2a_1 & a_2 a_2 & a_2a_3 & dots & a_2a_n \
vdots & vdots & vdots & vdots & vdots \
a_na_1 & a_na_2 & a_na_3 & dots & a_na_n
endpmatrix$$

All the columns can be obtained by mutiplying by a scalar the first column, so $rank(A^tA)=1$. So $0$ is an eigen value with multiplicity $n-1$. The sum of eigen values is equal to the trace of the matrix, thus you can easily find the last eigenvalue.

share|cite|improve this answer

answered Mar 12 at 17:49

JenniferJennifer

8,60521837

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much; this has been the most helpful!
  $endgroup$
  – Kamal Khalaily
  Mar 12 at 18:07
  

  
  
  
  

add a comment | 

$begingroup$

$$A^tA = beginpmatrix
a_1a_1 & a_1a_2 & a_1a_3 & dots & a_1a_n\
a_2a_1 & a_2 a_2 & a_2a_3 & dots & a_2a_n \
vdots & vdots & vdots & vdots & vdots \
a_na_1 & a_na_2 & a_na_3 & dots & a_na_n
endpmatrix$$

All the columns can be obtained by mutiplying by a scalar the first column, so $rank(A^tA)=1$. So $0$ is an eigen value with multiplicity $n-1$. The sum of eigen values is equal to the trace of the matrix, thus you can easily find the last eigenvalue.

share|cite|improve this answer

answered Mar 12 at 17:49

JenniferJennifer

8,60521837

$endgroup$

share|cite|improve this answer

answered Mar 12 at 17:49

JenniferJennifer

8,60521837

answered Mar 12 at 17:49

JenniferJennifer

8,60521837

answered Mar 12 at 17:49

JenniferJennifer

8,60521837

2

$begingroup$

Hint: Suppose that $v$ is an $n times 1$ vector orthogonal to $A$ (or more precisely, $w$ is a $1 times n$ vector orthogonal to $A$, and $v = w^t$). What do you get when you compute $A^*Av$?

share|cite|improve this answer

answered Mar 12 at 17:48

John HughesJohn Hughes

64.6k24191

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for your time! I managed to solve the problem :)
  $endgroup$
  – Kamal Khalaily
  Mar 12 at 18:08
  
  

  
  
  
  

add a comment | 

2

$begingroup$

Hint: Suppose that $v$ is an $n times 1$ vector orthogonal to $A$ (or more precisely, $w$ is a $1 times n$ vector orthogonal to $A$, and $v = w^t$). What do you get when you compute $A^*Av$?

share|cite|improve this answer

answered Mar 12 at 17:48

John HughesJohn Hughes

64.6k24191

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for your time! I managed to solve the problem :)
  $endgroup$
  – Kamal Khalaily
  Mar 12 at 18:08
  
  

  
  
  
  

add a comment | 

$begingroup$

Hint: Suppose that $v$ is an $n times 1$ vector orthogonal to $A$ (or more precisely, $w$ is a $1 times n$ vector orthogonal to $A$, and $v = w^t$). What do you get when you compute $A^*Av$?

share|cite|improve this answer

answered Mar 12 at 17:48

John HughesJohn Hughes

64.6k24191

$endgroup$

share|cite|improve this answer

answered Mar 12 at 17:48

John HughesJohn Hughes

64.6k24191

answered Mar 12 at 17:48

John HughesJohn Hughes

64.6k24191

answered Mar 12 at 17:48

John HughesJohn Hughes

64.6k24191

0

$begingroup$

Consider the definition of an eigenvalue / eigenvector pair,
$$
A^T A v = lambda v.
$$
Now extend $A$ to an orthogonal basis: so let $A_1, A_2, ldots, A_n$ be an orthogonal basis for $mathbbR^n$, where $A_1 = A^T$. Write $v$ in this basis:
$$
v = a_1 A^T + a_2 A_2 + a_3 A_3 + cdots + a_n A_n.
$$

Then,
$$
A^T Av = a_1 A^T A A^T + a_2 A^T A A_2 + a_3 A^T A A_3 + cdots + a_n A^T A A_n.
$$

Because the basis is orthogonal, $A A_k = 0$ for any $k ge 2$. Also $A A^T$ = $|A|^2$. So this just reduces to
$$
A^T A v = a_1 A^T |A|^2.
$$
The only way this can be equal to $lambda v$ is if $lambda = 0$ and the component $a_1 = 0$, or if $lambda = |A|^2$ and the components $a_2, a_3, ldots, a_n$ are all $0$.

So the only eigenvalues are $0$ and $|A|^2$, and we get the corresponding eigenvectors as well.

share|cite|improve this answer

answered Mar 12 at 18:44

60056005

37k751127

$endgroup$

add a comment | 

0

$begingroup$

Consider the definition of an eigenvalue / eigenvector pair,
$$
A^T A v = lambda v.
$$
Now extend $A$ to an orthogonal basis: so let $A_1, A_2, ldots, A_n$ be an orthogonal basis for $mathbbR^n$, where $A_1 = A^T$. Write $v$ in this basis:
$$
v = a_1 A^T + a_2 A_2 + a_3 A_3 + cdots + a_n A_n.
$$

Then,
$$
A^T Av = a_1 A^T A A^T + a_2 A^T A A_2 + a_3 A^T A A_3 + cdots + a_n A^T A A_n.
$$

Because the basis is orthogonal, $A A_k = 0$ for any $k ge 2$. Also $A A^T$ = $|A|^2$. So this just reduces to
$$
A^T A v = a_1 A^T |A|^2.
$$
The only way this can be equal to $lambda v$ is if $lambda = 0$ and the component $a_1 = 0$, or if $lambda = |A|^2$ and the components $a_2, a_3, ldots, a_n$ are all $0$.

So the only eigenvalues are $0$ and $|A|^2$, and we get the corresponding eigenvectors as well.

share|cite|improve this answer

answered Mar 12 at 18:44

60056005

37k751127

$endgroup$

add a comment | 

$begingroup$

Consider the definition of an eigenvalue / eigenvector pair,
$$
A^T A v = lambda v.
$$
Now extend $A$ to an orthogonal basis: so let $A_1, A_2, ldots, A_n$ be an orthogonal basis for $mathbbR^n$, where $A_1 = A^T$. Write $v$ in this basis:
$$
v = a_1 A^T + a_2 A_2 + a_3 A_3 + cdots + a_n A_n.
$$

Then,
$$
A^T Av = a_1 A^T A A^T + a_2 A^T A A_2 + a_3 A^T A A_3 + cdots + a_n A^T A A_n.
$$

Because the basis is orthogonal, $A A_k = 0$ for any $k ge 2$. Also $A A^T$ = $|A|^2$. So this just reduces to
$$
A^T A v = a_1 A^T |A|^2.
$$
The only way this can be equal to $lambda v$ is if $lambda = 0$ and the component $a_1 = 0$, or if $lambda = |A|^2$ and the components $a_2, a_3, ldots, a_n$ are all $0$.

So the only eigenvalues are $0$ and $|A|^2$, and we get the corresponding eigenvectors as well.

share|cite|improve this answer

answered Mar 12 at 18:44

60056005

37k751127

$endgroup$

share|cite|improve this answer

answered Mar 12 at 18:44

60056005

37k751127

answered Mar 12 at 18:44

60056005

37k751127

answered Mar 12 at 18:44

60056005

37k751127