Pseudoinverse of this matrix
If the matrix $U in R^{n times k}$ satisfies $UU^* = I$, how can I find pseudoniverse of $U^*U$ ?
I have tried solving Moore Penrose equations for pseudoinverse, but it didnt help.
linear-algebra matrices
asked Nov 21 '18 at 16:39
Studying Optimization
676
add a comment |
If the matrix $U in R^{n times k}$ satisfies $UU^* = I$, how can I find pseudoniverse of $U^*U$ ?
I have tried solving Moore Penrose equations for pseudoinverse, but it didnt help.
linear-algebra matrices
asked Nov 21 '18 at 16:39
Studying Optimization
676
Use Singular Value Decomposition.
– Jean Marie
Nov 21 '18 at 17:11
@JeanMarie Could you please elaborate on that?
– Studying Optimization
Nov 21 '18 at 17:17
A simple, direct, explanation : johndcook.com/blog/2018/05/05/svd about the connection between the two keywords ("pseudo-inverse" and SVD). But for your specific problem, sorry but I have no idea.
– Jean Marie
Nov 21 '18 at 17:24
add a comment |
If the matrix $U in R^{n times k}$ satisfies $UU^* = I$, how can I find pseudoniverse of $U^*U$ ?
I have tried solving Moore Penrose equations for pseudoinverse, but it didnt help.
linear-algebra matrices
asked Nov 21 '18 at 16:39
Studying Optimization
676
If the matrix $U in R^{n times k}$ satisfies $UU^* = I$, how can I find pseudoniverse of $U^*U$ ?
I have tried solving Moore Penrose equations for pseudoinverse, but it didnt help.
linear-algebra matrices
linear-algebra matrices
asked Nov 21 '18 at 16:39
Studying Optimization
676
asked Nov 21 '18 at 16:39
Studying Optimization
676
asked Nov 21 '18 at 16:39
Studying Optimization
676
asked Nov 21 '18 at 16:39
Studying Optimization
676
asked Nov 21 '18 at 16:39
Studying Optimization
676
676
Use Singular Value Decomposition.
– Jean Marie
Nov 21 '18 at 17:11
@JeanMarie Could you please elaborate on that?
– Studying Optimization
Nov 21 '18 at 17:17
A simple, direct, explanation : johndcook.com/blog/2018/05/05/svd about the connection between the two keywords ("pseudo-inverse" and SVD). But for your specific problem, sorry but I have no idea.
– Jean Marie
Nov 21 '18 at 17:24
add a comment |
Use Singular Value Decomposition.
– Jean Marie
Nov 21 '18 at 17:11
@JeanMarie Could you please elaborate on that?
– Studying Optimization
Nov 21 '18 at 17:17
A simple, direct, explanation : johndcook.com/blog/2018/05/05/svd about the connection between the two keywords ("pseudo-inverse" and SVD). But for your specific problem, sorry but I have no idea.
– Jean Marie
Nov 21 '18 at 17:24
Use Singular Value Decomposition.
– Jean Marie
Nov 21 '18 at 17:11
Use Singular Value Decomposition.
– Jean Marie
Nov 21 '18 at 17:11
@JeanMarie Could you please elaborate on that?
– Studying Optimization
Nov 21 '18 at 17:17
@JeanMarie Could you please elaborate on that?
– Studying Optimization
Nov 21 '18 at 17:17
A simple, direct, explanation : johndcook.com/blog/2018/05/05/svd about the connection between the two keywords ("pseudo-inverse" and SVD). But for your specific problem, sorry but I have no idea.
– Jean Marie
Nov 21 '18 at 17:24
A simple, direct, explanation : johndcook.com/blog/2018/05/05/svd about the connection between the two keywords ("pseudo-inverse" and SVD). But for your specific problem, sorry but I have no idea.
– Jean Marie
Nov 21 '18 at 17:24
add a comment |
1 Answer
1
active
oldest
votes
The pseudo inverse of $U^*U$ is itself.
Proof. Clearly, when $kgeq 1$, $(U^*U)^k=U^*U$.
On the other hand, $A^+$, the pseudo inverse of $A$, is uniquely defined by the relations
$AA^+A=A,A^+AA^+=A^+,(AA^+)^*=AA^+,(A^+A)^*=A^+A$.
It is easy to see that $U^*U$ satisfies the required relations.
answered Nov 21 '18 at 19:22
loup blanc
22.5k21750
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
The pseudo inverse of $U^*U$ is itself.
Proof. Clearly, when $kgeq 1$, $(U^*U)^k=U^*U$.
On the other hand, $A^+$, the pseudo inverse of $A$, is uniquely defined by the relations
$AA^+A=A,A^+AA^+=A^+,(AA^+)^*=AA^+,(A^+A)^*=A^+A$.
It is easy to see that $U^*U$ satisfies the required relations.
answered Nov 21 '18 at 19:22
loup blanc
22.5k21750
add a comment |
The pseudo inverse of $U^*U$ is itself.
Proof. Clearly, when $kgeq 1$, $(U^*U)^k=U^*U$.
On the other hand, $A^+$, the pseudo inverse of $A$, is uniquely defined by the relations
$AA^+A=A,A^+AA^+=A^+,(AA^+)^*=AA^+,(A^+A)^*=A^+A$.
It is easy to see that $U^*U$ satisfies the required relations.
answered Nov 21 '18 at 19:22
loup blanc
22.5k21750
add a comment |
The pseudo inverse of $U^*U$ is itself.
Proof. Clearly, when $kgeq 1$, $(U^*U)^k=U^*U$.
On the other hand, $A^+$, the pseudo inverse of $A$, is uniquely defined by the relations
$AA^+A=A,A^+AA^+=A^+,(AA^+)^*=AA^+,(A^+A)^*=A^+A$.
It is easy to see that $U^*U$ satisfies the required relations.
answered Nov 21 '18 at 19:22
loup blanc
22.5k21750
The pseudo inverse of $U^*U$ is itself.
Proof. Clearly, when $kgeq 1$, $(U^*U)^k=U^*U$.
On the other hand, $A^+$, the pseudo inverse of $A$, is uniquely defined by the relations
$AA^+A=A,A^+AA^+=A^+,(AA^+)^*=AA^+,(A^+A)^*=A^+A$.
It is easy to see that $U^*U$ satisfies the required relations.
answered Nov 21 '18 at 19:22
loup blanc
22.5k21750
answered Nov 21 '18 at 19:22
loup blanc
22.5k21750
answered Nov 21 '18 at 19:22
loup blanc
22.5k21750
answered Nov 21 '18 at 19:22
loup blanc
22.5k21750
22.5k21750
add a comment |
add a comment |
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Use Singular Value Decomposition.
– Jean Marie
Nov 21 '18 at 17:11
@JeanMarie Could you please elaborate on that?
– Studying Optimization
Nov 21 '18 at 17:17
A simple, direct, explanation : johndcook.com/blog/2018/05/05/svd about the connection between the two keywords ("pseudo-inverse" and SVD). But for your specific problem, sorry but I have no idea.
– Jean Marie
Nov 21 '18 at 17:24