Inner Product and Operators: Exercise with vectors with three indices
$begingroup$
I have the following question:
I started off as follows: Let $vin ell(mathbb{N}, mathbb{C})$, and $w,x$ defined as above. Then:
begin{align}
langle w, xrangle_J &= sum_{(i,j,k)in J}w_{[i, j, k]} bar{x}_{[i,j,k]}\
&=sum_{(i,j,k)in J} |v_{[j]}||v_{[k]}| && text{definition of $w,x$}\
end{align}
However I have literally no idea about how to go forward..
Ideas
My main ideas are that supposedly the matrix $A$ will have at most $Ntimes N$ non-zero entries. Also, the definition of $J$ is such that the first index of each triplet $(i,j,k)$ basically represents one of $N$ rows containing at most $N$ elements, and the indexes $j$ and $k$ represents one of $N$ columns having at most $N$ elements.
functional-analysis operator-theory hilbert-spaces lebesgue-integral inner-product-space
asked Dec 6 '18 at 22:56
Euler_SalterEuler_Salter
2,0671336
$endgroup$
add a comment |
$begingroup$
I have the following question:
I started off as follows: Let $vin ell(mathbb{N}, mathbb{C})$, and $w,x$ defined as above. Then:
begin{align}
langle w, xrangle_J &= sum_{(i,j,k)in J}w_{[i, j, k]} bar{x}_{[i,j,k]}\
&=sum_{(i,j,k)in J} |v_{[j]}||v_{[k]}| && text{definition of $w,x$}\
end{align}
However I have literally no idea about how to go forward..
Ideas
My main ideas are that supposedly the matrix $A$ will have at most $Ntimes N$ non-zero entries. Also, the definition of $J$ is such that the first index of each triplet $(i,j,k)$ basically represents one of $N$ rows containing at most $N$ elements, and the indexes $j$ and $k$ represents one of $N$ columns having at most $N$ elements.
functional-analysis operator-theory hilbert-spaces lebesgue-integral inner-product-space
asked Dec 6 '18 at 22:56
Euler_SalterEuler_Salter
2,0671336
$endgroup$
add a comment |
$begingroup$
I have the following question:
I started off as follows: Let $vin ell(mathbb{N}, mathbb{C})$, and $w,x$ defined as above. Then:
begin{align}
langle w, xrangle_J &= sum_{(i,j,k)in J}w_{[i, j, k]} bar{x}_{[i,j,k]}\
&=sum_{(i,j,k)in J} |v_{[j]}||v_{[k]}| && text{definition of $w,x$}\
end{align}
However I have literally no idea about how to go forward..
Ideas
My main ideas are that supposedly the matrix $A$ will have at most $Ntimes N$ non-zero entries. Also, the definition of $J$ is such that the first index of each triplet $(i,j,k)$ basically represents one of $N$ rows containing at most $N$ elements, and the indexes $j$ and $k$ represents one of $N$ columns having at most $N$ elements.
functional-analysis operator-theory hilbert-spaces lebesgue-integral inner-product-space
asked Dec 6 '18 at 22:56
Euler_SalterEuler_Salter
2,0671336
$endgroup$
I have the following question:
I started off as follows: Let $vin ell(mathbb{N}, mathbb{C})$, and $w,x$ defined as above. Then:
begin{align}
langle w, xrangle_J &= sum_{(i,j,k)in J}w_{[i, j, k]} bar{x}_{[i,j,k]}\
&=sum_{(i,j,k)in J} |v_{[j]}||v_{[k]}| && text{definition of $w,x$}\
end{align}
However I have literally no idea about how to go forward..
Ideas
My main ideas are that supposedly the matrix $A$ will have at most $Ntimes N$ non-zero entries. Also, the definition of $J$ is such that the first index of each triplet $(i,j,k)$ basically represents one of $N$ rows containing at most $N$ elements, and the indexes $j$ and $k$ represents one of $N$ columns having at most $N$ elements.
functional-analysis operator-theory hilbert-spaces lebesgue-integral inner-product-space
functional-analysis operator-theory hilbert-spaces lebesgue-integral inner-product-space
asked Dec 6 '18 at 22:56
Euler_SalterEuler_Salter
2,0671336
asked Dec 6 '18 at 22:56
Euler_SalterEuler_Salter
2,0671336
edited Dec 6 '18 at 23:01
Euler_Salter
asked Dec 6 '18 at 22:56
Euler_SalterEuler_Salter
2,0671336
asked Dec 6 '18 at 22:56
Euler_SalterEuler_Salter
2,0671336
asked Dec 6 '18 at 22:56
Euler_SalterEuler_Salter
2,0671336
2,0671336
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Observe the following estimate:$$begin{eqnarray}
langle w, xrangle_J &=& sum_{(i,j,k)in J}w_{[i, j, k]} bar{x}_{[i,j,k]}\
&=&sum_{(i,j,k)in J} |v_{[j]}||v_{[k]}| \
&=&sum_{i} sum_{(j,k):(i,j,k)in J}|v_{[j]}||v_{[k]}|\
&= &sum_{i} left(sum_{j:a_{ij}neq 0}|v_{[j]}|right)^2\
&leq& sum_{i} left(sum_{j:a_{ij}neq 0}|v_{[j]}|^2 sum_{j:a_{ij}neq 0}1right)quadcdots(*)\
&leq&Nsum_{i} sum_{j:a_{ij}neq 0}|v_{[j]}|^2 \
&=&Nsum_{(i,j):a_{ij}neq 0} |v_{[j]}|^2 = Nsum_{j} |v_{[j]}|^2sum_{i:a_{ij}neq 0}1\
&leq &N^2 sum_j |v_{[j]}|^2 = N^2||v||^2,
end{eqnarray}$$ where Cauchy-Schwarz inequality is used at $(*)$. It is a combination of changing the order of the summation and the assumption that
$$
sum_{i:a_{ij}neq 0}1leq N, quad forall j,
$$ and
$$
sum_{j:a_{ij}neq 0}1leq Nquadforall i.
$$
answered Dec 9 '18 at 10:58
SongSong
12.2k630
$endgroup$
$begingroup$
Thanks a lot for your answer! I am not sure about the third step where you obtain $sum_i left(sum_{j:a_{ij}neq 0} |v_[j]|right)^2$. How do you pass from the "cross-product" to that? I totally understand that both the $j$ and the $k$ coordinates represent when the value in column $j$ for row $i$ is not zero $a_{ij}neq0$, but how do you get that expression?
$endgroup$
– Euler_Salter
Dec 9 '18 at 11:11
1
$begingroup$
Note that $(j,k):(i,j,k)in J$ is equivalent to $(j,k):a_{ij}neq 0, a_{ik} neq 0$. Given that $i$ is fixed, it is just the product of the sets ${j;|;a_{ij}neq 0}$ and ${k;|;a_{ik}neq 0}$. Thus the sum is separately evaluated as $sum_{j:a_{ij}neq 0} |v_{[j]}|$ times $sum_{k:a_{ik}neq 0}|v_{[k]}| $.
$endgroup$
– Song
Dec 9 '18 at 11:15
$begingroup$
Thanks! Okay, sorry if I'm drilling this down so much but I really want to be sure to understand. So what you're saying is that if we define the set $P={j, :, a_{ij}neq 0}$ Then in $sum_{(j,k):(i,j,k)in J}|v_{[j]}||v_{[k]}|$ we are summing over the set $Ptimes P:={(j, k), : , jin P, kin P}$, i.e. the Carthesian Product of $P$ with itself? I think I understand this. However I am not sure how we can use this to say that we can evaluate the sums separately
$endgroup$
– Euler_Salter
Dec 9 '18 at 11:36
1
$begingroup$
Yes. If a double sum is taken over a finite set, it can be calculated by iterative summation.
$endgroup$
– Song
Dec 9 '18 at 11:47
1
$begingroup$
Yes. Write $|v_{[j]}| = |v_{[j]}|cdot 1$ and apply (finite version of) Cauchy-Schwarz there.
$endgroup$
– Song
Dec 9 '18 at 11:55
|
show 5 more comments
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Observe the following estimate:$$begin{eqnarray}
langle w, xrangle_J &=& sum_{(i,j,k)in J}w_{[i, j, k]} bar{x}_{[i,j,k]}\
&=&sum_{(i,j,k)in J} |v_{[j]}||v_{[k]}| \
&=&sum_{i} sum_{(j,k):(i,j,k)in J}|v_{[j]}||v_{[k]}|\
&= &sum_{i} left(sum_{j:a_{ij}neq 0}|v_{[j]}|right)^2\
&leq& sum_{i} left(sum_{j:a_{ij}neq 0}|v_{[j]}|^2 sum_{j:a_{ij}neq 0}1right)quadcdots(*)\
&leq&Nsum_{i} sum_{j:a_{ij}neq 0}|v_{[j]}|^2 \
&=&Nsum_{(i,j):a_{ij}neq 0} |v_{[j]}|^2 = Nsum_{j} |v_{[j]}|^2sum_{i:a_{ij}neq 0}1\
&leq &N^2 sum_j |v_{[j]}|^2 = N^2||v||^2,
end{eqnarray}$$ where Cauchy-Schwarz inequality is used at $(*)$. It is a combination of changing the order of the summation and the assumption that
$$
sum_{i:a_{ij}neq 0}1leq N, quad forall j,
$$ and
$$
sum_{j:a_{ij}neq 0}1leq Nquadforall i.
$$
answered Dec 9 '18 at 10:58
SongSong
12.2k630
$endgroup$
$begingroup$
Thanks a lot for your answer! I am not sure about the third step where you obtain $sum_i left(sum_{j:a_{ij}neq 0} |v_[j]|right)^2$. How do you pass from the "cross-product" to that? I totally understand that both the $j$ and the $k$ coordinates represent when the value in column $j$ for row $i$ is not zero $a_{ij}neq0$, but how do you get that expression?
$endgroup$
– Euler_Salter
Dec 9 '18 at 11:11
1
$begingroup$
Note that $(j,k):(i,j,k)in J$ is equivalent to $(j,k):a_{ij}neq 0, a_{ik} neq 0$. Given that $i$ is fixed, it is just the product of the sets ${j;|;a_{ij}neq 0}$ and ${k;|;a_{ik}neq 0}$. Thus the sum is separately evaluated as $sum_{j:a_{ij}neq 0} |v_{[j]}|$ times $sum_{k:a_{ik}neq 0}|v_{[k]}| $.
$endgroup$
– Song
Dec 9 '18 at 11:15
$begingroup$
Thanks! Okay, sorry if I'm drilling this down so much but I really want to be sure to understand. So what you're saying is that if we define the set $P={j, :, a_{ij}neq 0}$ Then in $sum_{(j,k):(i,j,k)in J}|v_{[j]}||v_{[k]}|$ we are summing over the set $Ptimes P:={(j, k), : , jin P, kin P}$, i.e. the Carthesian Product of $P$ with itself? I think I understand this. However I am not sure how we can use this to say that we can evaluate the sums separately
$endgroup$
– Euler_Salter
Dec 9 '18 at 11:36
1
$begingroup$
Yes. If a double sum is taken over a finite set, it can be calculated by iterative summation.
$endgroup$
– Song
Dec 9 '18 at 11:47
1
$begingroup$
Yes. Write $|v_{[j]}| = |v_{[j]}|cdot 1$ and apply (finite version of) Cauchy-Schwarz there.
$endgroup$
– Song
Dec 9 '18 at 11:55
|
show 5 more comments
$begingroup$
Observe the following estimate:$$begin{eqnarray}
langle w, xrangle_J &=& sum_{(i,j,k)in J}w_{[i, j, k]} bar{x}_{[i,j,k]}\
&=&sum_{(i,j,k)in J} |v_{[j]}||v_{[k]}| \
&=&sum_{i} sum_{(j,k):(i,j,k)in J}|v_{[j]}||v_{[k]}|\
&= &sum_{i} left(sum_{j:a_{ij}neq 0}|v_{[j]}|right)^2\
&leq& sum_{i} left(sum_{j:a_{ij}neq 0}|v_{[j]}|^2 sum_{j:a_{ij}neq 0}1right)quadcdots(*)\
&leq&Nsum_{i} sum_{j:a_{ij}neq 0}|v_{[j]}|^2 \
&=&Nsum_{(i,j):a_{ij}neq 0} |v_{[j]}|^2 = Nsum_{j} |v_{[j]}|^2sum_{i:a_{ij}neq 0}1\
&leq &N^2 sum_j |v_{[j]}|^2 = N^2||v||^2,
end{eqnarray}$$ where Cauchy-Schwarz inequality is used at $(*)$. It is a combination of changing the order of the summation and the assumption that
$$
sum_{i:a_{ij}neq 0}1leq N, quad forall j,
$$ and
$$
sum_{j:a_{ij}neq 0}1leq Nquadforall i.
$$
answered Dec 9 '18 at 10:58
SongSong
12.2k630
$endgroup$
$begingroup$
Thanks a lot for your answer! I am not sure about the third step where you obtain $sum_i left(sum_{j:a_{ij}neq 0} |v_[j]|right)^2$. How do you pass from the "cross-product" to that? I totally understand that both the $j$ and the $k$ coordinates represent when the value in column $j$ for row $i$ is not zero $a_{ij}neq0$, but how do you get that expression?
$endgroup$
– Euler_Salter
Dec 9 '18 at 11:11
1
$begingroup$
Note that $(j,k):(i,j,k)in J$ is equivalent to $(j,k):a_{ij}neq 0, a_{ik} neq 0$. Given that $i$ is fixed, it is just the product of the sets ${j;|;a_{ij}neq 0}$ and ${k;|;a_{ik}neq 0}$. Thus the sum is separately evaluated as $sum_{j:a_{ij}neq 0} |v_{[j]}|$ times $sum_{k:a_{ik}neq 0}|v_{[k]}| $.
$endgroup$
– Song
Dec 9 '18 at 11:15
$begingroup$
Thanks! Okay, sorry if I'm drilling this down so much but I really want to be sure to understand. So what you're saying is that if we define the set $P={j, :, a_{ij}neq 0}$ Then in $sum_{(j,k):(i,j,k)in J}|v_{[j]}||v_{[k]}|$ we are summing over the set $Ptimes P:={(j, k), : , jin P, kin P}$, i.e. the Carthesian Product of $P$ with itself? I think I understand this. However I am not sure how we can use this to say that we can evaluate the sums separately
$endgroup$
– Euler_Salter
Dec 9 '18 at 11:36
1
$begingroup$
Yes. If a double sum is taken over a finite set, it can be calculated by iterative summation.
$endgroup$
– Song
Dec 9 '18 at 11:47
1
$begingroup$
Yes. Write $|v_{[j]}| = |v_{[j]}|cdot 1$ and apply (finite version of) Cauchy-Schwarz there.
$endgroup$
– Song
Dec 9 '18 at 11:55
|
show 5 more comments
$begingroup$
Observe the following estimate:$$begin{eqnarray}
langle w, xrangle_J &=& sum_{(i,j,k)in J}w_{[i, j, k]} bar{x}_{[i,j,k]}\
&=&sum_{(i,j,k)in J} |v_{[j]}||v_{[k]}| \
&=&sum_{i} sum_{(j,k):(i,j,k)in J}|v_{[j]}||v_{[k]}|\
&= &sum_{i} left(sum_{j:a_{ij}neq 0}|v_{[j]}|right)^2\
&leq& sum_{i} left(sum_{j:a_{ij}neq 0}|v_{[j]}|^2 sum_{j:a_{ij}neq 0}1right)quadcdots(*)\
&leq&Nsum_{i} sum_{j:a_{ij}neq 0}|v_{[j]}|^2 \
&=&Nsum_{(i,j):a_{ij}neq 0} |v_{[j]}|^2 = Nsum_{j} |v_{[j]}|^2sum_{i:a_{ij}neq 0}1\
&leq &N^2 sum_j |v_{[j]}|^2 = N^2||v||^2,
end{eqnarray}$$ where Cauchy-Schwarz inequality is used at $(*)$. It is a combination of changing the order of the summation and the assumption that
$$
sum_{i:a_{ij}neq 0}1leq N, quad forall j,
$$ and
$$
sum_{j:a_{ij}neq 0}1leq Nquadforall i.
$$
answered Dec 9 '18 at 10:58
SongSong
12.2k630
$endgroup$
Observe the following estimate:$$begin{eqnarray}
langle w, xrangle_J &=& sum_{(i,j,k)in J}w_{[i, j, k]} bar{x}_{[i,j,k]}\
&=&sum_{(i,j,k)in J} |v_{[j]}||v_{[k]}| \
&=&sum_{i} sum_{(j,k):(i,j,k)in J}|v_{[j]}||v_{[k]}|\
&= &sum_{i} left(sum_{j:a_{ij}neq 0}|v_{[j]}|right)^2\
&leq& sum_{i} left(sum_{j:a_{ij}neq 0}|v_{[j]}|^2 sum_{j:a_{ij}neq 0}1right)quadcdots(*)\
&leq&Nsum_{i} sum_{j:a_{ij}neq 0}|v_{[j]}|^2 \
&=&Nsum_{(i,j):a_{ij}neq 0} |v_{[j]}|^2 = Nsum_{j} |v_{[j]}|^2sum_{i:a_{ij}neq 0}1\
&leq &N^2 sum_j |v_{[j]}|^2 = N^2||v||^2,
end{eqnarray}$$ where Cauchy-Schwarz inequality is used at $(*)$. It is a combination of changing the order of the summation and the assumption that
$$
sum_{i:a_{ij}neq 0}1leq N, quad forall j,
$$ and
$$
sum_{j:a_{ij}neq 0}1leq Nquadforall i.
$$
answered Dec 9 '18 at 10:58
SongSong
12.2k630
answered Dec 9 '18 at 10:58
SongSong
12.2k630
answered Dec 9 '18 at 10:58
SongSong
12.2k630
answered Dec 9 '18 at 10:58
SongSong
12.2k630
12.2k630
$begingroup$
Thanks a lot for your answer! I am not sure about the third step where you obtain $sum_i left(sum_{j:a_{ij}neq 0} |v_[j]|right)^2$. How do you pass from the "cross-product" to that? I totally understand that both the $j$ and the $k$ coordinates represent when the value in column $j$ for row $i$ is not zero $a_{ij}neq0$, but how do you get that expression?
$endgroup$
– Euler_Salter
Dec 9 '18 at 11:11
1
$begingroup$
Note that $(j,k):(i,j,k)in J$ is equivalent to $(j,k):a_{ij}neq 0, a_{ik} neq 0$. Given that $i$ is fixed, it is just the product of the sets ${j;|;a_{ij}neq 0}$ and ${k;|;a_{ik}neq 0}$. Thus the sum is separately evaluated as $sum_{j:a_{ij}neq 0} |v_{[j]}|$ times $sum_{k:a_{ik}neq 0}|v_{[k]}| $.
$endgroup$
– Song
Dec 9 '18 at 11:15
$begingroup$
Thanks! Okay, sorry if I'm drilling this down so much but I really want to be sure to understand. So what you're saying is that if we define the set $P={j, :, a_{ij}neq 0}$ Then in $sum_{(j,k):(i,j,k)in J}|v_{[j]}||v_{[k]}|$ we are summing over the set $Ptimes P:={(j, k), : , jin P, kin P}$, i.e. the Carthesian Product of $P$ with itself? I think I understand this. However I am not sure how we can use this to say that we can evaluate the sums separately
$endgroup$
– Euler_Salter
Dec 9 '18 at 11:36
1
$begingroup$
Yes. If a double sum is taken over a finite set, it can be calculated by iterative summation.
$endgroup$
– Song
Dec 9 '18 at 11:47
1
$begingroup$
Yes. Write $|v_{[j]}| = |v_{[j]}|cdot 1$ and apply (finite version of) Cauchy-Schwarz there.
$endgroup$
– Song
Dec 9 '18 at 11:55
|
show 5 more comments
$begingroup$
Thanks a lot for your answer! I am not sure about the third step where you obtain $sum_i left(sum_{j:a_{ij}neq 0} |v_[j]|right)^2$. How do you pass from the "cross-product" to that? I totally understand that both the $j$ and the $k$ coordinates represent when the value in column $j$ for row $i$ is not zero $a_{ij}neq0$, but how do you get that expression?
$endgroup$
– Euler_Salter
Dec 9 '18 at 11:11
1
$begingroup$
Note that $(j,k):(i,j,k)in J$ is equivalent to $(j,k):a_{ij}neq 0, a_{ik} neq 0$. Given that $i$ is fixed, it is just the product of the sets ${j;|;a_{ij}neq 0}$ and ${k;|;a_{ik}neq 0}$. Thus the sum is separately evaluated as $sum_{j:a_{ij}neq 0} |v_{[j]}|$ times $sum_{k:a_{ik}neq 0}|v_{[k]}| $.
$endgroup$
– Song
Dec 9 '18 at 11:15
$begingroup$
Thanks! Okay, sorry if I'm drilling this down so much but I really want to be sure to understand. So what you're saying is that if we define the set $P={j, :, a_{ij}neq 0}$ Then in $sum_{(j,k):(i,j,k)in J}|v_{[j]}||v_{[k]}|$ we are summing over the set $Ptimes P:={(j, k), : , jin P, kin P}$, i.e. the Carthesian Product of $P$ with itself? I think I understand this. However I am not sure how we can use this to say that we can evaluate the sums separately
$endgroup$
– Euler_Salter
Dec 9 '18 at 11:36
1
$begingroup$
Yes. If a double sum is taken over a finite set, it can be calculated by iterative summation.
$endgroup$
– Song
Dec 9 '18 at 11:47
1
$begingroup$
Yes. Write $|v_{[j]}| = |v_{[j]}|cdot 1$ and apply (finite version of) Cauchy-Schwarz there.
$endgroup$
– Song
Dec 9 '18 at 11:55
$begingroup$
Thanks a lot for your answer! I am not sure about the third step where you obtain $sum_i left(sum_{j:a_{ij}neq 0} |v_[j]|right)^2$. How do you pass from the "cross-product" to that? I totally understand that both the $j$ and the $k$ coordinates represent when the value in column $j$ for row $i$ is not zero $a_{ij}neq0$, but how do you get that expression?
$endgroup$
– Euler_Salter
Dec 9 '18 at 11:11
$begingroup$
Thanks a lot for your answer! I am not sure about the third step where you obtain $sum_i left(sum_{j:a_{ij}neq 0} |v_[j]|right)^2$. How do you pass from the "cross-product" to that? I totally understand that both the $j$ and the $k$ coordinates represent when the value in column $j$ for row $i$ is not zero $a_{ij}neq0$, but how do you get that expression?
$endgroup$
– Euler_Salter
Dec 9 '18 at 11:11
1
1
$begingroup$
Note that $(j,k):(i,j,k)in J$ is equivalent to $(j,k):a_{ij}neq 0, a_{ik} neq 0$. Given that $i$ is fixed, it is just the product of the sets ${j;|;a_{ij}neq 0}$ and ${k;|;a_{ik}neq 0}$. Thus the sum is separately evaluated as $sum_{j:a_{ij}neq 0} |v_{[j]}|$ times $sum_{k:a_{ik}neq 0}|v_{[k]}| $.
$endgroup$
– Song
Dec 9 '18 at 11:15
$begingroup$
Note that $(j,k):(i,j,k)in J$ is equivalent to $(j,k):a_{ij}neq 0, a_{ik} neq 0$. Given that $i$ is fixed, it is just the product of the sets ${j;|;a_{ij}neq 0}$ and ${k;|;a_{ik}neq 0}$. Thus the sum is separately evaluated as $sum_{j:a_{ij}neq 0} |v_{[j]}|$ times $sum_{k:a_{ik}neq 0}|v_{[k]}| $.
$endgroup$
– Song
Dec 9 '18 at 11:15
$begingroup$
Thanks! Okay, sorry if I'm drilling this down so much but I really want to be sure to understand. So what you're saying is that if we define the set $P={j, :, a_{ij}neq 0}$ Then in $sum_{(j,k):(i,j,k)in J}|v_{[j]}||v_{[k]}|$ we are summing over the set $Ptimes P:={(j, k), : , jin P, kin P}$, i.e. the Carthesian Product of $P$ with itself? I think I understand this. However I am not sure how we can use this to say that we can evaluate the sums separately
$endgroup$
– Euler_Salter
Dec 9 '18 at 11:36
$begingroup$
Thanks! Okay, sorry if I'm drilling this down so much but I really want to be sure to understand. So what you're saying is that if we define the set $P={j, :, a_{ij}neq 0}$ Then in $sum_{(j,k):(i,j,k)in J}|v_{[j]}||v_{[k]}|$ we are summing over the set $Ptimes P:={(j, k), : , jin P, kin P}$, i.e. the Carthesian Product of $P$ with itself? I think I understand this. However I am not sure how we can use this to say that we can evaluate the sums separately
$endgroup$
– Euler_Salter
Dec 9 '18 at 11:36
1
1
$begingroup$
Yes. If a double sum is taken over a finite set, it can be calculated by iterative summation.
$endgroup$
– Song
Dec 9 '18 at 11:47
$begingroup$
Yes. If a double sum is taken over a finite set, it can be calculated by iterative summation.
$endgroup$
– Song
Dec 9 '18 at 11:47
1
1
$begingroup$
Yes. Write $|v_{[j]}| = |v_{[j]}|cdot 1$ and apply (finite version of) Cauchy-Schwarz there.
$endgroup$
– Song
Dec 9 '18 at 11:55
$begingroup$
Yes. Write $|v_{[j]}| = |v_{[j]}|cdot 1$ and apply (finite version of) Cauchy-Schwarz there.
$endgroup$
– Song
Dec 9 '18 at 11:55
|
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