Mixing
Proof for Cauchy-Schwarz inequality for Trace [closed]
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 '18 at 9:23
Shew
563413
closed as off-topic by user21820, amWhy, KReiser, user10354138, Cesareo Nov 21 '18 at 2:11
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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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 '18 at 9:23
Shew
563413
closed as off-topic by user21820, amWhy, KReiser, user10354138, Cesareo Nov 21 '18 at 2:11
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 |
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 '18 at 9:23
Shew
563413
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 '18 at 9:23
Shew
563413
asked Nov 20 '18 at 9:23
Shew
563413
asked Nov 20 '18 at 9:23
Shew
563413
asked Nov 20 '18 at 9:23
Shew
563413
asked Nov 20 '18 at 9:23
Shew
563413
563413
closed as off-topic by user21820, amWhy, KReiser, user10354138, Cesareo Nov 21 '18 at 2:11
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.
closed as off-topic by user21820, amWhy, KReiser, user10354138, Cesareo Nov 21 '18 at 2:11
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 |
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2 Answers
2
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oldest
votes
$Tr(A-tB)'(A-tB) geq 0$ 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 '18 at 9:31
Kavi Rama Murthy
50.4k31854
add a comment |
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 '18 at 9:28
J.G.
22.8k22136
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 '18 at 19:13
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$Tr(A-tB)'(A-tB) geq 0$ 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 '18 at 9:31
Kavi Rama Murthy
50.4k31854
add a comment |
$Tr(A-tB)'(A-tB) geq 0$ 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 '18 at 9:31
Kavi Rama Murthy
50.4k31854
add a comment |
$Tr(A-tB)'(A-tB) geq 0$ 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 '18 at 9:31
Kavi Rama Murthy
50.4k31854
$Tr(A-tB)'(A-tB) geq 0$ 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 '18 at 9:31
Kavi Rama Murthy
50.4k31854
answered Nov 20 '18 at 9:31
Kavi Rama Murthy
50.4k31854
answered Nov 20 '18 at 9:31
Kavi Rama Murthy
50.4k31854
answered Nov 20 '18 at 9:31
Kavi Rama Murthy
50.4k31854
50.4k31854
add a comment |
add a comment |
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 '18 at 9:28
J.G.
22.8k22136
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 '18 at 19:13
add a comment |
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 '18 at 9:28
J.G.
22.8k22136
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 '18 at 19:13
add a comment |
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 '18 at 9:28
J.G.
22.8k22136
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 '18 at 9:28
J.G.
22.8k22136
answered Nov 20 '18 at 9:28
J.G.
22.8k22136
answered Nov 20 '18 at 9:28
J.G.
22.8k22136
answered Nov 20 '18 at 9:28
J.G.
22.8k22136
22.8k22136
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 '18 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 '18 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 '18 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 '18 at 19:13
add a comment |
