If the multiplication of a column independent matrix and its pseudo-inverse is positive definition?
$begingroup$
Let's consider a matrix J has the size of m by n (m>n) and all columns are linear independent. Therefore, $J^TJ$ is invertible. Then, the Moore–Penrose inverse of J is $J^+=(J^TJ)^{-1}J^T$. Obviously, there is $J^+J=I$.
My question is if $JJ^+$ is positive definition?
I test this in MATALB software and find this is true for several cases, but how to prove it rigorously?
Thank you!
linear-algebra
asked Dec 9 '18 at 3:04
DavidDavid
32
$endgroup$
add a comment |
$begingroup$
Let's consider a matrix J has the size of m by n (m>n) and all columns are linear independent. Therefore, $J^TJ$ is invertible. Then, the Moore–Penrose inverse of J is $J^+=(J^TJ)^{-1}J^T$. Obviously, there is $J^+J=I$.
My question is if $JJ^+$ is positive definition?
I test this in MATALB software and find this is true for several cases, but how to prove it rigorously?
Thank you!
linear-algebra
asked Dec 9 '18 at 3:04
DavidDavid
32
$endgroup$
add a comment |
$begingroup$
Let's consider a matrix J has the size of m by n (m>n) and all columns are linear independent. Therefore, $J^TJ$ is invertible. Then, the Moore–Penrose inverse of J is $J^+=(J^TJ)^{-1}J^T$. Obviously, there is $J^+J=I$.
My question is if $JJ^+$ is positive definition?
I test this in MATALB software and find this is true for several cases, but how to prove it rigorously?
Thank you!
linear-algebra
asked Dec 9 '18 at 3:04
DavidDavid
32
$endgroup$
Let's consider a matrix J has the size of m by n (m>n) and all columns are linear independent. Therefore, $J^TJ$ is invertible. Then, the Moore–Penrose inverse of J is $J^+=(J^TJ)^{-1}J^T$. Obviously, there is $J^+J=I$.
My question is if $JJ^+$ is positive definition?
I test this in MATALB software and find this is true for several cases, but how to prove it rigorously?
Thank you!
linear-algebra
linear-algebra
asked Dec 9 '18 at 3:04
DavidDavid
32
asked Dec 9 '18 at 3:04
DavidDavid
32
edited Dec 9 '18 at 4:34
David
asked Dec 9 '18 at 3:04
DavidDavid
32
asked Dec 9 '18 at 3:04
DavidDavid
32
asked Dec 9 '18 at 3:04
DavidDavid
32
32
add a comment |
add a comment |
1 Answer
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Welcome to the Math SE.
$JJ^+$ is not positive definite, but it is positive semi-definite, since it is not full rank. To see this first observe that if $P:=JJ^+$ then $P^2=P$, hence it is a projection matrix. A property of these matrices is that they have eigenvalues of either $1$ or $0$, which is easy to prove. Also, $P$ is symmetric, ie. $P=P^T$. These two facts are enough to conclude that $JJ^+$ is positive semi-definite.
answered Dec 9 '18 at 21:28
obareeyobareey
2,99411028
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$begingroup$
Welcome to the Math SE.
$JJ^+$ is not positive definite, but it is positive semi-definite, since it is not full rank. To see this first observe that if $P:=JJ^+$ then $P^2=P$, hence it is a projection matrix. A property of these matrices is that they have eigenvalues of either $1$ or $0$, which is easy to prove. Also, $P$ is symmetric, ie. $P=P^T$. These two facts are enough to conclude that $JJ^+$ is positive semi-definite.
answered Dec 9 '18 at 21:28
obareeyobareey
2,99411028
$endgroup$
add a comment |
$begingroup$
Welcome to the Math SE.
$JJ^+$ is not positive definite, but it is positive semi-definite, since it is not full rank. To see this first observe that if $P:=JJ^+$ then $P^2=P$, hence it is a projection matrix. A property of these matrices is that they have eigenvalues of either $1$ or $0$, which is easy to prove. Also, $P$ is symmetric, ie. $P=P^T$. These two facts are enough to conclude that $JJ^+$ is positive semi-definite.
answered Dec 9 '18 at 21:28
obareeyobareey
2,99411028
$endgroup$
add a comment |
$begingroup$
Welcome to the Math SE.
$JJ^+$ is not positive definite, but it is positive semi-definite, since it is not full rank. To see this first observe that if $P:=JJ^+$ then $P^2=P$, hence it is a projection matrix. A property of these matrices is that they have eigenvalues of either $1$ or $0$, which is easy to prove. Also, $P$ is symmetric, ie. $P=P^T$. These two facts are enough to conclude that $JJ^+$ is positive semi-definite.
answered Dec 9 '18 at 21:28
obareeyobareey
2,99411028
$endgroup$
Welcome to the Math SE.
$JJ^+$ is not positive definite, but it is positive semi-definite, since it is not full rank. To see this first observe that if $P:=JJ^+$ then $P^2=P$, hence it is a projection matrix. A property of these matrices is that they have eigenvalues of either $1$ or $0$, which is easy to prove. Also, $P$ is symmetric, ie. $P=P^T$. These two facts are enough to conclude that $JJ^+$ is positive semi-definite.
answered Dec 9 '18 at 21:28
obareeyobareey
2,99411028
answered Dec 9 '18 at 21:28
obareeyobareey
2,99411028
answered Dec 9 '18 at 21:28
obareeyobareey
2,99411028
answered Dec 9 '18 at 21:28
obareeyobareey
2,99411028
2,99411028
add a comment |
add a comment |
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