Mixing
Gram Schmidt Process for a Complex Vector Space
$begingroup$
Suppose I have certain independent vectors, say $lvert V_1rangle$ and $lvert V_2rangle$, which span a 2-dimensional subspace of a given Complex Vector Space on which inner product is defined, how is the standard Gram Schmidt Process extended?
Even though StackExchange has answers to related questions, I have a problem with how exactly the method works. Following the process, we get $$lvert v_1rangle = frac{lvert V_1 rangle}{sqrt{langle V_1 rvert V_1 rangle}}$$ and $$lvert V_2' rangle = lvert V_2 rangle - langle V_2 rvert v_1 rangle lvert v_1 rangle$$ and $$lvert v_2 rangle = frac{lvert V_2 rangle - langle V_2 rvert v_1 rangle lvert v_1 rangle}{sqrt{langle V_2' rvert V_2' rangle}}$$Now $lvert v_1 rangle$ and $lvert v_2 rangle$ form an orthonormal basis for the given subspace. If this is true, then $langle v_1 rvert v_2 rangle$ should be equal to 0.
But, $$langle v_1 rvert v_2 rangle = langle v_1 rvert left(frac{lvert V_2 rangle - langle V_2 rvert v_1 rangle lvert v_1 rangle}{sqrt{langle V_2' rvert V_2' rangle}}right)$$ Since $langle v_1 rvert v_1 rangle = 1$, $$langle v_1 rvert v_2 rangle = frac{langle v_1 rvert V_2 rangle - langle V_2 rvert v_1 rangle}{sqrt{langle V_2 rvert V_2 rangle}}$$ $$langle v_1 rvert v_2 rangle = frac{langle v_1 rvert V_2 rangle - {langle v_1 rvert V_2 rangle}^*}{sqrt{langle V_2 rvert V_2 rangle}}$$ But the final equation is not necessarily zero for a complex vector space. Am I going wrong somewhere?
linear-algebra vector-spaces inner-product-space orthonormal
asked Jan 28 at 22:12
Ajay Shanmuga SakthivasanAjay Shanmuga Sakthivasan
336
$endgroup$
add a comment |
$begingroup$
Suppose I have certain independent vectors, say $lvert V_1rangle$ and $lvert V_2rangle$, which span a 2-dimensional subspace of a given Complex Vector Space on which inner product is defined, how is the standard Gram Schmidt Process extended?
Even though StackExchange has answers to related questions, I have a problem with how exactly the method works. Following the process, we get $$lvert v_1rangle = frac{lvert V_1 rangle}{sqrt{langle V_1 rvert V_1 rangle}}$$ and $$lvert V_2' rangle = lvert V_2 rangle - langle V_2 rvert v_1 rangle lvert v_1 rangle$$ and $$lvert v_2 rangle = frac{lvert V_2 rangle - langle V_2 rvert v_1 rangle lvert v_1 rangle}{sqrt{langle V_2' rvert V_2' rangle}}$$Now $lvert v_1 rangle$ and $lvert v_2 rangle$ form an orthonormal basis for the given subspace. If this is true, then $langle v_1 rvert v_2 rangle$ should be equal to 0.
But, $$langle v_1 rvert v_2 rangle = langle v_1 rvert left(frac{lvert V_2 rangle - langle V_2 rvert v_1 rangle lvert v_1 rangle}{sqrt{langle V_2' rvert V_2' rangle}}right)$$ Since $langle v_1 rvert v_1 rangle = 1$, $$langle v_1 rvert v_2 rangle = frac{langle v_1 rvert V_2 rangle - langle V_2 rvert v_1 rangle}{sqrt{langle V_2 rvert V_2 rangle}}$$ $$langle v_1 rvert v_2 rangle = frac{langle v_1 rvert V_2 rangle - {langle v_1 rvert V_2 rangle}^*}{sqrt{langle V_2 rvert V_2 rangle}}$$ But the final equation is not necessarily zero for a complex vector space. Am I going wrong somewhere?
linear-algebra vector-spaces inner-product-space orthonormal
asked Jan 28 at 22:12
Ajay Shanmuga SakthivasanAjay Shanmuga Sakthivasan
336
$endgroup$
$begingroup$
Yes, but why don't you define $lvert v_2 rangle = frac{lvert V_2 rangle - langle v_1 rvert V_2 rangle lvert v_1 rangle}{sqrt{langle V_2' rvert V_2' rangle}}$?
$endgroup$
– Raskolnikov
Jan 28 at 22:16
$begingroup$
@Raskolnikov This would indeed solve the problem. Is the Gram Schmidt for complex vector spaces defined this way? Because the version of Gram Schmidt I studied didn't have this as the method.
$endgroup$
– Ajay Shanmuga Sakthivasan
Jan 28 at 22:30
add a comment |
$begingroup$
Suppose I have certain independent vectors, say $lvert V_1rangle$ and $lvert V_2rangle$, which span a 2-dimensional subspace of a given Complex Vector Space on which inner product is defined, how is the standard Gram Schmidt Process extended?
Even though StackExchange has answers to related questions, I have a problem with how exactly the method works. Following the process, we get $$lvert v_1rangle = frac{lvert V_1 rangle}{sqrt{langle V_1 rvert V_1 rangle}}$$ and $$lvert V_2' rangle = lvert V_2 rangle - langle V_2 rvert v_1 rangle lvert v_1 rangle$$ and $$lvert v_2 rangle = frac{lvert V_2 rangle - langle V_2 rvert v_1 rangle lvert v_1 rangle}{sqrt{langle V_2' rvert V_2' rangle}}$$Now $lvert v_1 rangle$ and $lvert v_2 rangle$ form an orthonormal basis for the given subspace. If this is true, then $langle v_1 rvert v_2 rangle$ should be equal to 0.
But, $$langle v_1 rvert v_2 rangle = langle v_1 rvert left(frac{lvert V_2 rangle - langle V_2 rvert v_1 rangle lvert v_1 rangle}{sqrt{langle V_2' rvert V_2' rangle}}right)$$ Since $langle v_1 rvert v_1 rangle = 1$, $$langle v_1 rvert v_2 rangle = frac{langle v_1 rvert V_2 rangle - langle V_2 rvert v_1 rangle}{sqrt{langle V_2 rvert V_2 rangle}}$$ $$langle v_1 rvert v_2 rangle = frac{langle v_1 rvert V_2 rangle - {langle v_1 rvert V_2 rangle}^*}{sqrt{langle V_2 rvert V_2 rangle}}$$ But the final equation is not necessarily zero for a complex vector space. Am I going wrong somewhere?
linear-algebra vector-spaces inner-product-space orthonormal
asked Jan 28 at 22:12
Ajay Shanmuga SakthivasanAjay Shanmuga Sakthivasan
336
$endgroup$
Suppose I have certain independent vectors, say $lvert V_1rangle$ and $lvert V_2rangle$, which span a 2-dimensional subspace of a given Complex Vector Space on which inner product is defined, how is the standard Gram Schmidt Process extended?
Even though StackExchange has answers to related questions, I have a problem with how exactly the method works. Following the process, we get $$lvert v_1rangle = frac{lvert V_1 rangle}{sqrt{langle V_1 rvert V_1 rangle}}$$ and $$lvert V_2' rangle = lvert V_2 rangle - langle V_2 rvert v_1 rangle lvert v_1 rangle$$ and $$lvert v_2 rangle = frac{lvert V_2 rangle - langle V_2 rvert v_1 rangle lvert v_1 rangle}{sqrt{langle V_2' rvert V_2' rangle}}$$Now $lvert v_1 rangle$ and $lvert v_2 rangle$ form an orthonormal basis for the given subspace. If this is true, then $langle v_1 rvert v_2 rangle$ should be equal to 0.
But, $$langle v_1 rvert v_2 rangle = langle v_1 rvert left(frac{lvert V_2 rangle - langle V_2 rvert v_1 rangle lvert v_1 rangle}{sqrt{langle V_2' rvert V_2' rangle}}right)$$ Since $langle v_1 rvert v_1 rangle = 1$, $$langle v_1 rvert v_2 rangle = frac{langle v_1 rvert V_2 rangle - langle V_2 rvert v_1 rangle}{sqrt{langle V_2 rvert V_2 rangle}}$$ $$langle v_1 rvert v_2 rangle = frac{langle v_1 rvert V_2 rangle - {langle v_1 rvert V_2 rangle}^*}{sqrt{langle V_2 rvert V_2 rangle}}$$ But the final equation is not necessarily zero for a complex vector space. Am I going wrong somewhere?
linear-algebra vector-spaces inner-product-space orthonormal
linear-algebra vector-spaces inner-product-space orthonormal
asked Jan 28 at 22:12
Ajay Shanmuga SakthivasanAjay Shanmuga Sakthivasan
336
asked Jan 28 at 22:12
Ajay Shanmuga SakthivasanAjay Shanmuga Sakthivasan
336
asked Jan 28 at 22:12
Ajay Shanmuga SakthivasanAjay Shanmuga Sakthivasan
336
asked Jan 28 at 22:12
Ajay Shanmuga SakthivasanAjay Shanmuga Sakthivasan
336
asked Jan 28 at 22:12
Ajay Shanmuga SakthivasanAjay Shanmuga Sakthivasan
336
336
$begingroup$
Yes, but why don't you define $lvert v_2 rangle = frac{lvert V_2 rangle - langle v_1 rvert V_2 rangle lvert v_1 rangle}{sqrt{langle V_2' rvert V_2' rangle}}$?
$endgroup$
– Raskolnikov
Jan 28 at 22:16
$begingroup$
@Raskolnikov This would indeed solve the problem. Is the Gram Schmidt for complex vector spaces defined this way? Because the version of Gram Schmidt I studied didn't have this as the method.
$endgroup$
– Ajay Shanmuga Sakthivasan
Jan 28 at 22:30
add a comment |
$begingroup$
Yes, but why don't you define $lvert v_2 rangle = frac{lvert V_2 rangle - langle v_1 rvert V_2 rangle lvert v_1 rangle}{sqrt{langle V_2' rvert V_2' rangle}}$?
$endgroup$
– Raskolnikov
Jan 28 at 22:16
$begingroup$
@Raskolnikov This would indeed solve the problem. Is the Gram Schmidt for complex vector spaces defined this way? Because the version of Gram Schmidt I studied didn't have this as the method.
$endgroup$
– Ajay Shanmuga Sakthivasan
Jan 28 at 22:30
$begingroup$
Yes, but why don't you define $lvert v_2 rangle = frac{lvert V_2 rangle - langle v_1 rvert V_2 rangle lvert v_1 rangle}{sqrt{langle V_2' rvert V_2' rangle}}$?
$endgroup$
– Raskolnikov
Jan 28 at 22:16
$begingroup$
Yes, but why don't you define $lvert v_2 rangle = frac{lvert V_2 rangle - langle v_1 rvert V_2 rangle lvert v_1 rangle}{sqrt{langle V_2' rvert V_2' rangle}}$?
$endgroup$
– Raskolnikov
Jan 28 at 22:16
$begingroup$
@Raskolnikov This would indeed solve the problem. Is the Gram Schmidt for complex vector spaces defined this way? Because the version of Gram Schmidt I studied didn't have this as the method.
$endgroup$
– Ajay Shanmuga Sakthivasan
Jan 28 at 22:30
$begingroup$
@Raskolnikov This would indeed solve the problem. Is the Gram Schmidt for complex vector spaces defined this way? Because the version of Gram Schmidt I studied didn't have this as the method.
$endgroup$
– Ajay Shanmuga Sakthivasan
Jan 28 at 22:30
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The problem lies in your definition of $lvert v_2rangle$. It should be$$lvert v_2 rangle = frac{lvert V_2'rangle - langle V_2'rvert v_1 rangle lvert v_1 rangle}{sqrt{langle V_2' rvert V_2' rangle}}.$$Using this definition, you will get that, indeed, $langle v_1|v_2rangle=0$. Unless your inner product is linear in the first variable and anti-linear in the second one. Then it should be$$lvert v_2 rangle = frac{lvert V_2'rangle - langle V_2'rvert v_1 rangle^*lvert v_1 rangle}{sqrt{langle V_2' rvert V_2' rangle}}.$$
answered Jan 28 at 22:19
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
$begingroup$
No, but $lvert v_2 rangle$ would be $frac{lvert V_2' rangle}{sqrt{langle V_2' rvert V_2' rangle}}$, where I have written $lvert V_2' rangle$ using the second equation. Even if I take $lvert v_2 rangle$ as given in your answer, I would end up in the same trouble.
$endgroup$
– Ajay Shanmuga Sakthivasan
Jan 28 at 22:27
$begingroup$
I see. You are working with an inner product which is linear in the first variable and anti-linear in the second one. The method that you described works when it is anti-linear in the first variable and linear in the second one. Just define$$lvert V_2' rangle=lvert V_2rangle-langle V_2 rvert v_1 rangle^*lvert v_1 rangle$$and all will be fine.
$endgroup$
– José Carlos Santos
Jan 28 at 22:48
$begingroup$
This would indeed work, Thank you.
$endgroup$
– Ajay Shanmuga Sakthivasan
Jan 28 at 23:20
add a comment |
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$begingroup$
The problem lies in your definition of $lvert v_2rangle$. It should be$$lvert v_2 rangle = frac{lvert V_2'rangle - langle V_2'rvert v_1 rangle lvert v_1 rangle}{sqrt{langle V_2' rvert V_2' rangle}}.$$Using this definition, you will get that, indeed, $langle v_1|v_2rangle=0$. Unless your inner product is linear in the first variable and anti-linear in the second one. Then it should be$$lvert v_2 rangle = frac{lvert V_2'rangle - langle V_2'rvert v_1 rangle^*lvert v_1 rangle}{sqrt{langle V_2' rvert V_2' rangle}}.$$
answered Jan 28 at 22:19
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
$begingroup$
No, but $lvert v_2 rangle$ would be $frac{lvert V_2' rangle}{sqrt{langle V_2' rvert V_2' rangle}}$, where I have written $lvert V_2' rangle$ using the second equation. Even if I take $lvert v_2 rangle$ as given in your answer, I would end up in the same trouble.
$endgroup$
– Ajay Shanmuga Sakthivasan
Jan 28 at 22:27
$begingroup$
I see. You are working with an inner product which is linear in the first variable and anti-linear in the second one. The method that you described works when it is anti-linear in the first variable and linear in the second one. Just define$$lvert V_2' rangle=lvert V_2rangle-langle V_2 rvert v_1 rangle^*lvert v_1 rangle$$and all will be fine.
$endgroup$
– José Carlos Santos
Jan 28 at 22:48
$begingroup$
This would indeed work, Thank you.
$endgroup$
– Ajay Shanmuga Sakthivasan
Jan 28 at 23:20
add a comment |
$begingroup$
The problem lies in your definition of $lvert v_2rangle$. It should be$$lvert v_2 rangle = frac{lvert V_2'rangle - langle V_2'rvert v_1 rangle lvert v_1 rangle}{sqrt{langle V_2' rvert V_2' rangle}}.$$Using this definition, you will get that, indeed, $langle v_1|v_2rangle=0$. Unless your inner product is linear in the first variable and anti-linear in the second one. Then it should be$$lvert v_2 rangle = frac{lvert V_2'rangle - langle V_2'rvert v_1 rangle^*lvert v_1 rangle}{sqrt{langle V_2' rvert V_2' rangle}}.$$
answered Jan 28 at 22:19
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
$begingroup$
No, but $lvert v_2 rangle$ would be $frac{lvert V_2' rangle}{sqrt{langle V_2' rvert V_2' rangle}}$, where I have written $lvert V_2' rangle$ using the second equation. Even if I take $lvert v_2 rangle$ as given in your answer, I would end up in the same trouble.
$endgroup$
– Ajay Shanmuga Sakthivasan
Jan 28 at 22:27
$begingroup$
I see. You are working with an inner product which is linear in the first variable and anti-linear in the second one. The method that you described works when it is anti-linear in the first variable and linear in the second one. Just define$$lvert V_2' rangle=lvert V_2rangle-langle V_2 rvert v_1 rangle^*lvert v_1 rangle$$and all will be fine.
$endgroup$
– José Carlos Santos
Jan 28 at 22:48
$begingroup$
This would indeed work, Thank you.
$endgroup$
– Ajay Shanmuga Sakthivasan
Jan 28 at 23:20
add a comment |
$begingroup$
The problem lies in your definition of $lvert v_2rangle$. It should be$$lvert v_2 rangle = frac{lvert V_2'rangle - langle V_2'rvert v_1 rangle lvert v_1 rangle}{sqrt{langle V_2' rvert V_2' rangle}}.$$Using this definition, you will get that, indeed, $langle v_1|v_2rangle=0$. Unless your inner product is linear in the first variable and anti-linear in the second one. Then it should be$$lvert v_2 rangle = frac{lvert V_2'rangle - langle V_2'rvert v_1 rangle^*lvert v_1 rangle}{sqrt{langle V_2' rvert V_2' rangle}}.$$
answered Jan 28 at 22:19
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
The problem lies in your definition of $lvert v_2rangle$. It should be$$lvert v_2 rangle = frac{lvert V_2'rangle - langle V_2'rvert v_1 rangle lvert v_1 rangle}{sqrt{langle V_2' rvert V_2' rangle}}.$$Using this definition, you will get that, indeed, $langle v_1|v_2rangle=0$. Unless your inner product is linear in the first variable and anti-linear in the second one. Then it should be$$lvert v_2 rangle = frac{lvert V_2'rangle - langle V_2'rvert v_1 rangle^*lvert v_1 rangle}{sqrt{langle V_2' rvert V_2' rangle}}.$$
answered Jan 28 at 22:19
José Carlos SantosJosé Carlos Santos
171k23132240
edited Jan 28 at 23:22
answered Jan 28 at 22:19
José Carlos SantosJosé Carlos Santos
171k23132240
answered Jan 28 at 22:19
José Carlos SantosJosé Carlos Santos
171k23132240
answered Jan 28 at 22:19
José Carlos SantosJosé Carlos Santos
171k23132240
171k23132240
$begingroup$
No, but $lvert v_2 rangle$ would be $frac{lvert V_2' rangle}{sqrt{langle V_2' rvert V_2' rangle}}$, where I have written $lvert V_2' rangle$ using the second equation. Even if I take $lvert v_2 rangle$ as given in your answer, I would end up in the same trouble.
$endgroup$
– Ajay Shanmuga Sakthivasan
Jan 28 at 22:27
$begingroup$
I see. You are working with an inner product which is linear in the first variable and anti-linear in the second one. The method that you described works when it is anti-linear in the first variable and linear in the second one. Just define$$lvert V_2' rangle=lvert V_2rangle-langle V_2 rvert v_1 rangle^*lvert v_1 rangle$$and all will be fine.
$endgroup$
– José Carlos Santos
Jan 28 at 22:48
$begingroup$
This would indeed work, Thank you.
$endgroup$
– Ajay Shanmuga Sakthivasan
Jan 28 at 23:20
add a comment |
$begingroup$
No, but $lvert v_2 rangle$ would be $frac{lvert V_2' rangle}{sqrt{langle V_2' rvert V_2' rangle}}$, where I have written $lvert V_2' rangle$ using the second equation. Even if I take $lvert v_2 rangle$ as given in your answer, I would end up in the same trouble.
$endgroup$
– Ajay Shanmuga Sakthivasan
Jan 28 at 22:27
$begingroup$
I see. You are working with an inner product which is linear in the first variable and anti-linear in the second one. The method that you described works when it is anti-linear in the first variable and linear in the second one. Just define$$lvert V_2' rangle=lvert V_2rangle-langle V_2 rvert v_1 rangle^*lvert v_1 rangle$$and all will be fine.
$endgroup$
– José Carlos Santos
Jan 28 at 22:48
$begingroup$
This would indeed work, Thank you.
$endgroup$
– Ajay Shanmuga Sakthivasan
Jan 28 at 23:20
$begingroup$
No, but $lvert v_2 rangle$ would be $frac{lvert V_2' rangle}{sqrt{langle V_2' rvert V_2' rangle}}$, where I have written $lvert V_2' rangle$ using the second equation. Even if I take $lvert v_2 rangle$ as given in your answer, I would end up in the same trouble.
$endgroup$
– Ajay Shanmuga Sakthivasan
Jan 28 at 22:27
$begingroup$
No, but $lvert v_2 rangle$ would be $frac{lvert V_2' rangle}{sqrt{langle V_2' rvert V_2' rangle}}$, where I have written $lvert V_2' rangle$ using the second equation. Even if I take $lvert v_2 rangle$ as given in your answer, I would end up in the same trouble.
$endgroup$
– Ajay Shanmuga Sakthivasan
Jan 28 at 22:27
$begingroup$
I see. You are working with an inner product which is linear in the first variable and anti-linear in the second one. The method that you described works when it is anti-linear in the first variable and linear in the second one. Just define$$lvert V_2' rangle=lvert V_2rangle-langle V_2 rvert v_1 rangle^*lvert v_1 rangle$$and all will be fine.
$endgroup$
– José Carlos Santos
Jan 28 at 22:48
$begingroup$
I see. You are working with an inner product which is linear in the first variable and anti-linear in the second one. The method that you described works when it is anti-linear in the first variable and linear in the second one. Just define$$lvert V_2' rangle=lvert V_2rangle-langle V_2 rvert v_1 rangle^*lvert v_1 rangle$$and all will be fine.
$endgroup$
– José Carlos Santos
Jan 28 at 22:48
$begingroup$
This would indeed work, Thank you.
$endgroup$
– Ajay Shanmuga Sakthivasan
Jan 28 at 23:20
$begingroup$
This would indeed work, Thank you.
$endgroup$
– Ajay Shanmuga Sakthivasan
Jan 28 at 23:20
add a comment |
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$begingroup$
Yes, but why don't you define $lvert v_2 rangle = frac{lvert V_2 rangle - langle v_1 rvert V_2 rangle lvert v_1 rangle}{sqrt{langle V_2' rvert V_2' rangle}}$?
$endgroup$
– Raskolnikov
Jan 28 at 22:16
$begingroup$
@Raskolnikov This would indeed solve the problem. Is the Gram Schmidt for complex vector spaces defined this way? Because the version of Gram Schmidt I studied didn't have this as the method.
$endgroup$
– Ajay Shanmuga Sakthivasan
Jan 28 at 22:30