Showing that $bigg| x - sum_{k=1}^n lambda_ke_kbigg| geq bigg|x-sum_{k=1}^nlangle x,e_krangle e_k bigg|$
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Exercise :
Let ${e_1,e_2,dots, e_n}$ be an orthonormal set over the Hilbert space $H$. Show that :
$$bigg| x - sum_{k=1}^n lambda_ke_kbigg| geq bigg|x-sum_{k=1}^nlangle x,e_krangle e_k bigg|$$
for every $lambda_1,lambda_2,dots,lambda_n in mathbb R$.
Attempt :
Let $x in H$ and $varepsilon > 0$. Then, we can find $lambda_1,dots, lambda_n in mathbb R$ such that :
$$bigg| x - sum_{k=1}^n lambda_ke_kbigg| < varepsilon$$
But the element $w = x-sum_{k=1}^nlangle x,e_krangle e_k$ achieves the minimum distance of $x$ from the finite dimensional space $ F = langle e_1,dots, e_n rangle$ and then it will be
$$bigg| x-sum_{k=1}^nlangle x,e_krangle e_k bigg| leq bigg| x - sum_{k=1}^n lambda_ke_kbigg| < varepsilon$$
and thus the inequality which we were asked to show is proven.
Question : Is my approach rigorous and legit enough ? If not, any recommendations, hints or elaborations will be greatly appreciated !
real-analysis functional-analysis hilbert-spaces normed-spaces inner-product-space
|
show 2 more comments
up vote
3
down vote
favorite
Exercise :
Let ${e_1,e_2,dots, e_n}$ be an orthonormal set over the Hilbert space $H$. Show that :
$$bigg| x - sum_{k=1}^n lambda_ke_kbigg| geq bigg|x-sum_{k=1}^nlangle x,e_krangle e_k bigg|$$
for every $lambda_1,lambda_2,dots,lambda_n in mathbb R$.
Attempt :
Let $x in H$ and $varepsilon > 0$. Then, we can find $lambda_1,dots, lambda_n in mathbb R$ such that :
$$bigg| x - sum_{k=1}^n lambda_ke_kbigg| < varepsilon$$
But the element $w = x-sum_{k=1}^nlangle x,e_krangle e_k$ achieves the minimum distance of $x$ from the finite dimensional space $ F = langle e_1,dots, e_n rangle$ and then it will be
$$bigg| x-sum_{k=1}^nlangle x,e_krangle e_k bigg| leq bigg| x - sum_{k=1}^n lambda_ke_kbigg| < varepsilon$$
and thus the inequality which we were asked to show is proven.
Question : Is my approach rigorous and legit enough ? If not, any recommendations, hints or elaborations will be greatly appreciated !
real-analysis functional-analysis hilbert-spaces normed-spaces inner-product-space
2
Just a remark: Since ${e_n}$ is a fixed orthonormal set, you might not be able to make the sum $sum_{k=1}^n lambda_k e_k$ arbitrarily close to $x$ in norm (which your solution seems to suggest since you're using $epsilon$).
– MisterRiemann
Dec 2 at 13:08
@MisterRiemann If that's the case, how would my proof be altered ? I assume that since ${e_n}$ is orthonormal then it would also be an orthonormal basis of the finite dimensional space $F$.
– Rebellos
Dec 2 at 13:11
I think that your idea is correct, i.e. you should use the fact that the projection of $x$ onto $F$ is indeed the closest point to $x$ in $F$. I'm just saying that you should remove $<epsilon$ in your argument (I don't even see the necessity for it). But let's wait for an answer from someone more knowledgeable than myself.
– MisterRiemann
Dec 2 at 13:13
@MisterRiemann I just elaborated using $varepsilon$ to show that the expression has some essence (meaning it wouldnt diverge).
– Rebellos
Dec 2 at 13:22
2
@MisterRiemann I get what you're saying, then it really is straightforward since the other distance is the smallest one anyway (or equal in case of being the same minimum distance).
– Rebellos
Dec 2 at 13:29
|
show 2 more comments
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Exercise :
Let ${e_1,e_2,dots, e_n}$ be an orthonormal set over the Hilbert space $H$. Show that :
$$bigg| x - sum_{k=1}^n lambda_ke_kbigg| geq bigg|x-sum_{k=1}^nlangle x,e_krangle e_k bigg|$$
for every $lambda_1,lambda_2,dots,lambda_n in mathbb R$.
Attempt :
Let $x in H$ and $varepsilon > 0$. Then, we can find $lambda_1,dots, lambda_n in mathbb R$ such that :
$$bigg| x - sum_{k=1}^n lambda_ke_kbigg| < varepsilon$$
But the element $w = x-sum_{k=1}^nlangle x,e_krangle e_k$ achieves the minimum distance of $x$ from the finite dimensional space $ F = langle e_1,dots, e_n rangle$ and then it will be
$$bigg| x-sum_{k=1}^nlangle x,e_krangle e_k bigg| leq bigg| x - sum_{k=1}^n lambda_ke_kbigg| < varepsilon$$
and thus the inequality which we were asked to show is proven.
Question : Is my approach rigorous and legit enough ? If not, any recommendations, hints or elaborations will be greatly appreciated !
real-analysis functional-analysis hilbert-spaces normed-spaces inner-product-space
Exercise :
Let ${e_1,e_2,dots, e_n}$ be an orthonormal set over the Hilbert space $H$. Show that :
$$bigg| x - sum_{k=1}^n lambda_ke_kbigg| geq bigg|x-sum_{k=1}^nlangle x,e_krangle e_k bigg|$$
for every $lambda_1,lambda_2,dots,lambda_n in mathbb R$.
Attempt :
Let $x in H$ and $varepsilon > 0$. Then, we can find $lambda_1,dots, lambda_n in mathbb R$ such that :
$$bigg| x - sum_{k=1}^n lambda_ke_kbigg| < varepsilon$$
But the element $w = x-sum_{k=1}^nlangle x,e_krangle e_k$ achieves the minimum distance of $x$ from the finite dimensional space $ F = langle e_1,dots, e_n rangle$ and then it will be
$$bigg| x-sum_{k=1}^nlangle x,e_krangle e_k bigg| leq bigg| x - sum_{k=1}^n lambda_ke_kbigg| < varepsilon$$
and thus the inequality which we were asked to show is proven.
Question : Is my approach rigorous and legit enough ? If not, any recommendations, hints or elaborations will be greatly appreciated !
real-analysis functional-analysis hilbert-spaces normed-spaces inner-product-space
real-analysis functional-analysis hilbert-spaces normed-spaces inner-product-space
edited Dec 2 at 13:08
asked Dec 2 at 13:05
Rebellos
13k21042
13k21042
2
Just a remark: Since ${e_n}$ is a fixed orthonormal set, you might not be able to make the sum $sum_{k=1}^n lambda_k e_k$ arbitrarily close to $x$ in norm (which your solution seems to suggest since you're using $epsilon$).
– MisterRiemann
Dec 2 at 13:08
@MisterRiemann If that's the case, how would my proof be altered ? I assume that since ${e_n}$ is orthonormal then it would also be an orthonormal basis of the finite dimensional space $F$.
– Rebellos
Dec 2 at 13:11
I think that your idea is correct, i.e. you should use the fact that the projection of $x$ onto $F$ is indeed the closest point to $x$ in $F$. I'm just saying that you should remove $<epsilon$ in your argument (I don't even see the necessity for it). But let's wait for an answer from someone more knowledgeable than myself.
– MisterRiemann
Dec 2 at 13:13
@MisterRiemann I just elaborated using $varepsilon$ to show that the expression has some essence (meaning it wouldnt diverge).
– Rebellos
Dec 2 at 13:22
2
@MisterRiemann I get what you're saying, then it really is straightforward since the other distance is the smallest one anyway (or equal in case of being the same minimum distance).
– Rebellos
Dec 2 at 13:29
|
show 2 more comments
2
Just a remark: Since ${e_n}$ is a fixed orthonormal set, you might not be able to make the sum $sum_{k=1}^n lambda_k e_k$ arbitrarily close to $x$ in norm (which your solution seems to suggest since you're using $epsilon$).
– MisterRiemann
Dec 2 at 13:08
@MisterRiemann If that's the case, how would my proof be altered ? I assume that since ${e_n}$ is orthonormal then it would also be an orthonormal basis of the finite dimensional space $F$.
– Rebellos
Dec 2 at 13:11
I think that your idea is correct, i.e. you should use the fact that the projection of $x$ onto $F$ is indeed the closest point to $x$ in $F$. I'm just saying that you should remove $<epsilon$ in your argument (I don't even see the necessity for it). But let's wait for an answer from someone more knowledgeable than myself.
– MisterRiemann
Dec 2 at 13:13
@MisterRiemann I just elaborated using $varepsilon$ to show that the expression has some essence (meaning it wouldnt diverge).
– Rebellos
Dec 2 at 13:22
2
@MisterRiemann I get what you're saying, then it really is straightforward since the other distance is the smallest one anyway (or equal in case of being the same minimum distance).
– Rebellos
Dec 2 at 13:29
2
2
Just a remark: Since ${e_n}$ is a fixed orthonormal set, you might not be able to make the sum $sum_{k=1}^n lambda_k e_k$ arbitrarily close to $x$ in norm (which your solution seems to suggest since you're using $epsilon$).
– MisterRiemann
Dec 2 at 13:08
Just a remark: Since ${e_n}$ is a fixed orthonormal set, you might not be able to make the sum $sum_{k=1}^n lambda_k e_k$ arbitrarily close to $x$ in norm (which your solution seems to suggest since you're using $epsilon$).
– MisterRiemann
Dec 2 at 13:08
@MisterRiemann If that's the case, how would my proof be altered ? I assume that since ${e_n}$ is orthonormal then it would also be an orthonormal basis of the finite dimensional space $F$.
– Rebellos
Dec 2 at 13:11
@MisterRiemann If that's the case, how would my proof be altered ? I assume that since ${e_n}$ is orthonormal then it would also be an orthonormal basis of the finite dimensional space $F$.
– Rebellos
Dec 2 at 13:11
I think that your idea is correct, i.e. you should use the fact that the projection of $x$ onto $F$ is indeed the closest point to $x$ in $F$. I'm just saying that you should remove $<epsilon$ in your argument (I don't even see the necessity for it). But let's wait for an answer from someone more knowledgeable than myself.
– MisterRiemann
Dec 2 at 13:13
I think that your idea is correct, i.e. you should use the fact that the projection of $x$ onto $F$ is indeed the closest point to $x$ in $F$. I'm just saying that you should remove $<epsilon$ in your argument (I don't even see the necessity for it). But let's wait for an answer from someone more knowledgeable than myself.
– MisterRiemann
Dec 2 at 13:13
@MisterRiemann I just elaborated using $varepsilon$ to show that the expression has some essence (meaning it wouldnt diverge).
– Rebellos
Dec 2 at 13:22
@MisterRiemann I just elaborated using $varepsilon$ to show that the expression has some essence (meaning it wouldnt diverge).
– Rebellos
Dec 2 at 13:22
2
2
@MisterRiemann I get what you're saying, then it really is straightforward since the other distance is the smallest one anyway (or equal in case of being the same minimum distance).
– Rebellos
Dec 2 at 13:29
@MisterRiemann I get what you're saying, then it really is straightforward since the other distance is the smallest one anyway (or equal in case of being the same minimum distance).
– Rebellos
Dec 2 at 13:29
|
show 2 more comments
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
Notice that $r = x - sum_{i=1}^n langle x, e_i rangle e_i$ is orthogonal to each vector $e_i$. If $lambda_1,lambda_2,ldots, lambda_n in mathbb R$, then
begin{align}
| x - sum_i lambda_i e_i |^2 &=
|x - sum_i langle x, e_i rangle e_i + sum_i langle x, e_i rangle e_i - sum_i lambda_i e_i |^2 \
&= | r + sum_i (langle x, e_i rangle - lambda_i) e_i |^2 \
&= |r|^2 + sum_i | langle x, e_i rangle - lambda_i | |e_i|^2 qquad text{(by Pythagorean theorem)}\
& geq |r|^2.
end{align}
Thanks a lot for the fast and swift algebraic proof.
– Rebellos
2 days ago
@Rebellos Sure. This proof is based in my mind on a picture. We know that the point $x^* = sum_i langle x, e_i rangle e_i$ has the property that $x - x^*$ is orthogonal to the subspace $S = text{span}(e_1,ldots,e_n)$. So, any other vector in $S$ must be further away from $x$, according to the Pythagorean theorem.
– littleO
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Notice that $r = x - sum_{i=1}^n langle x, e_i rangle e_i$ is orthogonal to each vector $e_i$. If $lambda_1,lambda_2,ldots, lambda_n in mathbb R$, then
begin{align}
| x - sum_i lambda_i e_i |^2 &=
|x - sum_i langle x, e_i rangle e_i + sum_i langle x, e_i rangle e_i - sum_i lambda_i e_i |^2 \
&= | r + sum_i (langle x, e_i rangle - lambda_i) e_i |^2 \
&= |r|^2 + sum_i | langle x, e_i rangle - lambda_i | |e_i|^2 qquad text{(by Pythagorean theorem)}\
& geq |r|^2.
end{align}
Thanks a lot for the fast and swift algebraic proof.
– Rebellos
2 days ago
@Rebellos Sure. This proof is based in my mind on a picture. We know that the point $x^* = sum_i langle x, e_i rangle e_i$ has the property that $x - x^*$ is orthogonal to the subspace $S = text{span}(e_1,ldots,e_n)$. So, any other vector in $S$ must be further away from $x$, according to the Pythagorean theorem.
– littleO
2 days ago
add a comment |
up vote
2
down vote
accepted
Notice that $r = x - sum_{i=1}^n langle x, e_i rangle e_i$ is orthogonal to each vector $e_i$. If $lambda_1,lambda_2,ldots, lambda_n in mathbb R$, then
begin{align}
| x - sum_i lambda_i e_i |^2 &=
|x - sum_i langle x, e_i rangle e_i + sum_i langle x, e_i rangle e_i - sum_i lambda_i e_i |^2 \
&= | r + sum_i (langle x, e_i rangle - lambda_i) e_i |^2 \
&= |r|^2 + sum_i | langle x, e_i rangle - lambda_i | |e_i|^2 qquad text{(by Pythagorean theorem)}\
& geq |r|^2.
end{align}
Thanks a lot for the fast and swift algebraic proof.
– Rebellos
2 days ago
@Rebellos Sure. This proof is based in my mind on a picture. We know that the point $x^* = sum_i langle x, e_i rangle e_i$ has the property that $x - x^*$ is orthogonal to the subspace $S = text{span}(e_1,ldots,e_n)$. So, any other vector in $S$ must be further away from $x$, according to the Pythagorean theorem.
– littleO
2 days ago
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Notice that $r = x - sum_{i=1}^n langle x, e_i rangle e_i$ is orthogonal to each vector $e_i$. If $lambda_1,lambda_2,ldots, lambda_n in mathbb R$, then
begin{align}
| x - sum_i lambda_i e_i |^2 &=
|x - sum_i langle x, e_i rangle e_i + sum_i langle x, e_i rangle e_i - sum_i lambda_i e_i |^2 \
&= | r + sum_i (langle x, e_i rangle - lambda_i) e_i |^2 \
&= |r|^2 + sum_i | langle x, e_i rangle - lambda_i | |e_i|^2 qquad text{(by Pythagorean theorem)}\
& geq |r|^2.
end{align}
Notice that $r = x - sum_{i=1}^n langle x, e_i rangle e_i$ is orthogonal to each vector $e_i$. If $lambda_1,lambda_2,ldots, lambda_n in mathbb R$, then
begin{align}
| x - sum_i lambda_i e_i |^2 &=
|x - sum_i langle x, e_i rangle e_i + sum_i langle x, e_i rangle e_i - sum_i lambda_i e_i |^2 \
&= | r + sum_i (langle x, e_i rangle - lambda_i) e_i |^2 \
&= |r|^2 + sum_i | langle x, e_i rangle - lambda_i | |e_i|^2 qquad text{(by Pythagorean theorem)}\
& geq |r|^2.
end{align}
edited Dec 2 at 14:32
answered Dec 2 at 14:09
littleO
27.6k642104
27.6k642104
Thanks a lot for the fast and swift algebraic proof.
– Rebellos
2 days ago
@Rebellos Sure. This proof is based in my mind on a picture. We know that the point $x^* = sum_i langle x, e_i rangle e_i$ has the property that $x - x^*$ is orthogonal to the subspace $S = text{span}(e_1,ldots,e_n)$. So, any other vector in $S$ must be further away from $x$, according to the Pythagorean theorem.
– littleO
2 days ago
add a comment |
Thanks a lot for the fast and swift algebraic proof.
– Rebellos
2 days ago
@Rebellos Sure. This proof is based in my mind on a picture. We know that the point $x^* = sum_i langle x, e_i rangle e_i$ has the property that $x - x^*$ is orthogonal to the subspace $S = text{span}(e_1,ldots,e_n)$. So, any other vector in $S$ must be further away from $x$, according to the Pythagorean theorem.
– littleO
2 days ago
Thanks a lot for the fast and swift algebraic proof.
– Rebellos
2 days ago
Thanks a lot for the fast and swift algebraic proof.
– Rebellos
2 days ago
@Rebellos Sure. This proof is based in my mind on a picture. We know that the point $x^* = sum_i langle x, e_i rangle e_i$ has the property that $x - x^*$ is orthogonal to the subspace $S = text{span}(e_1,ldots,e_n)$. So, any other vector in $S$ must be further away from $x$, according to the Pythagorean theorem.
– littleO
2 days ago
@Rebellos Sure. This proof is based in my mind on a picture. We know that the point $x^* = sum_i langle x, e_i rangle e_i$ has the property that $x - x^*$ is orthogonal to the subspace $S = text{span}(e_1,ldots,e_n)$. So, any other vector in $S$ must be further away from $x$, according to the Pythagorean theorem.
– littleO
2 days ago
add a comment |
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2
Just a remark: Since ${e_n}$ is a fixed orthonormal set, you might not be able to make the sum $sum_{k=1}^n lambda_k e_k$ arbitrarily close to $x$ in norm (which your solution seems to suggest since you're using $epsilon$).
– MisterRiemann
Dec 2 at 13:08
@MisterRiemann If that's the case, how would my proof be altered ? I assume that since ${e_n}$ is orthonormal then it would also be an orthonormal basis of the finite dimensional space $F$.
– Rebellos
Dec 2 at 13:11
I think that your idea is correct, i.e. you should use the fact that the projection of $x$ onto $F$ is indeed the closest point to $x$ in $F$. I'm just saying that you should remove $<epsilon$ in your argument (I don't even see the necessity for it). But let's wait for an answer from someone more knowledgeable than myself.
– MisterRiemann
Dec 2 at 13:13
@MisterRiemann I just elaborated using $varepsilon$ to show that the expression has some essence (meaning it wouldnt diverge).
– Rebellos
Dec 2 at 13:22
2
@MisterRiemann I get what you're saying, then it really is straightforward since the other distance is the smallest one anyway (or equal in case of being the same minimum distance).
– Rebellos
Dec 2 at 13:29