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Proving ${u_k}to u$ given $lim_{ktoinfty}langle u_k,vrangle=langle u,vrangle$ for $uin mathbb{R}^n,forall vin...
$begingroup$
I'm having trouble solving the following problem.
Problem. Prove ${u_k}to u$ given $lim_{ktoinfty}langle u_k,vrangle=langle u,vrangle$ for $uin mathbb{R}^n$, $forall vin mathbb{R}^n$
The textbook introduces the $i^{text{th}}$ component function $p_{i} : mathbb{R}^{n} to mathbb{R}$ for $1 leq i leq n$ by $p_{i}(u) = u_{i}$, where $u in mathbb{R}^{n}$.
Using this definition, I can express any vector $u in mathbb{R}^{n}$ by $u = (p_{1}(u), ldots, p_{n}(u))$. I know that this function is linear.
The book also tells us that a sequence ${u_{k}}$ converges to $u$ in $mathbb{R}^{n}$ if and only if it converges componentwise (i.e. for each $1 leq i leq n$, $lim_{ktoinfty} p_{i}(u_{k}) = p_{i}(u)).$
The book provides a hint to define the point $e_{i} in mathbb{R}^{n}$ whose $i^{text{ith}}$ component is equal to $1$ and every other component equals $0$. This way, $p_{i}(u) = langle u, e_{i}rangle$ for each point $u in mathbb{R}^{n}$. I've been working with this component function and don't seem to be making any progress. Any help is appreciated
Note:$langle u,vrangle$ denotes $ucdot v$, u and v are points in $mathbb{R}^n$, $u_k$ is a sequence in $mathbb{R}^n$.
My attempt:
Letting $u_k=(u_1^k,u_2^k,...u_n^k)$ with the $k$ representing the $k$-th term in the sequence $u=(u_1,u_2,...u_n)$. Then $lim_{ktoinfty}u_i^k=u_i$ $forall i$.
$lim_{ktoinfty}langle u_k,vrangle=langle u,vrangle$ for any given $vinmathbb{R}$
, let $v=e_i$ then $lim_{ktoinfty}langle u_k,e_irangle=langle u,e_irangle$.
Before proceeding we establish $langle e_i, e_irangle=1$ and $langle e_i,e_jrangle=0$ assuming $ineq j$.
begin{align*}
lim_{ktoinfty} langle u_k,e_irangle
&= Big< lim_{ktoinfty} u_k, e_i Big>
= Big< lim_{ktoinfty}(u_1^k, u_2^k, cdots, u_n^k),e_i Big> \
&= Big< Big( lim_{ktoinfty}u_1^k, lim_{ktoinfty}u_2^k, cdots, lim_{ktoinfty}u_n^k Big),e_i Big>
=lim_{ktoinfty} u_i^k
end{align*}
Now we have
$$lim_{ktoinfty}langle u_k,e_irangle
= lim_{ktoinfty}u_i^k
= langle u,e_irangle=langle (u_1,u_2,...u_n),e_irangle
= u_i, forall i quad Box$$
real-analysis linear-algebra sequences-and-series limits convergence
edited Feb 3 at 6:11
Sangchul Lee
96.6k12173283
asked Feb 3 at 5:52
Dillain SmithDillain Smith
848
$endgroup$
add a comment |
$begingroup$
I'm having trouble solving the following problem.
Problem. Prove ${u_k}to u$ given $lim_{ktoinfty}langle u_k,vrangle=langle u,vrangle$ for $uin mathbb{R}^n$, $forall vin mathbb{R}^n$
The textbook introduces the $i^{text{th}}$ component function $p_{i} : mathbb{R}^{n} to mathbb{R}$ for $1 leq i leq n$ by $p_{i}(u) = u_{i}$, where $u in mathbb{R}^{n}$.
Using this definition, I can express any vector $u in mathbb{R}^{n}$ by $u = (p_{1}(u), ldots, p_{n}(u))$. I know that this function is linear.
The book also tells us that a sequence ${u_{k}}$ converges to $u$ in $mathbb{R}^{n}$ if and only if it converges componentwise (i.e. for each $1 leq i leq n$, $lim_{ktoinfty} p_{i}(u_{k}) = p_{i}(u)).$
The book provides a hint to define the point $e_{i} in mathbb{R}^{n}$ whose $i^{text{ith}}$ component is equal to $1$ and every other component equals $0$. This way, $p_{i}(u) = langle u, e_{i}rangle$ for each point $u in mathbb{R}^{n}$. I've been working with this component function and don't seem to be making any progress. Any help is appreciated
Note:$langle u,vrangle$ denotes $ucdot v$, u and v are points in $mathbb{R}^n$, $u_k$ is a sequence in $mathbb{R}^n$.
My attempt:
Letting $u_k=(u_1^k,u_2^k,...u_n^k)$ with the $k$ representing the $k$-th term in the sequence $u=(u_1,u_2,...u_n)$. Then $lim_{ktoinfty}u_i^k=u_i$ $forall i$.
$lim_{ktoinfty}langle u_k,vrangle=langle u,vrangle$ for any given $vinmathbb{R}$
, let $v=e_i$ then $lim_{ktoinfty}langle u_k,e_irangle=langle u,e_irangle$.
Before proceeding we establish $langle e_i, e_irangle=1$ and $langle e_i,e_jrangle=0$ assuming $ineq j$.
begin{align*}
lim_{ktoinfty} langle u_k,e_irangle
&= Big< lim_{ktoinfty} u_k, e_i Big>
= Big< lim_{ktoinfty}(u_1^k, u_2^k, cdots, u_n^k),e_i Big> \
&= Big< Big( lim_{ktoinfty}u_1^k, lim_{ktoinfty}u_2^k, cdots, lim_{ktoinfty}u_n^k Big),e_i Big>
=lim_{ktoinfty} u_i^k
end{align*}
Now we have
$$lim_{ktoinfty}langle u_k,e_irangle
= lim_{ktoinfty}u_i^k
= langle u,e_irangle=langle (u_1,u_2,...u_n),e_irangle
= u_i, forall i quad Box$$
real-analysis linear-algebra sequences-and-series limits convergence
edited Feb 3 at 6:11
Sangchul Lee
96.6k12173283
asked Feb 3 at 5:52
Dillain SmithDillain Smith
848
$endgroup$
1
$begingroup$
You should uselangle$langle$ andrangle$rangle$ instead of<and>. The former are delimiters, so the spacing for them is similar to that given for parentheses. The spacing for the latter is that of operators. That’s why it looks so bad when you use them. Compare $<e_1,e_2>$ with $langle e_1,e_2rangle$.
$endgroup$
– Arturo Magidin
Feb 3 at 6:04
2
$begingroup$
Well, if you know that $u_i^k to u_i$ as $ktoinfty$ for each $i=1,cdots,n$, then the proof is done. And this follows from the assumption by plugging $v = e_i$ for each $i$. Your intermediate computation is puzzling me because $lim_k langle u_k, e_i rangle = lim_k u_i^k$ holds simply because $ langle u_k, e_i rangle = u_i^k $ for all $k$ and $i$, so I am not sure why you made a significant detour for it.
$endgroup$
– Sangchul Lee
Feb 3 at 6:14
$begingroup$
@Sangchul Lee How can you just assume $v = e_{i}$; wouldn't you lose generality? We are trying to prove it for all $v$
$endgroup$
– user614735
Feb 3 at 16:21
add a comment |
$begingroup$
I'm having trouble solving the following problem.
Problem. Prove ${u_k}to u$ given $lim_{ktoinfty}langle u_k,vrangle=langle u,vrangle$ for $uin mathbb{R}^n$, $forall vin mathbb{R}^n$
The textbook introduces the $i^{text{th}}$ component function $p_{i} : mathbb{R}^{n} to mathbb{R}$ for $1 leq i leq n$ by $p_{i}(u) = u_{i}$, where $u in mathbb{R}^{n}$.
Using this definition, I can express any vector $u in mathbb{R}^{n}$ by $u = (p_{1}(u), ldots, p_{n}(u))$. I know that this function is linear.
The book also tells us that a sequence ${u_{k}}$ converges to $u$ in $mathbb{R}^{n}$ if and only if it converges componentwise (i.e. for each $1 leq i leq n$, $lim_{ktoinfty} p_{i}(u_{k}) = p_{i}(u)).$
The book provides a hint to define the point $e_{i} in mathbb{R}^{n}$ whose $i^{text{ith}}$ component is equal to $1$ and every other component equals $0$. This way, $p_{i}(u) = langle u, e_{i}rangle$ for each point $u in mathbb{R}^{n}$. I've been working with this component function and don't seem to be making any progress. Any help is appreciated
Note:$langle u,vrangle$ denotes $ucdot v$, u and v are points in $mathbb{R}^n$, $u_k$ is a sequence in $mathbb{R}^n$.
My attempt:
Letting $u_k=(u_1^k,u_2^k,...u_n^k)$ with the $k$ representing the $k$-th term in the sequence $u=(u_1,u_2,...u_n)$. Then $lim_{ktoinfty}u_i^k=u_i$ $forall i$.
$lim_{ktoinfty}langle u_k,vrangle=langle u,vrangle$ for any given $vinmathbb{R}$
, let $v=e_i$ then $lim_{ktoinfty}langle u_k,e_irangle=langle u,e_irangle$.
Before proceeding we establish $langle e_i, e_irangle=1$ and $langle e_i,e_jrangle=0$ assuming $ineq j$.
begin{align*}
lim_{ktoinfty} langle u_k,e_irangle
&= Big< lim_{ktoinfty} u_k, e_i Big>
= Big< lim_{ktoinfty}(u_1^k, u_2^k, cdots, u_n^k),e_i Big> \
&= Big< Big( lim_{ktoinfty}u_1^k, lim_{ktoinfty}u_2^k, cdots, lim_{ktoinfty}u_n^k Big),e_i Big>
=lim_{ktoinfty} u_i^k
end{align*}
Now we have
$$lim_{ktoinfty}langle u_k,e_irangle
= lim_{ktoinfty}u_i^k
= langle u,e_irangle=langle (u_1,u_2,...u_n),e_irangle
= u_i, forall i quad Box$$
real-analysis linear-algebra sequences-and-series limits convergence
edited Feb 3 at 6:11
Sangchul Lee
96.6k12173283
asked Feb 3 at 5:52
Dillain SmithDillain Smith
848
$endgroup$
I'm having trouble solving the following problem.
Problem. Prove ${u_k}to u$ given $lim_{ktoinfty}langle u_k,vrangle=langle u,vrangle$ for $uin mathbb{R}^n$, $forall vin mathbb{R}^n$
The textbook introduces the $i^{text{th}}$ component function $p_{i} : mathbb{R}^{n} to mathbb{R}$ for $1 leq i leq n$ by $p_{i}(u) = u_{i}$, where $u in mathbb{R}^{n}$.
Using this definition, I can express any vector $u in mathbb{R}^{n}$ by $u = (p_{1}(u), ldots, p_{n}(u))$. I know that this function is linear.
The book also tells us that a sequence ${u_{k}}$ converges to $u$ in $mathbb{R}^{n}$ if and only if it converges componentwise (i.e. for each $1 leq i leq n$, $lim_{ktoinfty} p_{i}(u_{k}) = p_{i}(u)).$
The book provides a hint to define the point $e_{i} in mathbb{R}^{n}$ whose $i^{text{ith}}$ component is equal to $1$ and every other component equals $0$. This way, $p_{i}(u) = langle u, e_{i}rangle$ for each point $u in mathbb{R}^{n}$. I've been working with this component function and don't seem to be making any progress. Any help is appreciated
Note:$langle u,vrangle$ denotes $ucdot v$, u and v are points in $mathbb{R}^n$, $u_k$ is a sequence in $mathbb{R}^n$.
My attempt:
Letting $u_k=(u_1^k,u_2^k,...u_n^k)$ with the $k$ representing the $k$-th term in the sequence $u=(u_1,u_2,...u_n)$. Then $lim_{ktoinfty}u_i^k=u_i$ $forall i$.
$lim_{ktoinfty}langle u_k,vrangle=langle u,vrangle$ for any given $vinmathbb{R}$
, let $v=e_i$ then $lim_{ktoinfty}langle u_k,e_irangle=langle u,e_irangle$.
Before proceeding we establish $langle e_i, e_irangle=1$ and $langle e_i,e_jrangle=0$ assuming $ineq j$.
begin{align*}
lim_{ktoinfty} langle u_k,e_irangle
&= Big< lim_{ktoinfty} u_k, e_i Big>
= Big< lim_{ktoinfty}(u_1^k, u_2^k, cdots, u_n^k),e_i Big> \
&= Big< Big( lim_{ktoinfty}u_1^k, lim_{ktoinfty}u_2^k, cdots, lim_{ktoinfty}u_n^k Big),e_i Big>
=lim_{ktoinfty} u_i^k
end{align*}
Now we have
$$lim_{ktoinfty}langle u_k,e_irangle
= lim_{ktoinfty}u_i^k
= langle u,e_irangle=langle (u_1,u_2,...u_n),e_irangle
= u_i, forall i quad Box$$
real-analysis linear-algebra sequences-and-series limits convergence
real-analysis linear-algebra sequences-and-series limits convergence
edited Feb 3 at 6:11
Sangchul Lee
96.6k12173283
asked Feb 3 at 5:52
Dillain SmithDillain Smith
848
edited Feb 3 at 6:11
Sangchul Lee
96.6k12173283
asked Feb 3 at 5:52
Dillain SmithDillain Smith
848
edited Feb 3 at 6:11
Sangchul Lee
96.6k12173283
edited Feb 3 at 6:11
Sangchul Lee
96.6k12173283
edited Feb 3 at 6:11
Sangchul Lee
96.6k12173283
96.6k12173283
asked Feb 3 at 5:52
Dillain SmithDillain Smith
848
asked Feb 3 at 5:52
Dillain SmithDillain Smith
848
asked Feb 3 at 5:52
Dillain SmithDillain Smith
848
848
1
$begingroup$
You should uselangle$langle$ andrangle$rangle$ instead of<and>. The former are delimiters, so the spacing for them is similar to that given for parentheses. The spacing for the latter is that of operators. That’s why it looks so bad when you use them. Compare $<e_1,e_2>$ with $langle e_1,e_2rangle$.
$endgroup$
– Arturo Magidin
Feb 3 at 6:04
2
$begingroup$
Well, if you know that $u_i^k to u_i$ as $ktoinfty$ for each $i=1,cdots,n$, then the proof is done. And this follows from the assumption by plugging $v = e_i$ for each $i$. Your intermediate computation is puzzling me because $lim_k langle u_k, e_i rangle = lim_k u_i^k$ holds simply because $ langle u_k, e_i rangle = u_i^k $ for all $k$ and $i$, so I am not sure why you made a significant detour for it.
$endgroup$
– Sangchul Lee
Feb 3 at 6:14
$begingroup$
@Sangchul Lee How can you just assume $v = e_{i}$; wouldn't you lose generality? We are trying to prove it for all $v$
$endgroup$
– user614735
Feb 3 at 16:21
add a comment |
1
$begingroup$
You should uselangle$langle$ andrangle$rangle$ instead of<and>. The former are delimiters, so the spacing for them is similar to that given for parentheses. The spacing for the latter is that of operators. That’s why it looks so bad when you use them. Compare $<e_1,e_2>$ with $langle e_1,e_2rangle$.
$endgroup$
– Arturo Magidin
Feb 3 at 6:04
2
$begingroup$
Well, if you know that $u_i^k to u_i$ as $ktoinfty$ for each $i=1,cdots,n$, then the proof is done. And this follows from the assumption by plugging $v = e_i$ for each $i$. Your intermediate computation is puzzling me because $lim_k langle u_k, e_i rangle = lim_k u_i^k$ holds simply because $ langle u_k, e_i rangle = u_i^k $ for all $k$ and $i$, so I am not sure why you made a significant detour for it.
$endgroup$
– Sangchul Lee
Feb 3 at 6:14
$begingroup$
@Sangchul Lee How can you just assume $v = e_{i}$; wouldn't you lose generality? We are trying to prove it for all $v$
$endgroup$
– user614735
Feb 3 at 16:21
1
1
$begingroup$
You should use
langle $langle$ and rangle $rangle$ instead of < and >. The former are delimiters, so the spacing for them is similar to that given for parentheses. The spacing for the latter is that of operators. That’s why it looks so bad when you use them. Compare $<e_1,e_2>$ with $langle e_1,e_2rangle$.$endgroup$
– Arturo Magidin
Feb 3 at 6:04
$begingroup$
You should use
langle $langle$ and rangle $rangle$ instead of < and >. The former are delimiters, so the spacing for them is similar to that given for parentheses. The spacing for the latter is that of operators. That’s why it looks so bad when you use them. Compare $<e_1,e_2>$ with $langle e_1,e_2rangle$.$endgroup$
– Arturo Magidin
Feb 3 at 6:04
2
2
$begingroup$
Well, if you know that $u_i^k to u_i$ as $ktoinfty$ for each $i=1,cdots,n$, then the proof is done. And this follows from the assumption by plugging $v = e_i$ for each $i$. Your intermediate computation is puzzling me because $lim_k langle u_k, e_i rangle = lim_k u_i^k$ holds simply because $ langle u_k, e_i rangle = u_i^k $ for all $k$ and $i$, so I am not sure why you made a significant detour for it.
$endgroup$
– Sangchul Lee
Feb 3 at 6:14
$begingroup$
Well, if you know that $u_i^k to u_i$ as $ktoinfty$ for each $i=1,cdots,n$, then the proof is done. And this follows from the assumption by plugging $v = e_i$ for each $i$. Your intermediate computation is puzzling me because $lim_k langle u_k, e_i rangle = lim_k u_i^k$ holds simply because $ langle u_k, e_i rangle = u_i^k $ for all $k$ and $i$, so I am not sure why you made a significant detour for it.
$endgroup$
– Sangchul Lee
Feb 3 at 6:14
$begingroup$
@Sangchul Lee How can you just assume $v = e_{i}$; wouldn't you lose generality? We are trying to prove it for all $v$
$endgroup$
– user614735
Feb 3 at 16:21
$begingroup$
@Sangchul Lee How can you just assume $v = e_{i}$; wouldn't you lose generality? We are trying to prove it for all $v$
$endgroup$
– user614735
Feb 3 at 16:21
add a comment |
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1
$begingroup$
You should use
langle$langle$ andrangle$rangle$ instead of<and>. The former are delimiters, so the spacing for them is similar to that given for parentheses. The spacing for the latter is that of operators. That’s why it looks so bad when you use them. Compare $<e_1,e_2>$ with $langle e_1,e_2rangle$.$endgroup$
– Arturo Magidin
Feb 3 at 6:04
2
$begingroup$
Well, if you know that $u_i^k to u_i$ as $ktoinfty$ for each $i=1,cdots,n$, then the proof is done. And this follows from the assumption by plugging $v = e_i$ for each $i$. Your intermediate computation is puzzling me because $lim_k langle u_k, e_i rangle = lim_k u_i^k$ holds simply because $ langle u_k, e_i rangle = u_i^k $ for all $k$ and $i$, so I am not sure why you made a significant detour for it.
$endgroup$
– Sangchul Lee
Feb 3 at 6:14
$begingroup$
@Sangchul Lee How can you just assume $v = e_{i}$; wouldn't you lose generality? We are trying to prove it for all $v$
$endgroup$
– user614735
Feb 3 at 16:21