Computing the spectral norm of a projection matrix The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Is the rank of a matrix the same of its transpose? If yes, how can I prove it?Convexity of the squared Frobenius norm of a matrixTwo norm of the identity matrixMatrix norm proofLipschitz continuity for generalized inverse matrixMatrix factorization by a full row rank matrix in MATLABBounding the matrix $2$-norm of a Frobenius matrixUpper bound on norm of linear projection on column space of $X$Frobenius Norm Inequality; Spectral Radius is smaller than Frobenius NormProof of infinity matrix norm
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Computing the spectral norm of a projection matrix
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Is the rank of a matrix the same of its transpose? If yes, how can I prove it?Convexity of the squared Frobenius norm of a matrixTwo norm of the identity matrixMatrix norm proofLipschitz continuity for generalized inverse matrixMatrix factorization by a full row rank matrix in MATLABBounding the matrix $2$-norm of a Frobenius matrixUpper bound on norm of linear projection on column space of $X$Frobenius Norm Inequality; Spectral Radius is smaller than Frobenius NormProof of infinity matrix norm
$begingroup$
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
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
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
edited Mar 31 at 7:42
Rodrigo de Azevedo
13.2k41962
asked Jul 24 '18 at 10:02
Majid MohammadiMajid Mohammadi
275
$endgroup$
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
linear-algebra matrices projection-matrices matrix-norms spectral-norm
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
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
edited Mar 31 at 7:42
Rodrigo de Azevedo
13.2k41962
13.2k41962
asked Jul 24 '18 at 10:02
Majid MohammadiMajid Mohammadi
275
asked Jul 24 '18 at 10:02
Majid MohammadiMajid Mohammadi
275
asked Jul 24 '18 at 10:02
Majid MohammadiMajid Mohammadi
275
275
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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,
$$| rm P |_2 = sigma_max (rm P) = lambda_max (rm P) = 1$$
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 |
$begingroup$
We recognize in that expression a projection matrix onto $operatornameRow(A)$ and since $A$ is a full row rank we have that
$$P=A^T(AA^T)^-1A implies supleftfrac,, xneq 0right=1$$
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 |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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,
$$| rm P |_2 = sigma_max (rm P) = lambda_max (rm P) = 1$$
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,
$$| rm P |_2 = sigma_max (rm P) = lambda_max (rm P) = 1$$
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,
$$| rm P |_2 = sigma_max (rm P) = lambda_max (rm P) = 1$$
answered Jul 24 '18 at 10:23
Rodrigo de AzevedoRodrigo de Azevedo
13.2k41962
$endgroup$
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,
$$| rm P |_2 = sigma_max (rm P) = lambda_max (rm P) = 1$$
answered Jul 24 '18 at 10:23
Rodrigo de AzevedoRodrigo de Azevedo
13.2k41962
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
13.2k41962
$begingroup$
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 |
$begingroup$
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
$begingroup$
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
$begingroup$
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
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
$begingroup$
@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 |
$begingroup$
We recognize in that expression a projection matrix onto $operatornameRow(A)$ and since $A$ is a full row rank we have that
$$P=A^T(AA^T)^-1A implies supleftfrac,, xneq 0right=1$$
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
$$P=A^T(AA^T)^-1A implies supleftfrac,, xneq 0right=1$$
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
$$P=A^T(AA^T)^-1A implies supleftfrac,, xneq 0right=1$$
answered Jul 24 '18 at 10:07
gimusigimusi
92.9k84594
$endgroup$
We recognize in that expression a projection matrix onto $operatornameRow(A)$ and since $A$ is a full row rank we have that
$$P=A^T(AA^T)^-1A implies supleftfrac,, xneq 0right=1$$
answered Jul 24 '18 at 10:07
gimusigimusi
92.9k84594
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
92.9k84594
$begingroup$
@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 |
$begingroup$
@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
$begingroup$
@RodrigodeAzevedo A is declared a full rank matrix.
$endgroup$
– gimusi
Jul 24 '18 at 10:17
$begingroup$
@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
$begingroup$
@RodrigodeAzevedo Opsssss...Thanks!
$endgroup$
– gimusi
Jul 24 '18 at 10:20
add a comment |
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