Mixing
Riesz Representation Theorem and Uniqueness
$begingroup$
I'm working through the proof of the Reisz Representation Theorem in Axler's Linear Algebra Done Right and am confused about the uniqueness portion of the proof.
Axler claims:
"Suppose $V$ is finite-dimensional and $phi $ is a linear functional on V. Then there is a unique vector $u in V$ such $$ phi(v) = langle v,u rangle$$ for every $v in V$."
The text then proceeds to prove existence (which I understand, so am not going to elaborate here) and then uniqueness of $ u$.
To prove uniqueness, the text claims the following:
Now we prove that only one vector $ u in V $ has the desired behavior.
Suppose $ u_1, u_2 in V $ are such that $$ phi(v) = langle v,u_1 rangle = langle v,u_2 rangle $$ for every $v in V$. Then $$ 0 = langle v,u_1 rangle - langle v,u_2 rangle = langle v,u_1-u_2 rangle $$ for every $ v in V$. Taking $ v = u_1 - u_2 $ shows that $ u_1 - u_2 = 0 $. In other words, $u_1 = u_2$, completing the proof.
Some questions about this:
By "taking $ v = u_1 - u_2 $", I assume the text means that we set $ v = u_1 - u_2 $. However, why does $ 0 = langle u_1 - u_2 , u_1 - u_2 rangle implies u_1 - u_2 = 0 $?
Isn't $langle u_1 - u_2 , u_1 - u_2 rangle = (u_1 - u_2)^T (u_1 - u_2) $? As a result, isn't $langle u_1 - u_2 , u_1 - u_2 rangle = 0 $ iff $V$ is a vector space over $mathbb{R^n}$? (The text assumes that $V$ is a vector space over $mathbb{R^n}$ or $mathbb{C^n}$.)
(I realize that this might be more of a question about whether $AA^T = 0 implies A = 0$, and I've already looked at a similar question about that. However, I've included details of the proof, just in case my interpretation of what it was saying was wrong. I've also looked at other questions that also discuss this line of the proof, but am still confused about the calculation above.)
Thanks!
linear-algebra
asked Jan 6 at 21:59
GinaGina
183
$endgroup$
add a comment |
$begingroup$
I'm working through the proof of the Reisz Representation Theorem in Axler's Linear Algebra Done Right and am confused about the uniqueness portion of the proof.
Axler claims:
"Suppose $V$ is finite-dimensional and $phi $ is a linear functional on V. Then there is a unique vector $u in V$ such $$ phi(v) = langle v,u rangle$$ for every $v in V$."
The text then proceeds to prove existence (which I understand, so am not going to elaborate here) and then uniqueness of $ u$.
To prove uniqueness, the text claims the following:
Now we prove that only one vector $ u in V $ has the desired behavior.
Suppose $ u_1, u_2 in V $ are such that $$ phi(v) = langle v,u_1 rangle = langle v,u_2 rangle $$ for every $v in V$. Then $$ 0 = langle v,u_1 rangle - langle v,u_2 rangle = langle v,u_1-u_2 rangle $$ for every $ v in V$. Taking $ v = u_1 - u_2 $ shows that $ u_1 - u_2 = 0 $. In other words, $u_1 = u_2$, completing the proof.
Some questions about this:
By "taking $ v = u_1 - u_2 $", I assume the text means that we set $ v = u_1 - u_2 $. However, why does $ 0 = langle u_1 - u_2 , u_1 - u_2 rangle implies u_1 - u_2 = 0 $?
Isn't $langle u_1 - u_2 , u_1 - u_2 rangle = (u_1 - u_2)^T (u_1 - u_2) $? As a result, isn't $langle u_1 - u_2 , u_1 - u_2 rangle = 0 $ iff $V$ is a vector space over $mathbb{R^n}$? (The text assumes that $V$ is a vector space over $mathbb{R^n}$ or $mathbb{C^n}$.)
(I realize that this might be more of a question about whether $AA^T = 0 implies A = 0$, and I've already looked at a similar question about that. However, I've included details of the proof, just in case my interpretation of what it was saying was wrong. I've also looked at other questions that also discuss this line of the proof, but am still confused about the calculation above.)
Thanks!
linear-algebra
asked Jan 6 at 21:59
GinaGina
183
$endgroup$
add a comment |
$begingroup$
I'm working through the proof of the Reisz Representation Theorem in Axler's Linear Algebra Done Right and am confused about the uniqueness portion of the proof.
Axler claims:
"Suppose $V$ is finite-dimensional and $phi $ is a linear functional on V. Then there is a unique vector $u in V$ such $$ phi(v) = langle v,u rangle$$ for every $v in V$."
The text then proceeds to prove existence (which I understand, so am not going to elaborate here) and then uniqueness of $ u$.
To prove uniqueness, the text claims the following:
Now we prove that only one vector $ u in V $ has the desired behavior.
Suppose $ u_1, u_2 in V $ are such that $$ phi(v) = langle v,u_1 rangle = langle v,u_2 rangle $$ for every $v in V$. Then $$ 0 = langle v,u_1 rangle - langle v,u_2 rangle = langle v,u_1-u_2 rangle $$ for every $ v in V$. Taking $ v = u_1 - u_2 $ shows that $ u_1 - u_2 = 0 $. In other words, $u_1 = u_2$, completing the proof.
Some questions about this:
By "taking $ v = u_1 - u_2 $", I assume the text means that we set $ v = u_1 - u_2 $. However, why does $ 0 = langle u_1 - u_2 , u_1 - u_2 rangle implies u_1 - u_2 = 0 $?
Isn't $langle u_1 - u_2 , u_1 - u_2 rangle = (u_1 - u_2)^T (u_1 - u_2) $? As a result, isn't $langle u_1 - u_2 , u_1 - u_2 rangle = 0 $ iff $V$ is a vector space over $mathbb{R^n}$? (The text assumes that $V$ is a vector space over $mathbb{R^n}$ or $mathbb{C^n}$.)
(I realize that this might be more of a question about whether $AA^T = 0 implies A = 0$, and I've already looked at a similar question about that. However, I've included details of the proof, just in case my interpretation of what it was saying was wrong. I've also looked at other questions that also discuss this line of the proof, but am still confused about the calculation above.)
Thanks!
linear-algebra
asked Jan 6 at 21:59
GinaGina
183
$endgroup$
I'm working through the proof of the Reisz Representation Theorem in Axler's Linear Algebra Done Right and am confused about the uniqueness portion of the proof.
Axler claims:
"Suppose $V$ is finite-dimensional and $phi $ is a linear functional on V. Then there is a unique vector $u in V$ such $$ phi(v) = langle v,u rangle$$ for every $v in V$."
The text then proceeds to prove existence (which I understand, so am not going to elaborate here) and then uniqueness of $ u$.
To prove uniqueness, the text claims the following:
Now we prove that only one vector $ u in V $ has the desired behavior.
Suppose $ u_1, u_2 in V $ are such that $$ phi(v) = langle v,u_1 rangle = langle v,u_2 rangle $$ for every $v in V$. Then $$ 0 = langle v,u_1 rangle - langle v,u_2 rangle = langle v,u_1-u_2 rangle $$ for every $ v in V$. Taking $ v = u_1 - u_2 $ shows that $ u_1 - u_2 = 0 $. In other words, $u_1 = u_2$, completing the proof.
Some questions about this:
By "taking $ v = u_1 - u_2 $", I assume the text means that we set $ v = u_1 - u_2 $. However, why does $ 0 = langle u_1 - u_2 , u_1 - u_2 rangle implies u_1 - u_2 = 0 $?
Isn't $langle u_1 - u_2 , u_1 - u_2 rangle = (u_1 - u_2)^T (u_1 - u_2) $? As a result, isn't $langle u_1 - u_2 , u_1 - u_2 rangle = 0 $ iff $V$ is a vector space over $mathbb{R^n}$? (The text assumes that $V$ is a vector space over $mathbb{R^n}$ or $mathbb{C^n}$.)
(I realize that this might be more of a question about whether $AA^T = 0 implies A = 0$, and I've already looked at a similar question about that. However, I've included details of the proof, just in case my interpretation of what it was saying was wrong. I've also looked at other questions that also discuss this line of the proof, but am still confused about the calculation above.)
Thanks!
linear-algebra
linear-algebra
asked Jan 6 at 21:59
GinaGina
183
asked Jan 6 at 21:59
GinaGina
183
asked Jan 6 at 21:59
GinaGina
183
asked Jan 6 at 21:59
GinaGina
183
asked Jan 6 at 21:59
GinaGina
183
183
add a comment |
add a comment |
2 Answers
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$begingroup$
If $phi(v)=langle v,u_1rangle=langle v,u_2rangle$, then $langle v,u_1-u_2rangle=0$ for any choice of $vin V$. So, cleverly choose $v=u_1-u_2$. Then
$$ 0=langle u_1-u_2,u_1-u_2rangle.$$
One of the conditions in the definition of an inner product is that $langle w,wrangle=0$ if and only $w=0$. So, $u_1-u_2=0$ and $u_1=u_2$. As a concrete example, let's examine the standard inner product $langle x,yrangle=xcdot y$ on $mathbb{R}^3$. If $x_i$ denote the components of $x$ and $y_i$ those of $y$, then
$$ xcdot y=sum_{i=1}^3 x_iy_i=x^Ty.$$
Note that
$$ langle x,xrangle=xcdot x=sum_{i=1}^3x_i^2ge 0$$
and equals zero exactly when $x_1=x_2=x_3=0$.
answered Jan 6 at 22:10
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,91741640
$endgroup$
$begingroup$
Thank you -- that's very helpful. I completely forgot about this property. As a followup, suppose we have a vector v = [1+i; -1+i]; in this case, <v,v> = adj(v)*v = 0, which violates definiteness. Does this mean that v is not in an inner product space?
$endgroup$
– Gina
Jan 6 at 22:38
$begingroup$
Once you work over the complex field, the appropriate generalization of the standard dot product is $langle z,wrangle =overline{z}^Tcdot w$.
$endgroup$
– Antonios-Alexandros Robotis
Jan 6 at 22:39
$begingroup$
Right, so $ langle v,v rangle = adj(v)*v = overline{v}^T v = [1-i, -1-i] * begin{bmatrix}1+i; \ -1+i end{bmatrix} = 0 $. But doesn't this violate definiteness since $ v neq 0 $?
$endgroup$
– Gina
Jan 6 at 22:45
1
$begingroup$
Check your calculations: $overline{v}^Tv=4$ here.
$endgroup$
– Antonios-Alexandros Robotis
Jan 6 at 22:48
$begingroup$
My bad -- thanks!
$endgroup$
– Gina
Jan 6 at 23:02
add a comment |
$begingroup$
A scalar product $langlecdot,cdotrangle$ induces on a vector space the norm $$ |v|=sqrt{langle v,vrangle}. $$ One of the properties of the norm is that $$ |v|=0Leftrightarrow v=0. $$ Then, specialising this property to the proof of the theorem, we have $$ langle u_1-u_2, u_1-u_2rangle=0Leftrightarrow |u_1-u_2|=0 Leftrightarrow u_1=u_2. $$ This is a trick used very often in Hilbert spaces.
answered Jan 6 at 22:23
diec_diec_
1253
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
If $phi(v)=langle v,u_1rangle=langle v,u_2rangle$, then $langle v,u_1-u_2rangle=0$ for any choice of $vin V$. So, cleverly choose $v=u_1-u_2$. Then
$$ 0=langle u_1-u_2,u_1-u_2rangle.$$
One of the conditions in the definition of an inner product is that $langle w,wrangle=0$ if and only $w=0$. So, $u_1-u_2=0$ and $u_1=u_2$. As a concrete example, let's examine the standard inner product $langle x,yrangle=xcdot y$ on $mathbb{R}^3$. If $x_i$ denote the components of $x$ and $y_i$ those of $y$, then
$$ xcdot y=sum_{i=1}^3 x_iy_i=x^Ty.$$
Note that
$$ langle x,xrangle=xcdot x=sum_{i=1}^3x_i^2ge 0$$
and equals zero exactly when $x_1=x_2=x_3=0$.
answered Jan 6 at 22:10
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,91741640
$endgroup$
$begingroup$
Thank you -- that's very helpful. I completely forgot about this property. As a followup, suppose we have a vector v = [1+i; -1+i]; in this case, <v,v> = adj(v)*v = 0, which violates definiteness. Does this mean that v is not in an inner product space?
$endgroup$
– Gina
Jan 6 at 22:38
$begingroup$
Once you work over the complex field, the appropriate generalization of the standard dot product is $langle z,wrangle =overline{z}^Tcdot w$.
$endgroup$
– Antonios-Alexandros Robotis
Jan 6 at 22:39
$begingroup$
Right, so $ langle v,v rangle = adj(v)*v = overline{v}^T v = [1-i, -1-i] * begin{bmatrix}1+i; \ -1+i end{bmatrix} = 0 $. But doesn't this violate definiteness since $ v neq 0 $?
$endgroup$
– Gina
Jan 6 at 22:45
1
$begingroup$
Check your calculations: $overline{v}^Tv=4$ here.
$endgroup$
– Antonios-Alexandros Robotis
Jan 6 at 22:48
$begingroup$
My bad -- thanks!
$endgroup$
– Gina
Jan 6 at 23:02
add a comment |
$begingroup$
If $phi(v)=langle v,u_1rangle=langle v,u_2rangle$, then $langle v,u_1-u_2rangle=0$ for any choice of $vin V$. So, cleverly choose $v=u_1-u_2$. Then
$$ 0=langle u_1-u_2,u_1-u_2rangle.$$
One of the conditions in the definition of an inner product is that $langle w,wrangle=0$ if and only $w=0$. So, $u_1-u_2=0$ and $u_1=u_2$. As a concrete example, let's examine the standard inner product $langle x,yrangle=xcdot y$ on $mathbb{R}^3$. If $x_i$ denote the components of $x$ and $y_i$ those of $y$, then
$$ xcdot y=sum_{i=1}^3 x_iy_i=x^Ty.$$
Note that
$$ langle x,xrangle=xcdot x=sum_{i=1}^3x_i^2ge 0$$
and equals zero exactly when $x_1=x_2=x_3=0$.
answered Jan 6 at 22:10
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,91741640
$endgroup$
$begingroup$
Thank you -- that's very helpful. I completely forgot about this property. As a followup, suppose we have a vector v = [1+i; -1+i]; in this case, <v,v> = adj(v)*v = 0, which violates definiteness. Does this mean that v is not in an inner product space?
$endgroup$
– Gina
Jan 6 at 22:38
$begingroup$
Once you work over the complex field, the appropriate generalization of the standard dot product is $langle z,wrangle =overline{z}^Tcdot w$.
$endgroup$
– Antonios-Alexandros Robotis
Jan 6 at 22:39
$begingroup$
Right, so $ langle v,v rangle = adj(v)*v = overline{v}^T v = [1-i, -1-i] * begin{bmatrix}1+i; \ -1+i end{bmatrix} = 0 $. But doesn't this violate definiteness since $ v neq 0 $?
$endgroup$
– Gina
Jan 6 at 22:45
1
$begingroup$
Check your calculations: $overline{v}^Tv=4$ here.
$endgroup$
– Antonios-Alexandros Robotis
Jan 6 at 22:48
$begingroup$
My bad -- thanks!
$endgroup$
– Gina
Jan 6 at 23:02
add a comment |
$begingroup$
If $phi(v)=langle v,u_1rangle=langle v,u_2rangle$, then $langle v,u_1-u_2rangle=0$ for any choice of $vin V$. So, cleverly choose $v=u_1-u_2$. Then
$$ 0=langle u_1-u_2,u_1-u_2rangle.$$
One of the conditions in the definition of an inner product is that $langle w,wrangle=0$ if and only $w=0$. So, $u_1-u_2=0$ and $u_1=u_2$. As a concrete example, let's examine the standard inner product $langle x,yrangle=xcdot y$ on $mathbb{R}^3$. If $x_i$ denote the components of $x$ and $y_i$ those of $y$, then
$$ xcdot y=sum_{i=1}^3 x_iy_i=x^Ty.$$
Note that
$$ langle x,xrangle=xcdot x=sum_{i=1}^3x_i^2ge 0$$
and equals zero exactly when $x_1=x_2=x_3=0$.
answered Jan 6 at 22:10
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,91741640
$endgroup$
If $phi(v)=langle v,u_1rangle=langle v,u_2rangle$, then $langle v,u_1-u_2rangle=0$ for any choice of $vin V$. So, cleverly choose $v=u_1-u_2$. Then
$$ 0=langle u_1-u_2,u_1-u_2rangle.$$
One of the conditions in the definition of an inner product is that $langle w,wrangle=0$ if and only $w=0$. So, $u_1-u_2=0$ and $u_1=u_2$. As a concrete example, let's examine the standard inner product $langle x,yrangle=xcdot y$ on $mathbb{R}^3$. If $x_i$ denote the components of $x$ and $y_i$ those of $y$, then
$$ xcdot y=sum_{i=1}^3 x_iy_i=x^Ty.$$
Note that
$$ langle x,xrangle=xcdot x=sum_{i=1}^3x_i^2ge 0$$
and equals zero exactly when $x_1=x_2=x_3=0$.
answered Jan 6 at 22:10
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,91741640
answered Jan 6 at 22:10
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,91741640
answered Jan 6 at 22:10
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,91741640
answered Jan 6 at 22:10
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
9,91741640
9,91741640
$begingroup$
Thank you -- that's very helpful. I completely forgot about this property. As a followup, suppose we have a vector v = [1+i; -1+i]; in this case, <v,v> = adj(v)*v = 0, which violates definiteness. Does this mean that v is not in an inner product space?
$endgroup$
– Gina
Jan 6 at 22:38
$begingroup$
Once you work over the complex field, the appropriate generalization of the standard dot product is $langle z,wrangle =overline{z}^Tcdot w$.
$endgroup$
– Antonios-Alexandros Robotis
Jan 6 at 22:39
$begingroup$
Right, so $ langle v,v rangle = adj(v)*v = overline{v}^T v = [1-i, -1-i] * begin{bmatrix}1+i; \ -1+i end{bmatrix} = 0 $. But doesn't this violate definiteness since $ v neq 0 $?
$endgroup$
– Gina
Jan 6 at 22:45
1
$begingroup$
Check your calculations: $overline{v}^Tv=4$ here.
$endgroup$
– Antonios-Alexandros Robotis
Jan 6 at 22:48
$begingroup$
My bad -- thanks!
$endgroup$
– Gina
Jan 6 at 23:02
add a comment |
$begingroup$
Thank you -- that's very helpful. I completely forgot about this property. As a followup, suppose we have a vector v = [1+i; -1+i]; in this case, <v,v> = adj(v)*v = 0, which violates definiteness. Does this mean that v is not in an inner product space?
$endgroup$
– Gina
Jan 6 at 22:38
$begingroup$
Once you work over the complex field, the appropriate generalization of the standard dot product is $langle z,wrangle =overline{z}^Tcdot w$.
$endgroup$
– Antonios-Alexandros Robotis
Jan 6 at 22:39
$begingroup$
Right, so $ langle v,v rangle = adj(v)*v = overline{v}^T v = [1-i, -1-i] * begin{bmatrix}1+i; \ -1+i end{bmatrix} = 0 $. But doesn't this violate definiteness since $ v neq 0 $?
$endgroup$
– Gina
Jan 6 at 22:45
1
$begingroup$
Check your calculations: $overline{v}^Tv=4$ here.
$endgroup$
– Antonios-Alexandros Robotis
Jan 6 at 22:48
$begingroup$
My bad -- thanks!
$endgroup$
– Gina
Jan 6 at 23:02
$begingroup$
Thank you -- that's very helpful. I completely forgot about this property. As a followup, suppose we have a vector v = [1+i; -1+i]; in this case, <v,v> = adj(v)*v = 0, which violates definiteness. Does this mean that v is not in an inner product space?
$endgroup$
– Gina
Jan 6 at 22:38
$begingroup$
Thank you -- that's very helpful. I completely forgot about this property. As a followup, suppose we have a vector v = [1+i; -1+i]; in this case, <v,v> = adj(v)*v = 0, which violates definiteness. Does this mean that v is not in an inner product space?
$endgroup$
– Gina
Jan 6 at 22:38
$begingroup$
Once you work over the complex field, the appropriate generalization of the standard dot product is $langle z,wrangle =overline{z}^Tcdot w$.
$endgroup$
– Antonios-Alexandros Robotis
Jan 6 at 22:39
$begingroup$
Once you work over the complex field, the appropriate generalization of the standard dot product is $langle z,wrangle =overline{z}^Tcdot w$.
$endgroup$
– Antonios-Alexandros Robotis
Jan 6 at 22:39
$begingroup$
Right, so $ langle v,v rangle = adj(v)*v = overline{v}^T v = [1-i, -1-i] * begin{bmatrix}1+i; \ -1+i end{bmatrix} = 0 $. But doesn't this violate definiteness since $ v neq 0 $?
$endgroup$
– Gina
Jan 6 at 22:45
$begingroup$
Right, so $ langle v,v rangle = adj(v)*v = overline{v}^T v = [1-i, -1-i] * begin{bmatrix}1+i; \ -1+i end{bmatrix} = 0 $. But doesn't this violate definiteness since $ v neq 0 $?
$endgroup$
– Gina
Jan 6 at 22:45
1
1
$begingroup$
Check your calculations: $overline{v}^Tv=4$ here.
$endgroup$
– Antonios-Alexandros Robotis
Jan 6 at 22:48
$begingroup$
Check your calculations: $overline{v}^Tv=4$ here.
$endgroup$
– Antonios-Alexandros Robotis
Jan 6 at 22:48
$begingroup$
My bad -- thanks!
$endgroup$
– Gina
Jan 6 at 23:02
$begingroup$
My bad -- thanks!
$endgroup$
– Gina
Jan 6 at 23:02
add a comment |
$begingroup$
A scalar product $langlecdot,cdotrangle$ induces on a vector space the norm $$ |v|=sqrt{langle v,vrangle}. $$ One of the properties of the norm is that $$ |v|=0Leftrightarrow v=0. $$ Then, specialising this property to the proof of the theorem, we have $$ langle u_1-u_2, u_1-u_2rangle=0Leftrightarrow |u_1-u_2|=0 Leftrightarrow u_1=u_2. $$ This is a trick used very often in Hilbert spaces.
answered Jan 6 at 22:23
diec_diec_
1253
$endgroup$
add a comment |
$begingroup$
A scalar product $langlecdot,cdotrangle$ induces on a vector space the norm $$ |v|=sqrt{langle v,vrangle}. $$ One of the properties of the norm is that $$ |v|=0Leftrightarrow v=0. $$ Then, specialising this property to the proof of the theorem, we have $$ langle u_1-u_2, u_1-u_2rangle=0Leftrightarrow |u_1-u_2|=0 Leftrightarrow u_1=u_2. $$ This is a trick used very often in Hilbert spaces.
answered Jan 6 at 22:23
diec_diec_
1253
$endgroup$
add a comment |
$begingroup$
A scalar product $langlecdot,cdotrangle$ induces on a vector space the norm $$ |v|=sqrt{langle v,vrangle}. $$ One of the properties of the norm is that $$ |v|=0Leftrightarrow v=0. $$ Then, specialising this property to the proof of the theorem, we have $$ langle u_1-u_2, u_1-u_2rangle=0Leftrightarrow |u_1-u_2|=0 Leftrightarrow u_1=u_2. $$ This is a trick used very often in Hilbert spaces.
answered Jan 6 at 22:23
diec_diec_
1253
$endgroup$
A scalar product $langlecdot,cdotrangle$ induces on a vector space the norm $$ |v|=sqrt{langle v,vrangle}. $$ One of the properties of the norm is that $$ |v|=0Leftrightarrow v=0. $$ Then, specialising this property to the proof of the theorem, we have $$ langle u_1-u_2, u_1-u_2rangle=0Leftrightarrow |u_1-u_2|=0 Leftrightarrow u_1=u_2. $$ This is a trick used very often in Hilbert spaces.
answered Jan 6 at 22:23
diec_diec_
1253
answered Jan 6 at 22:23
diec_diec_
1253
answered Jan 6 at 22:23
diec_diec_
1253
answered Jan 6 at 22:23
diec_diec_
1253
1253
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