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Computing the spectral norm of a projection matrix

1

$begingroup$

I was reading a paper in which there was an argument as trivial, but could not make myself sure about it. It is said that given a full row-rank matrix $A$, the norm (probably $ell_2$-induced matrix norm) of $A^T(AA^T)^-1A$ is one. Is that trivial, and correct for any given matrix $A$?

linear-algebra matrices projection-matrices matrix-norms spectral-norm

share|cite|improve this question

edited Mar 31 at 7:42

Rodrigo de Azevedo

13.2k41962

asked Jul 24 '18 at 10:02

Majid MohammadiMajid Mohammadi

275

$endgroup$

add a comment | 

1

$begingroup$

I was reading a paper in which there was an argument as trivial, but could not make myself sure about it. It is said that given a full row-rank matrix $A$, the norm (probably $ell_2$-induced matrix norm) of $A^T(AA^T)^-1A$ is one. Is that trivial, and correct for any given matrix $A$?

linear-algebra matrices projection-matrices matrix-norms spectral-norm

share|cite|improve this question

edited Mar 31 at 7:42

Rodrigo de Azevedo

13.2k41962

asked Jul 24 '18 at 10:02

Majid MohammadiMajid Mohammadi

275

$endgroup$

add a comment | 

$begingroup$

I was reading a paper in which there was an argument as trivial, but could not make myself sure about it. It is said that given a full row-rank matrix $A$, the norm (probably $ell_2$-induced matrix norm) of $A^T(AA^T)^-1A$ is one. Is that trivial, and correct for any given matrix $A$?

linear-algebra matrices projection-matrices matrix-norms spectral-norm

share|cite|improve this question

edited Mar 31 at 7:42

Rodrigo de Azevedo

13.2k41962

asked Jul 24 '18 at 10:02

Majid MohammadiMajid Mohammadi

275

$endgroup$

share|cite|improve this question

edited Mar 31 at 7:42

Rodrigo de Azevedo

13.2k41962

asked Jul 24 '18 at 10:02

Majid MohammadiMajid Mohammadi

275

share|cite|improve this question

edited Mar 31 at 7:42

Rodrigo de Azevedo

13.2k41962

asked Jul 24 '18 at 10:02

Majid MohammadiMajid Mohammadi

275

edited Mar 31 at 7:42

Rodrigo de Azevedo

13.2k41962

edited Mar 31 at 7:42

Rodrigo de Azevedo

13.2k41962

asked Jul 24 '18 at 10:02

Majid MohammadiMajid Mohammadi

275

asked Jul 24 '18 at 10:02

Majid MohammadiMajid Mohammadi

275

2 Answers
2

active

oldest

votes

2

$begingroup$

Given a fat matrix $rm A$ with full row rank,

$$rm P := A^top (A A^top)^-1 A$$

is the (symmetric) projection matrix that projects onto the row space of $rm A$. Every projection matrix has eigenvalues $0$ and $1$. Since $rm P$ is symmetric and positive semidefinite,

$$| rm P |_2 = sigma_max (rm P) = lambda_max (rm P) = 1$$

share|cite|improve this answer

edited Jul 24 '18 at 10:44

answered Jul 24 '18 at 10:23

Rodrigo de AzevedoRodrigo de Azevedo

13.2k41962

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  what's the relation with this please? looks like transpose since it's instead $A(A^TA)^-1A^T$
  $endgroup$
  – BCLC
  Jul 24 '18 at 11:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @BCLC Your projection matrix requires full column rank and projects onto the column space of $rm A$. Note that the column space of $rm A$ is the row space of $rm A^top$.
  $endgroup$
  – Rodrigo de Azevedo
  Jul 24 '18 at 11:34
  
  

  
  
  
  

add a comment | 

0

$begingroup$

We recognize in that expression a projection matrix onto $operatornameRow(A)$ and since $A$ is a full row rank we have that

$$P=A^T(AA^T)^-1A implies supleftfrac,, xneq 0right=1$$

share|cite|improve this answer

edited Jul 24 '18 at 10:26

answered Jul 24 '18 at 10:07

gimusigimusi

92.9k84594

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RodrigodeAzevedo A is declared a full rank matrix.
  $endgroup$
  – gimusi
  Jul 24 '18 at 10:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RodrigodeAzevedo Opsssss...Thanks!
  $endgroup$
  – gimusi
  Jul 24 '18 at 10:20
  

  
  
  
  

add a comment | 

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2

$begingroup$

Given a fat matrix $rm A$ with full row rank,

$$rm P := A^top (A A^top)^-1 A$$

is the (symmetric) projection matrix that projects onto the row space of $rm A$. Every projection matrix has eigenvalues $0$ and $1$. Since $rm P$ is symmetric and positive semidefinite,

$$| rm P |_2 = sigma_max (rm P) = lambda_max (rm P) = 1$$

share|cite|improve this answer

edited Jul 24 '18 at 10:44

answered Jul 24 '18 at 10:23

Rodrigo de AzevedoRodrigo de Azevedo

13.2k41962

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  what's the relation with this please? looks like transpose since it's instead $A(A^TA)^-1A^T$
  $endgroup$
  – BCLC
  Jul 24 '18 at 11:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @BCLC Your projection matrix requires full column rank and projects onto the column space of $rm A$. Note that the column space of $rm A$ is the row space of $rm A^top$.
  $endgroup$
  – Rodrigo de Azevedo
  Jul 24 '18 at 11:34
  
  

  
  
  
  

add a comment | 

2

$begingroup$

Given a fat matrix $rm A$ with full row rank,

$$rm P := A^top (A A^top)^-1 A$$

is the (symmetric) projection matrix that projects onto the row space of $rm A$. Every projection matrix has eigenvalues $0$ and $1$. Since $rm P$ is symmetric and positive semidefinite,

$$| rm P |_2 = sigma_max (rm P) = lambda_max (rm P) = 1$$

share|cite|improve this answer

edited Jul 24 '18 at 10:44

answered Jul 24 '18 at 10:23

Rodrigo de AzevedoRodrigo de Azevedo

13.2k41962

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  what's the relation with this please? looks like transpose since it's instead $A(A^TA)^-1A^T$
  $endgroup$
  – BCLC
  Jul 24 '18 at 11:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @BCLC Your projection matrix requires full column rank and projects onto the column space of $rm A$. Note that the column space of $rm A$ is the row space of $rm A^top$.
  $endgroup$
  – Rodrigo de Azevedo
  Jul 24 '18 at 11:34
  
  

  
  
  
  

add a comment | 

$begingroup$

Given a fat matrix $rm A$ with full row rank,

$$rm P := A^top (A A^top)^-1 A$$

is the (symmetric) projection matrix that projects onto the row space of $rm A$. Every projection matrix has eigenvalues $0$ and $1$. Since $rm P$ is symmetric and positive semidefinite,

$$| rm P |_2 = sigma_max (rm P) = lambda_max (rm P) = 1$$

share|cite|improve this answer

edited Jul 24 '18 at 10:44

answered Jul 24 '18 at 10:23

Rodrigo de AzevedoRodrigo de Azevedo

13.2k41962

$endgroup$

share|cite|improve this answer

edited Jul 24 '18 at 10:44

answered Jul 24 '18 at 10:23

Rodrigo de AzevedoRodrigo de Azevedo

13.2k41962

answered Jul 24 '18 at 10:23

Rodrigo de AzevedoRodrigo de Azevedo

13.2k41962

answered Jul 24 '18 at 10:23

Rodrigo de AzevedoRodrigo de Azevedo

13.2k41962

0

$begingroup$

We recognize in that expression a projection matrix onto $operatornameRow(A)$ and since $A$ is a full row rank we have that

$$P=A^T(AA^T)^-1A implies supleftfrac,, xneq 0right=1$$

share|cite|improve this answer

edited Jul 24 '18 at 10:26

answered Jul 24 '18 at 10:07

gimusigimusi

92.9k84594

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RodrigodeAzevedo A is declared a full rank matrix.
  $endgroup$
  – gimusi
  Jul 24 '18 at 10:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RodrigodeAzevedo Opsssss...Thanks!
  $endgroup$
  – gimusi
  Jul 24 '18 at 10:20
  

  
  
  
  

add a comment | 

0

$begingroup$

We recognize in that expression a projection matrix onto $operatornameRow(A)$ and since $A$ is a full row rank we have that

$$P=A^T(AA^T)^-1A implies supleftfrac,, xneq 0right=1$$

share|cite|improve this answer

edited Jul 24 '18 at 10:26

answered Jul 24 '18 at 10:07

gimusigimusi

92.9k84594

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RodrigodeAzevedo A is declared a full rank matrix.
  $endgroup$
  – gimusi
  Jul 24 '18 at 10:17
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RodrigodeAzevedo Opsssss...Thanks!
  $endgroup$
  – gimusi
  Jul 24 '18 at 10:20
  

  
  
  
  

add a comment | 

$begingroup$

We recognize in that expression a projection matrix onto $operatornameRow(A)$ and since $A$ is a full row rank we have that

$$P=A^T(AA^T)^-1A implies supleftfrac,, xneq 0right=1$$

share|cite|improve this answer

edited Jul 24 '18 at 10:26

answered Jul 24 '18 at 10:07

gimusigimusi

92.9k84594

$endgroup$

share|cite|improve this answer

edited Jul 24 '18 at 10:26

answered Jul 24 '18 at 10:07

gimusigimusi

92.9k84594

answered Jul 24 '18 at 10:07

gimusigimusi

92.9k84594

answered Jul 24 '18 at 10:07

gimusigimusi

92.9k84594