How to show $u_1^TA^TAu_1+u_2^TA^TAu_2+cdots+u_n^TA^TAu_n$ as a matrix multiplication? [closed]
$begingroup$
Let $u_i$'s in $mathbb{R}^m$ be a set of orthogonal vectors and $A in mathbb{R}^{n times m}$.
Consider the following matrix multiplication
$$u_1^TA^TAu_1+u_2^TA^TAu_2+cdots+u_n^TA^TAu_n$$
which is a scalar.
How can we abbreviate this matrix summation into just one matrix multiplication?
linear-algebra matrices
edited Jan 3 at 15:17
JimmyK4542
41.3k245107
asked Jan 3 at 15:01
SaeedSaeed
1,124310
$endgroup$
closed as off-topic by Saad, Dando18, Namaste, Shailesh, KReiser Jan 4 at 0:09
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Namaste, Shailesh, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Let $u_i$'s in $mathbb{R}^m$ be a set of orthogonal vectors and $A in mathbb{R}^{n times m}$.
Consider the following matrix multiplication
$$u_1^TA^TAu_1+u_2^TA^TAu_2+cdots+u_n^TA^TAu_n$$
which is a scalar.
How can we abbreviate this matrix summation into just one matrix multiplication?
linear-algebra matrices
edited Jan 3 at 15:17
JimmyK4542
41.3k245107
asked Jan 3 at 15:01
SaeedSaeed
1,124310
$endgroup$
closed as off-topic by Saad, Dando18, Namaste, Shailesh, KReiser Jan 4 at 0:09
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Namaste, Shailesh, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Let $u_i$'s in $mathbb{R}^m$ be a set of orthogonal vectors and $A in mathbb{R}^{n times m}$.
Consider the following matrix multiplication
$$u_1^TA^TAu_1+u_2^TA^TAu_2+cdots+u_n^TA^TAu_n$$
which is a scalar.
How can we abbreviate this matrix summation into just one matrix multiplication?
linear-algebra matrices
edited Jan 3 at 15:17
JimmyK4542
41.3k245107
asked Jan 3 at 15:01
SaeedSaeed
1,124310
$endgroup$
Let $u_i$'s in $mathbb{R}^m$ be a set of orthogonal vectors and $A in mathbb{R}^{n times m}$.
Consider the following matrix multiplication
$$u_1^TA^TAu_1+u_2^TA^TAu_2+cdots+u_n^TA^TAu_n$$
which is a scalar.
How can we abbreviate this matrix summation into just one matrix multiplication?
linear-algebra matrices
linear-algebra matrices
edited Jan 3 at 15:17
JimmyK4542
41.3k245107
asked Jan 3 at 15:01
SaeedSaeed
1,124310
edited Jan 3 at 15:17
JimmyK4542
41.3k245107
asked Jan 3 at 15:01
SaeedSaeed
1,124310
edited Jan 3 at 15:17
JimmyK4542
41.3k245107
edited Jan 3 at 15:17
JimmyK4542
41.3k245107
edited Jan 3 at 15:17
JimmyK4542
41.3k245107
41.3k245107
asked Jan 3 at 15:01
SaeedSaeed
1,124310
asked Jan 3 at 15:01
SaeedSaeed
1,124310
asked Jan 3 at 15:01
SaeedSaeed
1,124310
1,124310
closed as off-topic by Saad, Dando18, Namaste, Shailesh, KReiser Jan 4 at 0:09
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Namaste, Shailesh, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, Dando18, Namaste, Shailesh, KReiser Jan 4 at 0:09
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Namaste, Shailesh, KReiser
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
If you set $U = begin{bmatrix}u_1 & u_2 & cdots & u_nend{bmatrix} in mathbb{R}^{mtimes n}$, then $AU = begin{bmatrix}Au_1 & Au_2 & cdots & Au_nend{bmatrix}$.
So $|AU|_F^2 = |Au_1|_2^2+|Au_2^2|+cdots+|Au_n|_2^2 = u_1^TA^TAu_1+u_2^TA^TAu_2+cdots+u_n^TA^TAu_n$.
answered Jan 3 at 15:14
JimmyK4542JimmyK4542
41.3k245107
$endgroup$
add a comment |
$begingroup$
You've got
begin{align}
sum_{i=1}^{m}u_i^{top}A^{top}Au_i &= sum_{i=1}^{m}operatorname{Tr}(u_iu_i^{T}A^{T}A)=operatorname{Tr}left(sum_{i=1}^m u_i u_i^{top} cdot A^{top}Aright)
end{align}
answered Jan 3 at 15:14
NadielsNadiels
2,385413
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If you set $U = begin{bmatrix}u_1 & u_2 & cdots & u_nend{bmatrix} in mathbb{R}^{mtimes n}$, then $AU = begin{bmatrix}Au_1 & Au_2 & cdots & Au_nend{bmatrix}$.
So $|AU|_F^2 = |Au_1|_2^2+|Au_2^2|+cdots+|Au_n|_2^2 = u_1^TA^TAu_1+u_2^TA^TAu_2+cdots+u_n^TA^TAu_n$.
answered Jan 3 at 15:14
JimmyK4542JimmyK4542
41.3k245107
$endgroup$
add a comment |
$begingroup$
If you set $U = begin{bmatrix}u_1 & u_2 & cdots & u_nend{bmatrix} in mathbb{R}^{mtimes n}$, then $AU = begin{bmatrix}Au_1 & Au_2 & cdots & Au_nend{bmatrix}$.
So $|AU|_F^2 = |Au_1|_2^2+|Au_2^2|+cdots+|Au_n|_2^2 = u_1^TA^TAu_1+u_2^TA^TAu_2+cdots+u_n^TA^TAu_n$.
answered Jan 3 at 15:14
JimmyK4542JimmyK4542
41.3k245107
$endgroup$
add a comment |
$begingroup$
If you set $U = begin{bmatrix}u_1 & u_2 & cdots & u_nend{bmatrix} in mathbb{R}^{mtimes n}$, then $AU = begin{bmatrix}Au_1 & Au_2 & cdots & Au_nend{bmatrix}$.
So $|AU|_F^2 = |Au_1|_2^2+|Au_2^2|+cdots+|Au_n|_2^2 = u_1^TA^TAu_1+u_2^TA^TAu_2+cdots+u_n^TA^TAu_n$.
answered Jan 3 at 15:14
JimmyK4542JimmyK4542
41.3k245107
$endgroup$
If you set $U = begin{bmatrix}u_1 & u_2 & cdots & u_nend{bmatrix} in mathbb{R}^{mtimes n}$, then $AU = begin{bmatrix}Au_1 & Au_2 & cdots & Au_nend{bmatrix}$.
So $|AU|_F^2 = |Au_1|_2^2+|Au_2^2|+cdots+|Au_n|_2^2 = u_1^TA^TAu_1+u_2^TA^TAu_2+cdots+u_n^TA^TAu_n$.
answered Jan 3 at 15:14
JimmyK4542JimmyK4542
41.3k245107
answered Jan 3 at 15:14
JimmyK4542JimmyK4542
41.3k245107
answered Jan 3 at 15:14
JimmyK4542JimmyK4542
41.3k245107
answered Jan 3 at 15:14
JimmyK4542JimmyK4542
41.3k245107
41.3k245107
add a comment |
add a comment |
$begingroup$
You've got
begin{align}
sum_{i=1}^{m}u_i^{top}A^{top}Au_i &= sum_{i=1}^{m}operatorname{Tr}(u_iu_i^{T}A^{T}A)=operatorname{Tr}left(sum_{i=1}^m u_i u_i^{top} cdot A^{top}Aright)
end{align}
answered Jan 3 at 15:14
NadielsNadiels
2,385413
$endgroup$
add a comment |
$begingroup$
You've got
begin{align}
sum_{i=1}^{m}u_i^{top}A^{top}Au_i &= sum_{i=1}^{m}operatorname{Tr}(u_iu_i^{T}A^{T}A)=operatorname{Tr}left(sum_{i=1}^m u_i u_i^{top} cdot A^{top}Aright)
end{align}
answered Jan 3 at 15:14
NadielsNadiels
2,385413
$endgroup$
add a comment |
$begingroup$
You've got
begin{align}
sum_{i=1}^{m}u_i^{top}A^{top}Au_i &= sum_{i=1}^{m}operatorname{Tr}(u_iu_i^{T}A^{T}A)=operatorname{Tr}left(sum_{i=1}^m u_i u_i^{top} cdot A^{top}Aright)
end{align}
answered Jan 3 at 15:14
NadielsNadiels
2,385413
$endgroup$
You've got
begin{align}
sum_{i=1}^{m}u_i^{top}A^{top}Au_i &= sum_{i=1}^{m}operatorname{Tr}(u_iu_i^{T}A^{T}A)=operatorname{Tr}left(sum_{i=1}^m u_i u_i^{top} cdot A^{top}Aright)
end{align}
answered Jan 3 at 15:14
NadielsNadiels
2,385413
answered Jan 3 at 15:14
NadielsNadiels
2,385413
answered Jan 3 at 15:14
NadielsNadiels
2,385413
answered Jan 3 at 15:14
NadielsNadiels
2,385413
2,385413
add a comment |
add a comment |
