A matrix inequality involving pseudoinverse
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I am trying to solve a problem in the context of signal processing, which leads to the following question. Consider two positive definite Hermitian matrices $A$ and $B$, and a full cloumn rank matrix $W$ with more rows than columns. Assume $B succ A$, which means $B-A$ is positive definite. Let $C=AW$, and $D=BW$. I want to prove the following inequality $$CC^+ succ DD^+.$$ Here $C^+$ refers to $C^+=(C^HC)^{-1}C^H$. However, I didn't find any good approach to this proof. Is this inequality true? In addition, I also wonder if other necessary conditions on the involving matrices are needed.
linear-algebra matrix-analysis
asked Dec 25 '18 at 11:19
abao5887abao5887
133
$endgroup$
add a comment |
$begingroup$
I am trying to solve a problem in the context of signal processing, which leads to the following question. Consider two positive definite Hermitian matrices $A$ and $B$, and a full cloumn rank matrix $W$ with more rows than columns. Assume $B succ A$, which means $B-A$ is positive definite. Let $C=AW$, and $D=BW$. I want to prove the following inequality $$CC^+ succ DD^+.$$ Here $C^+$ refers to $C^+=(C^HC)^{-1}C^H$. However, I didn't find any good approach to this proof. Is this inequality true? In addition, I also wonder if other necessary conditions on the involving matrices are needed.
linear-algebra matrix-analysis
asked Dec 25 '18 at 11:19
abao5887abao5887
133
$endgroup$
$begingroup$
Thanks for this information, @AVK. I forgot to mention that W is not a square matrix. Now I've added it into the question statement.
$endgroup$
– abao5887
Dec 25 '18 at 12:10
add a comment |
$begingroup$
I am trying to solve a problem in the context of signal processing, which leads to the following question. Consider two positive definite Hermitian matrices $A$ and $B$, and a full cloumn rank matrix $W$ with more rows than columns. Assume $B succ A$, which means $B-A$ is positive definite. Let $C=AW$, and $D=BW$. I want to prove the following inequality $$CC^+ succ DD^+.$$ Here $C^+$ refers to $C^+=(C^HC)^{-1}C^H$. However, I didn't find any good approach to this proof. Is this inequality true? In addition, I also wonder if other necessary conditions on the involving matrices are needed.
linear-algebra matrix-analysis
asked Dec 25 '18 at 11:19
abao5887abao5887
133
$endgroup$
I am trying to solve a problem in the context of signal processing, which leads to the following question. Consider two positive definite Hermitian matrices $A$ and $B$, and a full cloumn rank matrix $W$ with more rows than columns. Assume $B succ A$, which means $B-A$ is positive definite. Let $C=AW$, and $D=BW$. I want to prove the following inequality $$CC^+ succ DD^+.$$ Here $C^+$ refers to $C^+=(C^HC)^{-1}C^H$. However, I didn't find any good approach to this proof. Is this inequality true? In addition, I also wonder if other necessary conditions on the involving matrices are needed.
linear-algebra matrix-analysis
linear-algebra matrix-analysis
asked Dec 25 '18 at 11:19
abao5887abao5887
133
asked Dec 25 '18 at 11:19
abao5887abao5887
133
edited Dec 25 '18 at 12:41
abao5887
asked Dec 25 '18 at 11:19
abao5887abao5887
133
asked Dec 25 '18 at 11:19
abao5887abao5887
133
asked Dec 25 '18 at 11:19
abao5887abao5887
133
133
$begingroup$
Thanks for this information, @AVK. I forgot to mention that W is not a square matrix. Now I've added it into the question statement.
$endgroup$
– abao5887
Dec 25 '18 at 12:10
add a comment |
$begingroup$
Thanks for this information, @AVK. I forgot to mention that W is not a square matrix. Now I've added it into the question statement.
$endgroup$
– abao5887
Dec 25 '18 at 12:10
$begingroup$
Thanks for this information, @AVK. I forgot to mention that W is not a square matrix. Now I've added it into the question statement.
$endgroup$
– abao5887
Dec 25 '18 at 12:10
$begingroup$
Thanks for this information, @AVK. I forgot to mention that W is not a square matrix. Now I've added it into the question statement.
$endgroup$
– abao5887
Dec 25 '18 at 12:10
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Consider the case
$$
A=left(begin{array}{cc}
2&0\0&2
end{array}right),quad
B=left(begin{array}{cc}
1&0\0&1
end{array}right),quad
W=left(begin{array}{c}
1\0
end{array}right).
$$
We have
$$
C=AW=left(begin{array}{c}
2\0
end{array}right),quad
D=BW=left(begin{array}{c}
1\0
end{array}right),
$$
$$
C^{+}=left(begin{array}{cc}
frac12 & 0
end{array}right),quad
D^{+}=left(begin{array}{cc}
1&0
end{array}right),
$$
$$
CC^{+}=DD^{+}=left(begin{array}{cc}
1&0\0&0
end{array}right).
$$
answered Dec 25 '18 at 12:35
AVKAVK
2,1111517
$endgroup$
$begingroup$
Thanks for this example. Actually, I found that there was something wrong in the previous question, the condition should be $B succ A$. Afterall, this example still illustrates the problem. Furthermore, I might trace back to the proof of a more general question, which is the origin of the stated one, see (math.stackexchange.com/questions/3052118/…).
$endgroup$
– abao5887
Dec 25 '18 at 13:52
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Consider the case
$$
A=left(begin{array}{cc}
2&0\0&2
end{array}right),quad
B=left(begin{array}{cc}
1&0\0&1
end{array}right),quad
W=left(begin{array}{c}
1\0
end{array}right).
$$
We have
$$
C=AW=left(begin{array}{c}
2\0
end{array}right),quad
D=BW=left(begin{array}{c}
1\0
end{array}right),
$$
$$
C^{+}=left(begin{array}{cc}
frac12 & 0
end{array}right),quad
D^{+}=left(begin{array}{cc}
1&0
end{array}right),
$$
$$
CC^{+}=DD^{+}=left(begin{array}{cc}
1&0\0&0
end{array}right).
$$
answered Dec 25 '18 at 12:35
AVKAVK
2,1111517
$endgroup$
$begingroup$
Thanks for this example. Actually, I found that there was something wrong in the previous question, the condition should be $B succ A$. Afterall, this example still illustrates the problem. Furthermore, I might trace back to the proof of a more general question, which is the origin of the stated one, see (math.stackexchange.com/questions/3052118/…).
$endgroup$
– abao5887
Dec 25 '18 at 13:52
add a comment |
$begingroup$
Consider the case
$$
A=left(begin{array}{cc}
2&0\0&2
end{array}right),quad
B=left(begin{array}{cc}
1&0\0&1
end{array}right),quad
W=left(begin{array}{c}
1\0
end{array}right).
$$
We have
$$
C=AW=left(begin{array}{c}
2\0
end{array}right),quad
D=BW=left(begin{array}{c}
1\0
end{array}right),
$$
$$
C^{+}=left(begin{array}{cc}
frac12 & 0
end{array}right),quad
D^{+}=left(begin{array}{cc}
1&0
end{array}right),
$$
$$
CC^{+}=DD^{+}=left(begin{array}{cc}
1&0\0&0
end{array}right).
$$
answered Dec 25 '18 at 12:35
AVKAVK
2,1111517
$endgroup$
$begingroup$
Thanks for this example. Actually, I found that there was something wrong in the previous question, the condition should be $B succ A$. Afterall, this example still illustrates the problem. Furthermore, I might trace back to the proof of a more general question, which is the origin of the stated one, see (math.stackexchange.com/questions/3052118/…).
$endgroup$
– abao5887
Dec 25 '18 at 13:52
add a comment |
$begingroup$
Consider the case
$$
A=left(begin{array}{cc}
2&0\0&2
end{array}right),quad
B=left(begin{array}{cc}
1&0\0&1
end{array}right),quad
W=left(begin{array}{c}
1\0
end{array}right).
$$
We have
$$
C=AW=left(begin{array}{c}
2\0
end{array}right),quad
D=BW=left(begin{array}{c}
1\0
end{array}right),
$$
$$
C^{+}=left(begin{array}{cc}
frac12 & 0
end{array}right),quad
D^{+}=left(begin{array}{cc}
1&0
end{array}right),
$$
$$
CC^{+}=DD^{+}=left(begin{array}{cc}
1&0\0&0
end{array}right).
$$
answered Dec 25 '18 at 12:35
AVKAVK
2,1111517
$endgroup$
Consider the case
$$
A=left(begin{array}{cc}
2&0\0&2
end{array}right),quad
B=left(begin{array}{cc}
1&0\0&1
end{array}right),quad
W=left(begin{array}{c}
1\0
end{array}right).
$$
We have
$$
C=AW=left(begin{array}{c}
2\0
end{array}right),quad
D=BW=left(begin{array}{c}
1\0
end{array}right),
$$
$$
C^{+}=left(begin{array}{cc}
frac12 & 0
end{array}right),quad
D^{+}=left(begin{array}{cc}
1&0
end{array}right),
$$
$$
CC^{+}=DD^{+}=left(begin{array}{cc}
1&0\0&0
end{array}right).
$$
answered Dec 25 '18 at 12:35
AVKAVK
2,1111517
answered Dec 25 '18 at 12:35
AVKAVK
2,1111517
answered Dec 25 '18 at 12:35
AVKAVK
2,1111517
answered Dec 25 '18 at 12:35
AVKAVK
2,1111517
2,1111517
$begingroup$
Thanks for this example. Actually, I found that there was something wrong in the previous question, the condition should be $B succ A$. Afterall, this example still illustrates the problem. Furthermore, I might trace back to the proof of a more general question, which is the origin of the stated one, see (math.stackexchange.com/questions/3052118/…).
$endgroup$
– abao5887
Dec 25 '18 at 13:52
add a comment |
$begingroup$
Thanks for this example. Actually, I found that there was something wrong in the previous question, the condition should be $B succ A$. Afterall, this example still illustrates the problem. Furthermore, I might trace back to the proof of a more general question, which is the origin of the stated one, see (math.stackexchange.com/questions/3052118/…).
$endgroup$
– abao5887
Dec 25 '18 at 13:52
$begingroup$
Thanks for this example. Actually, I found that there was something wrong in the previous question, the condition should be $B succ A$. Afterall, this example still illustrates the problem. Furthermore, I might trace back to the proof of a more general question, which is the origin of the stated one, see (math.stackexchange.com/questions/3052118/…).
$endgroup$
– abao5887
Dec 25 '18 at 13:52
$begingroup$
Thanks for this example. Actually, I found that there was something wrong in the previous question, the condition should be $B succ A$. Afterall, this example still illustrates the problem. Furthermore, I might trace back to the proof of a more general question, which is the origin of the stated one, see (math.stackexchange.com/questions/3052118/…).
$endgroup$
– abao5887
Dec 25 '18 at 13:52
add a comment |
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$begingroup$
Thanks for this information, @AVK. I forgot to mention that W is not a square matrix. Now I've added it into the question statement.
$endgroup$
– abao5887
Dec 25 '18 at 12:10