Mixing
Is there any generalized version of Cauchy-Schwarz inequality?
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
all. I want to ask your opinion about the following question:
Let $mathbfA in mathbbR^m times n$ be a matrix having full column-rank, and let $ mathbfu_1, cdots, mathbfu_r $ be the set of vectors in $mathbbR^n$ such that $ mathbfA mathbfu_1, cdots, mathbfA mathbfu_r $ is orthonormal. Then, does the following inequality hold true? Here, $mathbfA^T$ denotes the transpose of $mathbfA$.
beginalign
underseti=1oversetrsum | mathbfA^T mathbfA mathbfu_i |_2^2
&ge fracr^2
endalign
When $r=1$, then the above inequality clearly holds true by the well-known Cauchy-Schwarz inequality ($| mathbfA^T mathbfA mathbfu_1 |_2 cdot | mathbfu_1 |_2 ge langle mathbfA^T mathbfA mathbfu_1, mathbfu_1 rangle = | mathbfA mathbfu_1 |_2^2 = 1$).
How about the case for $r neq 1$?
linear-algebra inequality orthonormal
asked Aug 3 at 1:49
user580055
213
add a comment |Â
up vote
1
down vote
favorite
all. I want to ask your opinion about the following question:
Let $mathbfA in mathbbR^m times n$ be a matrix having full column-rank, and let $ mathbfu_1, cdots, mathbfu_r $ be the set of vectors in $mathbbR^n$ such that $ mathbfA mathbfu_1, cdots, mathbfA mathbfu_r $ is orthonormal. Then, does the following inequality hold true? Here, $mathbfA^T$ denotes the transpose of $mathbfA$.
beginalign
underseti=1oversetrsum | mathbfA^T mathbfA mathbfu_i |_2^2
&ge fracr^2
endalign
When $r=1$, then the above inequality clearly holds true by the well-known Cauchy-Schwarz inequality ($| mathbfA^T mathbfA mathbfu_1 |_2 cdot | mathbfu_1 |_2 ge langle mathbfA^T mathbfA mathbfu_1, mathbfu_1 rangle = | mathbfA mathbfu_1 |_2^2 = 1$).
How about the case for $r neq 1$?
linear-algebra inequality orthonormal
asked Aug 3 at 1:49
user580055
213
1
If you just want my opinion, I'd say it's fun to read.
– mathworker21
Aug 3 at 1:53
Hint: If $p_1, p_2, ldots, p_r, q_1, q_2, ldots, q_r$ are any $2r$ vectors in $mathbbR^n$, then $left(sum_i=1^r left|left|p_iright|right|_2^2right) left(sum_i=1^r left|left|q_iright|right|_2^2right) geq left(sum_i=1^r left< p_i, q_iright> right)^2$. This is a kind of vectorial version of the Cauchy-Schwarz inequality and follows easily from the usual Cauchy-Schwarz inequality. Now, apply this inequality to $p_i = mathbfA^T mathbfA u_i$ and $q_i = sum_j=1^r mathbfu_j$.
– darij grinberg
Aug 3 at 2:29
I appreciate your answer, but in this way, we only get $underseti=1oversetrsum | mathbfA^T mathbfA mathbfu_i |_2^2 ge fracr$.
– user580055
Aug 3 at 2:52
Ah, you're right.
– darij grinberg
Aug 4 at 16:34
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
all. I want to ask your opinion about the following question:
Let $mathbfA in mathbbR^m times n$ be a matrix having full column-rank, and let $ mathbfu_1, cdots, mathbfu_r $ be the set of vectors in $mathbbR^n$ such that $ mathbfA mathbfu_1, cdots, mathbfA mathbfu_r $ is orthonormal. Then, does the following inequality hold true? Here, $mathbfA^T$ denotes the transpose of $mathbfA$.
beginalign
underseti=1oversetrsum | mathbfA^T mathbfA mathbfu_i |_2^2
&ge fracr^2
endalign
When $r=1$, then the above inequality clearly holds true by the well-known Cauchy-Schwarz inequality ($| mathbfA^T mathbfA mathbfu_1 |_2 cdot | mathbfu_1 |_2 ge langle mathbfA^T mathbfA mathbfu_1, mathbfu_1 rangle = | mathbfA mathbfu_1 |_2^2 = 1$).
How about the case for $r neq 1$?
linear-algebra inequality orthonormal
asked Aug 3 at 1:49
user580055
213
all. I want to ask your opinion about the following question:
Let $mathbfA in mathbbR^m times n$ be a matrix having full column-rank, and let $ mathbfu_1, cdots, mathbfu_r $ be the set of vectors in $mathbbR^n$ such that $ mathbfA mathbfu_1, cdots, mathbfA mathbfu_r $ is orthonormal. Then, does the following inequality hold true? Here, $mathbfA^T$ denotes the transpose of $mathbfA$.
beginalign
underseti=1oversetrsum | mathbfA^T mathbfA mathbfu_i |_2^2
&ge fracr^2
endalign
When $r=1$, then the above inequality clearly holds true by the well-known Cauchy-Schwarz inequality ($| mathbfA^T mathbfA mathbfu_1 |_2 cdot | mathbfu_1 |_2 ge langle mathbfA^T mathbfA mathbfu_1, mathbfu_1 rangle = | mathbfA mathbfu_1 |_2^2 = 1$).
How about the case for $r neq 1$?
linear-algebra inequality orthonormal
asked Aug 3 at 1:49
user580055
213
asked Aug 3 at 1:49
user580055
213
asked Aug 3 at 1:49
user580055
213
asked Aug 3 at 1:49
user580055
213
213
1
If you just want my opinion, I'd say it's fun to read.
– mathworker21
Aug 3 at 1:53
Hint: If $p_1, p_2, ldots, p_r, q_1, q_2, ldots, q_r$ are any $2r$ vectors in $mathbbR^n$, then $left(sum_i=1^r left|left|p_iright|right|_2^2right) left(sum_i=1^r left|left|q_iright|right|_2^2right) geq left(sum_i=1^r left< p_i, q_iright> right)^2$. This is a kind of vectorial version of the Cauchy-Schwarz inequality and follows easily from the usual Cauchy-Schwarz inequality. Now, apply this inequality to $p_i = mathbfA^T mathbfA u_i$ and $q_i = sum_j=1^r mathbfu_j$.
– darij grinberg
Aug 3 at 2:29
I appreciate your answer, but in this way, we only get $underseti=1oversetrsum | mathbfA^T mathbfA mathbfu_i |_2^2 ge fracr$.
– user580055
Aug 3 at 2:52
Ah, you're right.
– darij grinberg
Aug 4 at 16:34
add a comment |Â
1
If you just want my opinion, I'd say it's fun to read.
– mathworker21
Aug 3 at 1:53
Hint: If $p_1, p_2, ldots, p_r, q_1, q_2, ldots, q_r$ are any $2r$ vectors in $mathbbR^n$, then $left(sum_i=1^r left|left|p_iright|right|_2^2right) left(sum_i=1^r left|left|q_iright|right|_2^2right) geq left(sum_i=1^r left< p_i, q_iright> right)^2$. This is a kind of vectorial version of the Cauchy-Schwarz inequality and follows easily from the usual Cauchy-Schwarz inequality. Now, apply this inequality to $p_i = mathbfA^T mathbfA u_i$ and $q_i = sum_j=1^r mathbfu_j$.
– darij grinberg
Aug 3 at 2:29
I appreciate your answer, but in this way, we only get $underseti=1oversetrsum | mathbfA^T mathbfA mathbfu_i |_2^2 ge fracr$.
– user580055
Aug 3 at 2:52
Ah, you're right.
– darij grinberg
Aug 4 at 16:34
1
1
If you just want my opinion, I'd say it's fun to read.
– mathworker21
Aug 3 at 1:53
If you just want my opinion, I'd say it's fun to read.
– mathworker21
Aug 3 at 1:53
Hint: If $p_1, p_2, ldots, p_r, q_1, q_2, ldots, q_r$ are any $2r$ vectors in $mathbbR^n$, then $left(sum_i=1^r left|left|p_iright|right|_2^2right) left(sum_i=1^r left|left|q_iright|right|_2^2right) geq left(sum_i=1^r left< p_i, q_iright> right)^2$. This is a kind of vectorial version of the Cauchy-Schwarz inequality and follows easily from the usual Cauchy-Schwarz inequality. Now, apply this inequality to $p_i = mathbfA^T mathbfA u_i$ and $q_i = sum_j=1^r mathbfu_j$.
– darij grinberg
Aug 3 at 2:29
Hint: If $p_1, p_2, ldots, p_r, q_1, q_2, ldots, q_r$ are any $2r$ vectors in $mathbbR^n$, then $left(sum_i=1^r left|left|p_iright|right|_2^2right) left(sum_i=1^r left|left|q_iright|right|_2^2right) geq left(sum_i=1^r left< p_i, q_iright> right)^2$. This is a kind of vectorial version of the Cauchy-Schwarz inequality and follows easily from the usual Cauchy-Schwarz inequality. Now, apply this inequality to $p_i = mathbfA^T mathbfA u_i$ and $q_i = sum_j=1^r mathbfu_j$.
– darij grinberg
Aug 3 at 2:29
I appreciate your answer, but in this way, we only get $underseti=1oversetrsum | mathbfA^T mathbfA mathbfu_i |_2^2 ge fracr$.
– user580055
Aug 3 at 2:52
I appreciate your answer, but in this way, we only get $underseti=1oversetrsum | mathbfA^T mathbfA mathbfu_i |_2^2 ge fracr$.
– user580055
Aug 3 at 2:52
Ah, you're right.
– darij grinberg
Aug 4 at 16:34
Ah, you're right.
– darij grinberg
Aug 4 at 16:34
add a comment |Â
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2870660%2fis-there-any-generalized-version-of-cauchy-schwarz-inequality%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
If you just want my opinion, I'd say it's fun to read.
– mathworker21
Aug 3 at 1:53
Hint: If $p_1, p_2, ldots, p_r, q_1, q_2, ldots, q_r$ are any $2r$ vectors in $mathbbR^n$, then $left(sum_i=1^r left|left|p_iright|right|_2^2right) left(sum_i=1^r left|left|q_iright|right|_2^2right) geq left(sum_i=1^r left< p_i, q_iright> right)^2$. This is a kind of vectorial version of the Cauchy-Schwarz inequality and follows easily from the usual Cauchy-Schwarz inequality. Now, apply this inequality to $p_i = mathbfA^T mathbfA u_i$ and $q_i = sum_j=1^r mathbfu_j$.
– darij grinberg
Aug 3 at 2:29
I appreciate your answer, but in this way, we only get $underseti=1oversetrsum | mathbfA^T mathbfA mathbfu_i |_2^2 ge fracr$.
– user580055
Aug 3 at 2:52
Ah, you're right.
– darij grinberg
Aug 4 at 16:34