Proof for Cauchy-Schwarz inequality for Trace [on hold]
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Cauchy-Schwarz inequality applied to Trace of two products $mathbf{Tr}(A'B)$ has the form
$$
mathbf{Tr}(A'B) leq sqrt{mathbf{Tr}(A'A)} sqrt{mathbf{Tr}(B'B)}
$$
I saw many places where people use this inequality. But did not see a formal proof. Is it difficult to prove ? Anyone can give a simple proof ?
linear-algebra trace cauchy-schwarz-inequality
asked Nov 20 at 9:23
Shew
553413
put on hold as off-topic by user21820, amWhy, KReiser, user10354138, Cesareo 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, amWhy, KReiser, user10354138, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
10
down vote
favorite
Cauchy-Schwarz inequality applied to Trace of two products $mathbf{Tr}(A'B)$ has the form
$$
mathbf{Tr}(A'B) leq sqrt{mathbf{Tr}(A'A)} sqrt{mathbf{Tr}(B'B)}
$$
I saw many places where people use this inequality. But did not see a formal proof. Is it difficult to prove ? Anyone can give a simple proof ?
linear-algebra trace cauchy-schwarz-inequality
asked Nov 20 at 9:23
Shew
553413
put on hold as off-topic by user21820, amWhy, KReiser, user10354138, Cesareo 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, amWhy, KReiser, user10354138, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
10
down vote
favorite
up vote
10
down vote
favorite
Cauchy-Schwarz inequality applied to Trace of two products $mathbf{Tr}(A'B)$ has the form
$$
mathbf{Tr}(A'B) leq sqrt{mathbf{Tr}(A'A)} sqrt{mathbf{Tr}(B'B)}
$$
I saw many places where people use this inequality. But did not see a formal proof. Is it difficult to prove ? Anyone can give a simple proof ?
linear-algebra trace cauchy-schwarz-inequality
asked Nov 20 at 9:23
Shew
553413
Cauchy-Schwarz inequality applied to Trace of two products $mathbf{Tr}(A'B)$ has the form
$$
mathbf{Tr}(A'B) leq sqrt{mathbf{Tr}(A'A)} sqrt{mathbf{Tr}(B'B)}
$$
I saw many places where people use this inequality. But did not see a formal proof. Is it difficult to prove ? Anyone can give a simple proof ?
linear-algebra trace cauchy-schwarz-inequality
linear-algebra trace cauchy-schwarz-inequality
asked Nov 20 at 9:23
Shew
553413
asked Nov 20 at 9:23
Shew
553413
asked Nov 20 at 9:23
Shew
553413
asked Nov 20 at 9:23
Shew
553413
asked Nov 20 at 9:23
Shew
553413
553413
put on hold as off-topic by user21820, amWhy, KReiser, user10354138, Cesareo 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, amWhy, KReiser, user10354138, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by user21820, amWhy, KReiser, user10354138, Cesareo 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user21820, amWhy, KReiser, user10354138, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
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2 Answers
2
active
oldest
votes
up vote
15
down vote
accepted
$Tr(A-tB)'(A-tB) geq 0$ for all $t$ real. Expand this, use the fact that $Tr(M')=Tr(M)$ (so $Tr(A'B)=Tr(B'A))$ and minimize the left side over $t$. You will get the inequality you want. (This is the standard proof of C-S inequality)
answered Nov 20 at 9:31
Kavi Rama Murthy
41.3k31751
add a comment |
up vote
31
down vote
The Cauchy-Schwarz inequality is valid for any inner product, so you just need to show $operatorname{textbf{Tr}}A'B$ is an inner product. It's clearly bilinear (or sesquilinear if by $'$ you meant a complex adjoint), with $$operatorname{textbf{Tr}}A'A=sum_i (A'A)_{i}=sum_{ij}A'_{ij}A_{ji}.$$Depending on whether you're working with the real or complex case, this quantity is either $sum_{ij}A_{ji}^2$ or $sum_{ij}|A_{ji}|^2$. Either way it's non-negative, completing the proof.
answered Nov 20 at 9:28
J.G.
18.6k21932
3
The point here is that while $mathbf{Tr}(A,B)$ looks complicated, it is really nothing more than a plain vanilla inner product with n^2-element vectors, so there's nothing special in proving C-S here.
– einpoklum
Nov 20 at 19:13
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
15
down vote
accepted
$Tr(A-tB)'(A-tB) geq 0$ for all $t$ real. Expand this, use the fact that $Tr(M')=Tr(M)$ (so $Tr(A'B)=Tr(B'A))$ and minimize the left side over $t$. You will get the inequality you want. (This is the standard proof of C-S inequality)
answered Nov 20 at 9:31
Kavi Rama Murthy
41.3k31751
add a comment |
up vote
15
down vote
accepted
$Tr(A-tB)'(A-tB) geq 0$ for all $t$ real. Expand this, use the fact that $Tr(M')=Tr(M)$ (so $Tr(A'B)=Tr(B'A))$ and minimize the left side over $t$. You will get the inequality you want. (This is the standard proof of C-S inequality)
answered Nov 20 at 9:31
Kavi Rama Murthy
41.3k31751
add a comment |
up vote
15
down vote
accepted
up vote
15
down vote
accepted
$Tr(A-tB)'(A-tB) geq 0$ for all $t$ real. Expand this, use the fact that $Tr(M')=Tr(M)$ (so $Tr(A'B)=Tr(B'A))$ and minimize the left side over $t$. You will get the inequality you want. (This is the standard proof of C-S inequality)
answered Nov 20 at 9:31
Kavi Rama Murthy
41.3k31751
$Tr(A-tB)'(A-tB) geq 0$ for all $t$ real. Expand this, use the fact that $Tr(M')=Tr(M)$ (so $Tr(A'B)=Tr(B'A))$ and minimize the left side over $t$. You will get the inequality you want. (This is the standard proof of C-S inequality)
answered Nov 20 at 9:31
Kavi Rama Murthy
41.3k31751
answered Nov 20 at 9:31
Kavi Rama Murthy
41.3k31751
answered Nov 20 at 9:31
Kavi Rama Murthy
41.3k31751
answered Nov 20 at 9:31
Kavi Rama Murthy
41.3k31751
41.3k31751
add a comment |
add a comment |
up vote
31
down vote
The Cauchy-Schwarz inequality is valid for any inner product, so you just need to show $operatorname{textbf{Tr}}A'B$ is an inner product. It's clearly bilinear (or sesquilinear if by $'$ you meant a complex adjoint), with $$operatorname{textbf{Tr}}A'A=sum_i (A'A)_{i}=sum_{ij}A'_{ij}A_{ji}.$$Depending on whether you're working with the real or complex case, this quantity is either $sum_{ij}A_{ji}^2$ or $sum_{ij}|A_{ji}|^2$. Either way it's non-negative, completing the proof.
answered Nov 20 at 9:28
J.G.
18.6k21932
3
The point here is that while $mathbf{Tr}(A,B)$ looks complicated, it is really nothing more than a plain vanilla inner product with n^2-element vectors, so there's nothing special in proving C-S here.
– einpoklum
Nov 20 at 19:13
add a comment |
up vote
31
down vote
The Cauchy-Schwarz inequality is valid for any inner product, so you just need to show $operatorname{textbf{Tr}}A'B$ is an inner product. It's clearly bilinear (or sesquilinear if by $'$ you meant a complex adjoint), with $$operatorname{textbf{Tr}}A'A=sum_i (A'A)_{i}=sum_{ij}A'_{ij}A_{ji}.$$Depending on whether you're working with the real or complex case, this quantity is either $sum_{ij}A_{ji}^2$ or $sum_{ij}|A_{ji}|^2$. Either way it's non-negative, completing the proof.
answered Nov 20 at 9:28
J.G.
18.6k21932
3
The point here is that while $mathbf{Tr}(A,B)$ looks complicated, it is really nothing more than a plain vanilla inner product with n^2-element vectors, so there's nothing special in proving C-S here.
– einpoklum
Nov 20 at 19:13
add a comment |
up vote
31
down vote
up vote
31
down vote
The Cauchy-Schwarz inequality is valid for any inner product, so you just need to show $operatorname{textbf{Tr}}A'B$ is an inner product. It's clearly bilinear (or sesquilinear if by $'$ you meant a complex adjoint), with $$operatorname{textbf{Tr}}A'A=sum_i (A'A)_{i}=sum_{ij}A'_{ij}A_{ji}.$$Depending on whether you're working with the real or complex case, this quantity is either $sum_{ij}A_{ji}^2$ or $sum_{ij}|A_{ji}|^2$. Either way it's non-negative, completing the proof.
answered Nov 20 at 9:28
J.G.
18.6k21932
The Cauchy-Schwarz inequality is valid for any inner product, so you just need to show $operatorname{textbf{Tr}}A'B$ is an inner product. It's clearly bilinear (or sesquilinear if by $'$ you meant a complex adjoint), with $$operatorname{textbf{Tr}}A'A=sum_i (A'A)_{i}=sum_{ij}A'_{ij}A_{ji}.$$Depending on whether you're working with the real or complex case, this quantity is either $sum_{ij}A_{ji}^2$ or $sum_{ij}|A_{ji}|^2$. Either way it's non-negative, completing the proof.
answered Nov 20 at 9:28
J.G.
18.6k21932
answered Nov 20 at 9:28
J.G.
18.6k21932
answered Nov 20 at 9:28
J.G.
18.6k21932
answered Nov 20 at 9:28
J.G.
18.6k21932
18.6k21932
3
The point here is that while $mathbf{Tr}(A,B)$ looks complicated, it is really nothing more than a plain vanilla inner product with n^2-element vectors, so there's nothing special in proving C-S here.
– einpoklum
Nov 20 at 19:13
add a comment |
3
The point here is that while $mathbf{Tr}(A,B)$ looks complicated, it is really nothing more than a plain vanilla inner product with n^2-element vectors, so there's nothing special in proving C-S here.
– einpoklum
Nov 20 at 19:13
3
3
The point here is that while $mathbf{Tr}(A,B)$ looks complicated, it is really nothing more than a plain vanilla inner product with n^2-element vectors, so there's nothing special in proving C-S here.
– einpoklum
Nov 20 at 19:13
The point here is that while $mathbf{Tr}(A,B)$ looks complicated, it is really nothing more than a plain vanilla inner product with n^2-element vectors, so there's nothing special in proving C-S here.
– einpoklum
Nov 20 at 19:13
add a comment |
