Mixing
Show that this matrix is unitary
$begingroup$
I'm currently working on quantum computing.
In a literature source that I have, it is said that the following matrix $U$ is unitary.
My question is, how can we show that in this matrix? I know that a matrix is unitary if:
$U^{*}U=I$
The matrix is an NxN matrix:
$U=begin{pmatrix} -1+frac{2}{N} & frac{2}{N} & ...& frac{2}{N} \
frac{2}{N} & -1+frac{2}{N} & ...& frac{2}{N} \ vdots & vdots & ddots & vdots \
frac{2}{N} & frac{2}{N} & ...& -1+frac{2}{N} end{pmatrix} $
my first thought would be to pull the 1 out:
$U=begin{pmatrix} frac{2}{N} & frac{2}{N} & ... & frac{2}{N} \
frac{2}{N} & frac{2}{N} & ... & frac{2}{N} \ vdots & vdots & ddots & vdots \
frac{2}{N} & frac{2}{N} & ... & frac{2}{N} end{pmatrix} - I $
Then you would have to show now that the matrix is unitary, but I can not get any further here.
I would be very happy if someone could help me to show that the matrix is unitary. I hope that my question is clear and understandable.
linear-algebra matrices
asked Jan 24 at 13:44
QuantaMagQuantaMag
134
$endgroup$
add a comment |
$begingroup$
I'm currently working on quantum computing.
In a literature source that I have, it is said that the following matrix $U$ is unitary.
My question is, how can we show that in this matrix? I know that a matrix is unitary if:
$U^{*}U=I$
The matrix is an NxN matrix:
$U=begin{pmatrix} -1+frac{2}{N} & frac{2}{N} & ...& frac{2}{N} \
frac{2}{N} & -1+frac{2}{N} & ...& frac{2}{N} \ vdots & vdots & ddots & vdots \
frac{2}{N} & frac{2}{N} & ...& -1+frac{2}{N} end{pmatrix} $
my first thought would be to pull the 1 out:
$U=begin{pmatrix} frac{2}{N} & frac{2}{N} & ... & frac{2}{N} \
frac{2}{N} & frac{2}{N} & ... & frac{2}{N} \ vdots & vdots & ddots & vdots \
frac{2}{N} & frac{2}{N} & ... & frac{2}{N} end{pmatrix} - I $
Then you would have to show now that the matrix is unitary, but I can not get any further here.
I would be very happy if someone could help me to show that the matrix is unitary. I hope that my question is clear and understandable.
linear-algebra matrices
asked Jan 24 at 13:44
QuantaMagQuantaMag
134
$endgroup$
add a comment |
$begingroup$
I'm currently working on quantum computing.
In a literature source that I have, it is said that the following matrix $U$ is unitary.
My question is, how can we show that in this matrix? I know that a matrix is unitary if:
$U^{*}U=I$
The matrix is an NxN matrix:
$U=begin{pmatrix} -1+frac{2}{N} & frac{2}{N} & ...& frac{2}{N} \
frac{2}{N} & -1+frac{2}{N} & ...& frac{2}{N} \ vdots & vdots & ddots & vdots \
frac{2}{N} & frac{2}{N} & ...& -1+frac{2}{N} end{pmatrix} $
my first thought would be to pull the 1 out:
$U=begin{pmatrix} frac{2}{N} & frac{2}{N} & ... & frac{2}{N} \
frac{2}{N} & frac{2}{N} & ... & frac{2}{N} \ vdots & vdots & ddots & vdots \
frac{2}{N} & frac{2}{N} & ... & frac{2}{N} end{pmatrix} - I $
Then you would have to show now that the matrix is unitary, but I can not get any further here.
I would be very happy if someone could help me to show that the matrix is unitary. I hope that my question is clear and understandable.
linear-algebra matrices
asked Jan 24 at 13:44
QuantaMagQuantaMag
134
$endgroup$
I'm currently working on quantum computing.
In a literature source that I have, it is said that the following matrix $U$ is unitary.
My question is, how can we show that in this matrix? I know that a matrix is unitary if:
$U^{*}U=I$
The matrix is an NxN matrix:
$U=begin{pmatrix} -1+frac{2}{N} & frac{2}{N} & ...& frac{2}{N} \
frac{2}{N} & -1+frac{2}{N} & ...& frac{2}{N} \ vdots & vdots & ddots & vdots \
frac{2}{N} & frac{2}{N} & ...& -1+frac{2}{N} end{pmatrix} $
my first thought would be to pull the 1 out:
$U=begin{pmatrix} frac{2}{N} & frac{2}{N} & ... & frac{2}{N} \
frac{2}{N} & frac{2}{N} & ... & frac{2}{N} \ vdots & vdots & ddots & vdots \
frac{2}{N} & frac{2}{N} & ... & frac{2}{N} end{pmatrix} - I $
Then you would have to show now that the matrix is unitary, but I can not get any further here.
I would be very happy if someone could help me to show that the matrix is unitary. I hope that my question is clear and understandable.
linear-algebra matrices
linear-algebra matrices
asked Jan 24 at 13:44
QuantaMagQuantaMag
134
asked Jan 24 at 13:44
QuantaMagQuantaMag
134
asked Jan 24 at 13:44
QuantaMagQuantaMag
134
asked Jan 24 at 13:44
QuantaMagQuantaMag
134
asked Jan 24 at 13:44
QuantaMagQuantaMag
134
134
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$ U^{*} = U = begin{pmatrix} frac{2}{N} & frac{2}{N} & ... & frac{2}{N} \
frac{2}{N} & frac{2}{N} & ... & frac{2}{N} \ vdots & vdots & ddots & vdots \
frac{2}{N} & frac{2}{N} & ... & frac{2}{N} end{pmatrix} - I $
(conjugate transpose is just transpose here, as conjugate of a real is the real number itself).
Let us take a new matrix $A$,
$ A= begin{pmatrix} frac{2}{N} & frac{2}{N} & ... & frac{2}{N} \
frac{2}{N} & frac{2}{N} & ... & frac{2}{N} \ vdots & vdots & ddots & vdots \
frac{2}{N} & frac{2}{N} & ... & frac{2}{N} end{pmatrix} $
So $U^{*}U= U^2 = (A-I)*(A-I) = A^2 - 2AI + I = A^2 - 2A + I$
$ A^2= begin{pmatrix} frac{4}{N} & frac{4}{N} & ... & frac{4}{N} \
frac{4}{N} & frac{4}{N} & ... & frac{4}{N} \ vdots & vdots & ddots & vdots \
frac{4}{N} & frac{4}{N} & ... & frac{4}{N} end{pmatrix} = 2A $
So $A^2-2A$ is null matrix (all entries $0$)
And so $U^{*}U= I$
answered Jan 24 at 13:55
Swapnil RustagiSwapnil Rustagi
758721
$endgroup$
$begingroup$
Thank you for your answer, I would like to reward you with an upvote, but I do not have enough reputation ... How did you come to the $4/N$ in the $A^2$ matrix? Can you explain that a little bit more in detail? My guess is that it's because you run the $((2/N)*(2/N))$ N times.
$endgroup$
– QuantaMag
Jan 24 at 14:37
$begingroup$
@QuantaMag Yes, exactly, I assume you can see how each entry of $A^2$ is obtained just by matrix multiplication, just like the way you described. $((2/N)*(2/N)) + ((2/N)*(2/N)) + ... $ (N times) = $ N*(4/N^2) = 4/N $. If there is still any confusion, I will edit my answer to clarify further. Also, if you found my answer satisfactorily answers your question, you can accept my answer, even if you do not have reputation to upvote.
$endgroup$
– Swapnil Rustagi
Jan 24 at 15:25
add a comment |
$begingroup$
Note that
$$
U=frac{2}{N}vv^T -I
$$ where $v=(1,1,ldots,1)^TinBbb R^N$. Hence $U=U^T$ and
$$
UU^T =U^2 =I-frac{4}{N}vv^T +frac{4}{N^2}(v^Tv)vv^T=I-frac{4}{N}vv^T +frac{4}{N}vv^T=I
$$ since $v^Tv = |v|^2=N$.
answered Jan 24 at 13:51
SongSong
18.1k21449
$endgroup$
$begingroup$
Thanks for your answer, I'm just trying to understand your answer. One more question, how did you come to this $v^Tv = |v|^2=N$ ?
$endgroup$
– QuantaMag
Jan 24 at 14:45
$begingroup$
@QuantaMag That is because $v^T v =vcdot v = |v|^2=1^2+1^2+cdots +1^2=N$ where $cdot$ is standard inner(scalar) product, i.e. $vec{a}cdot vec{b} = a_1b_1+a_2b_2+cdots+ a_Nb_N$.
$endgroup$
– Song
Jan 24 at 14:46
add a comment |
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2 Answers
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2 Answers
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active
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votes
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$begingroup$
$ U^{*} = U = begin{pmatrix} frac{2}{N} & frac{2}{N} & ... & frac{2}{N} \
frac{2}{N} & frac{2}{N} & ... & frac{2}{N} \ vdots & vdots & ddots & vdots \
frac{2}{N} & frac{2}{N} & ... & frac{2}{N} end{pmatrix} - I $
(conjugate transpose is just transpose here, as conjugate of a real is the real number itself).
Let us take a new matrix $A$,
$ A= begin{pmatrix} frac{2}{N} & frac{2}{N} & ... & frac{2}{N} \
frac{2}{N} & frac{2}{N} & ... & frac{2}{N} \ vdots & vdots & ddots & vdots \
frac{2}{N} & frac{2}{N} & ... & frac{2}{N} end{pmatrix} $
So $U^{*}U= U^2 = (A-I)*(A-I) = A^2 - 2AI + I = A^2 - 2A + I$
$ A^2= begin{pmatrix} frac{4}{N} & frac{4}{N} & ... & frac{4}{N} \
frac{4}{N} & frac{4}{N} & ... & frac{4}{N} \ vdots & vdots & ddots & vdots \
frac{4}{N} & frac{4}{N} & ... & frac{4}{N} end{pmatrix} = 2A $
So $A^2-2A$ is null matrix (all entries $0$)
And so $U^{*}U= I$
answered Jan 24 at 13:55
Swapnil RustagiSwapnil Rustagi
758721
$endgroup$
$begingroup$
Thank you for your answer, I would like to reward you with an upvote, but I do not have enough reputation ... How did you come to the $4/N$ in the $A^2$ matrix? Can you explain that a little bit more in detail? My guess is that it's because you run the $((2/N)*(2/N))$ N times.
$endgroup$
– QuantaMag
Jan 24 at 14:37
$begingroup$
@QuantaMag Yes, exactly, I assume you can see how each entry of $A^2$ is obtained just by matrix multiplication, just like the way you described. $((2/N)*(2/N)) + ((2/N)*(2/N)) + ... $ (N times) = $ N*(4/N^2) = 4/N $. If there is still any confusion, I will edit my answer to clarify further. Also, if you found my answer satisfactorily answers your question, you can accept my answer, even if you do not have reputation to upvote.
$endgroup$
– Swapnil Rustagi
Jan 24 at 15:25
add a comment |
$begingroup$
$ U^{*} = U = begin{pmatrix} frac{2}{N} & frac{2}{N} & ... & frac{2}{N} \
frac{2}{N} & frac{2}{N} & ... & frac{2}{N} \ vdots & vdots & ddots & vdots \
frac{2}{N} & frac{2}{N} & ... & frac{2}{N} end{pmatrix} - I $
(conjugate transpose is just transpose here, as conjugate of a real is the real number itself).
Let us take a new matrix $A$,
$ A= begin{pmatrix} frac{2}{N} & frac{2}{N} & ... & frac{2}{N} \
frac{2}{N} & frac{2}{N} & ... & frac{2}{N} \ vdots & vdots & ddots & vdots \
frac{2}{N} & frac{2}{N} & ... & frac{2}{N} end{pmatrix} $
So $U^{*}U= U^2 = (A-I)*(A-I) = A^2 - 2AI + I = A^2 - 2A + I$
$ A^2= begin{pmatrix} frac{4}{N} & frac{4}{N} & ... & frac{4}{N} \
frac{4}{N} & frac{4}{N} & ... & frac{4}{N} \ vdots & vdots & ddots & vdots \
frac{4}{N} & frac{4}{N} & ... & frac{4}{N} end{pmatrix} = 2A $
So $A^2-2A$ is null matrix (all entries $0$)
And so $U^{*}U= I$
answered Jan 24 at 13:55
Swapnil RustagiSwapnil Rustagi
758721
$endgroup$
$begingroup$
Thank you for your answer, I would like to reward you with an upvote, but I do not have enough reputation ... How did you come to the $4/N$ in the $A^2$ matrix? Can you explain that a little bit more in detail? My guess is that it's because you run the $((2/N)*(2/N))$ N times.
$endgroup$
– QuantaMag
Jan 24 at 14:37
$begingroup$
@QuantaMag Yes, exactly, I assume you can see how each entry of $A^2$ is obtained just by matrix multiplication, just like the way you described. $((2/N)*(2/N)) + ((2/N)*(2/N)) + ... $ (N times) = $ N*(4/N^2) = 4/N $. If there is still any confusion, I will edit my answer to clarify further. Also, if you found my answer satisfactorily answers your question, you can accept my answer, even if you do not have reputation to upvote.
$endgroup$
– Swapnil Rustagi
Jan 24 at 15:25
add a comment |
$begingroup$
$ U^{*} = U = begin{pmatrix} frac{2}{N} & frac{2}{N} & ... & frac{2}{N} \
frac{2}{N} & frac{2}{N} & ... & frac{2}{N} \ vdots & vdots & ddots & vdots \
frac{2}{N} & frac{2}{N} & ... & frac{2}{N} end{pmatrix} - I $
(conjugate transpose is just transpose here, as conjugate of a real is the real number itself).
Let us take a new matrix $A$,
$ A= begin{pmatrix} frac{2}{N} & frac{2}{N} & ... & frac{2}{N} \
frac{2}{N} & frac{2}{N} & ... & frac{2}{N} \ vdots & vdots & ddots & vdots \
frac{2}{N} & frac{2}{N} & ... & frac{2}{N} end{pmatrix} $
So $U^{*}U= U^2 = (A-I)*(A-I) = A^2 - 2AI + I = A^2 - 2A + I$
$ A^2= begin{pmatrix} frac{4}{N} & frac{4}{N} & ... & frac{4}{N} \
frac{4}{N} & frac{4}{N} & ... & frac{4}{N} \ vdots & vdots & ddots & vdots \
frac{4}{N} & frac{4}{N} & ... & frac{4}{N} end{pmatrix} = 2A $
So $A^2-2A$ is null matrix (all entries $0$)
And so $U^{*}U= I$
answered Jan 24 at 13:55
Swapnil RustagiSwapnil Rustagi
758721
$endgroup$
$ U^{*} = U = begin{pmatrix} frac{2}{N} & frac{2}{N} & ... & frac{2}{N} \
frac{2}{N} & frac{2}{N} & ... & frac{2}{N} \ vdots & vdots & ddots & vdots \
frac{2}{N} & frac{2}{N} & ... & frac{2}{N} end{pmatrix} - I $
(conjugate transpose is just transpose here, as conjugate of a real is the real number itself).
Let us take a new matrix $A$,
$ A= begin{pmatrix} frac{2}{N} & frac{2}{N} & ... & frac{2}{N} \
frac{2}{N} & frac{2}{N} & ... & frac{2}{N} \ vdots & vdots & ddots & vdots \
frac{2}{N} & frac{2}{N} & ... & frac{2}{N} end{pmatrix} $
So $U^{*}U= U^2 = (A-I)*(A-I) = A^2 - 2AI + I = A^2 - 2A + I$
$ A^2= begin{pmatrix} frac{4}{N} & frac{4}{N} & ... & frac{4}{N} \
frac{4}{N} & frac{4}{N} & ... & frac{4}{N} \ vdots & vdots & ddots & vdots \
frac{4}{N} & frac{4}{N} & ... & frac{4}{N} end{pmatrix} = 2A $
So $A^2-2A$ is null matrix (all entries $0$)
And so $U^{*}U= I$
answered Jan 24 at 13:55
Swapnil RustagiSwapnil Rustagi
758721
answered Jan 24 at 13:55
Swapnil RustagiSwapnil Rustagi
758721
answered Jan 24 at 13:55
Swapnil RustagiSwapnil Rustagi
758721
answered Jan 24 at 13:55
Swapnil RustagiSwapnil Rustagi
758721
758721
$begingroup$
Thank you for your answer, I would like to reward you with an upvote, but I do not have enough reputation ... How did you come to the $4/N$ in the $A^2$ matrix? Can you explain that a little bit more in detail? My guess is that it's because you run the $((2/N)*(2/N))$ N times.
$endgroup$
– QuantaMag
Jan 24 at 14:37
$begingroup$
@QuantaMag Yes, exactly, I assume you can see how each entry of $A^2$ is obtained just by matrix multiplication, just like the way you described. $((2/N)*(2/N)) + ((2/N)*(2/N)) + ... $ (N times) = $ N*(4/N^2) = 4/N $. If there is still any confusion, I will edit my answer to clarify further. Also, if you found my answer satisfactorily answers your question, you can accept my answer, even if you do not have reputation to upvote.
$endgroup$
– Swapnil Rustagi
Jan 24 at 15:25
add a comment |
$begingroup$
Thank you for your answer, I would like to reward you with an upvote, but I do not have enough reputation ... How did you come to the $4/N$ in the $A^2$ matrix? Can you explain that a little bit more in detail? My guess is that it's because you run the $((2/N)*(2/N))$ N times.
$endgroup$
– QuantaMag
Jan 24 at 14:37
$begingroup$
@QuantaMag Yes, exactly, I assume you can see how each entry of $A^2$ is obtained just by matrix multiplication, just like the way you described. $((2/N)*(2/N)) + ((2/N)*(2/N)) + ... $ (N times) = $ N*(4/N^2) = 4/N $. If there is still any confusion, I will edit my answer to clarify further. Also, if you found my answer satisfactorily answers your question, you can accept my answer, even if you do not have reputation to upvote.
$endgroup$
– Swapnil Rustagi
Jan 24 at 15:25
$begingroup$
Thank you for your answer, I would like to reward you with an upvote, but I do not have enough reputation ... How did you come to the $4/N$ in the $A^2$ matrix? Can you explain that a little bit more in detail? My guess is that it's because you run the $((2/N)*(2/N))$ N times.
$endgroup$
– QuantaMag
Jan 24 at 14:37
$begingroup$
Thank you for your answer, I would like to reward you with an upvote, but I do not have enough reputation ... How did you come to the $4/N$ in the $A^2$ matrix? Can you explain that a little bit more in detail? My guess is that it's because you run the $((2/N)*(2/N))$ N times.
$endgroup$
– QuantaMag
Jan 24 at 14:37
$begingroup$
@QuantaMag Yes, exactly, I assume you can see how each entry of $A^2$ is obtained just by matrix multiplication, just like the way you described. $((2/N)*(2/N)) + ((2/N)*(2/N)) + ... $ (N times) = $ N*(4/N^2) = 4/N $. If there is still any confusion, I will edit my answer to clarify further. Also, if you found my answer satisfactorily answers your question, you can accept my answer, even if you do not have reputation to upvote.
$endgroup$
– Swapnil Rustagi
Jan 24 at 15:25
$begingroup$
@QuantaMag Yes, exactly, I assume you can see how each entry of $A^2$ is obtained just by matrix multiplication, just like the way you described. $((2/N)*(2/N)) + ((2/N)*(2/N)) + ... $ (N times) = $ N*(4/N^2) = 4/N $. If there is still any confusion, I will edit my answer to clarify further. Also, if you found my answer satisfactorily answers your question, you can accept my answer, even if you do not have reputation to upvote.
$endgroup$
– Swapnil Rustagi
Jan 24 at 15:25
add a comment |
$begingroup$
Note that
$$
U=frac{2}{N}vv^T -I
$$ where $v=(1,1,ldots,1)^TinBbb R^N$. Hence $U=U^T$ and
$$
UU^T =U^2 =I-frac{4}{N}vv^T +frac{4}{N^2}(v^Tv)vv^T=I-frac{4}{N}vv^T +frac{4}{N}vv^T=I
$$ since $v^Tv = |v|^2=N$.
answered Jan 24 at 13:51
SongSong
18.1k21449
$endgroup$
$begingroup$
Thanks for your answer, I'm just trying to understand your answer. One more question, how did you come to this $v^Tv = |v|^2=N$ ?
$endgroup$
– QuantaMag
Jan 24 at 14:45
$begingroup$
@QuantaMag That is because $v^T v =vcdot v = |v|^2=1^2+1^2+cdots +1^2=N$ where $cdot$ is standard inner(scalar) product, i.e. $vec{a}cdot vec{b} = a_1b_1+a_2b_2+cdots+ a_Nb_N$.
$endgroup$
– Song
Jan 24 at 14:46
add a comment |
$begingroup$
Note that
$$
U=frac{2}{N}vv^T -I
$$ where $v=(1,1,ldots,1)^TinBbb R^N$. Hence $U=U^T$ and
$$
UU^T =U^2 =I-frac{4}{N}vv^T +frac{4}{N^2}(v^Tv)vv^T=I-frac{4}{N}vv^T +frac{4}{N}vv^T=I
$$ since $v^Tv = |v|^2=N$.
answered Jan 24 at 13:51
SongSong
18.1k21449
$endgroup$
$begingroup$
Thanks for your answer, I'm just trying to understand your answer. One more question, how did you come to this $v^Tv = |v|^2=N$ ?
$endgroup$
– QuantaMag
Jan 24 at 14:45
$begingroup$
@QuantaMag That is because $v^T v =vcdot v = |v|^2=1^2+1^2+cdots +1^2=N$ where $cdot$ is standard inner(scalar) product, i.e. $vec{a}cdot vec{b} = a_1b_1+a_2b_2+cdots+ a_Nb_N$.
$endgroup$
– Song
Jan 24 at 14:46
add a comment |
$begingroup$
Note that
$$
U=frac{2}{N}vv^T -I
$$ where $v=(1,1,ldots,1)^TinBbb R^N$. Hence $U=U^T$ and
$$
UU^T =U^2 =I-frac{4}{N}vv^T +frac{4}{N^2}(v^Tv)vv^T=I-frac{4}{N}vv^T +frac{4}{N}vv^T=I
$$ since $v^Tv = |v|^2=N$.
answered Jan 24 at 13:51
SongSong
18.1k21449
$endgroup$
Note that
$$
U=frac{2}{N}vv^T -I
$$ where $v=(1,1,ldots,1)^TinBbb R^N$. Hence $U=U^T$ and
$$
UU^T =U^2 =I-frac{4}{N}vv^T +frac{4}{N^2}(v^Tv)vv^T=I-frac{4}{N}vv^T +frac{4}{N}vv^T=I
$$ since $v^Tv = |v|^2=N$.
answered Jan 24 at 13:51
SongSong
18.1k21449
answered Jan 24 at 13:51
SongSong
18.1k21449
answered Jan 24 at 13:51
SongSong
18.1k21449
answered Jan 24 at 13:51
SongSong
18.1k21449
18.1k21449
$begingroup$
Thanks for your answer, I'm just trying to understand your answer. One more question, how did you come to this $v^Tv = |v|^2=N$ ?
$endgroup$
– QuantaMag
Jan 24 at 14:45
$begingroup$
@QuantaMag That is because $v^T v =vcdot v = |v|^2=1^2+1^2+cdots +1^2=N$ where $cdot$ is standard inner(scalar) product, i.e. $vec{a}cdot vec{b} = a_1b_1+a_2b_2+cdots+ a_Nb_N$.
$endgroup$
– Song
Jan 24 at 14:46
add a comment |
$begingroup$
Thanks for your answer, I'm just trying to understand your answer. One more question, how did you come to this $v^Tv = |v|^2=N$ ?
$endgroup$
– QuantaMag
Jan 24 at 14:45
$begingroup$
@QuantaMag That is because $v^T v =vcdot v = |v|^2=1^2+1^2+cdots +1^2=N$ where $cdot$ is standard inner(scalar) product, i.e. $vec{a}cdot vec{b} = a_1b_1+a_2b_2+cdots+ a_Nb_N$.
$endgroup$
– Song
Jan 24 at 14:46
$begingroup$
Thanks for your answer, I'm just trying to understand your answer. One more question, how did you come to this $v^Tv = |v|^2=N$ ?
$endgroup$
– QuantaMag
Jan 24 at 14:45
$begingroup$
Thanks for your answer, I'm just trying to understand your answer. One more question, how did you come to this $v^Tv = |v|^2=N$ ?
$endgroup$
– QuantaMag
Jan 24 at 14:45
$begingroup$
@QuantaMag That is because $v^T v =vcdot v = |v|^2=1^2+1^2+cdots +1^2=N$ where $cdot$ is standard inner(scalar) product, i.e. $vec{a}cdot vec{b} = a_1b_1+a_2b_2+cdots+ a_Nb_N$.
$endgroup$
– Song
Jan 24 at 14:46
$begingroup$
@QuantaMag That is because $v^T v =vcdot v = |v|^2=1^2+1^2+cdots +1^2=N$ where $cdot$ is standard inner(scalar) product, i.e. $vec{a}cdot vec{b} = a_1b_1+a_2b_2+cdots+ a_Nb_N$.
$endgroup$
– Song
Jan 24 at 14:46
add a comment |
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