What is the householder reflector's singular value? [closed]
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A householder reflector is $I-2ww^*$ where $ w$ is an unit vector.
linear-algebra
asked Dec 17 '18 at 13:09
KristyKristy
204
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closed as off-topic by Arnaud D., GNUSupporter 8964民主女神 地下教會, Adrian Keister, Paul Frost, José Carlos Santos Dec 17 '18 at 17:19
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$begingroup$
A householder reflector is $I-2ww^*$ where $ w$ is an unit vector.
linear-algebra
asked Dec 17 '18 at 13:09
KristyKristy
204
$endgroup$
closed as off-topic by Arnaud D., GNUSupporter 8964民主女神 地下教會, Adrian Keister, Paul Frost, José Carlos Santos Dec 17 '18 at 17:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Arnaud D., GNUSupporter 8964民主女神 地下教會, Adrian Keister, Paul Frost, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
A householder reflector is $I-2ww^*$ where $ w$ is an unit vector.
linear-algebra
asked Dec 17 '18 at 13:09
KristyKristy
204
$endgroup$
A householder reflector is $I-2ww^*$ where $ w$ is an unit vector.
linear-algebra
linear-algebra
asked Dec 17 '18 at 13:09
KristyKristy
204
asked Dec 17 '18 at 13:09
KristyKristy
204
edited Dec 17 '18 at 13:23
Kristy
asked Dec 17 '18 at 13:09
KristyKristy
204
asked Dec 17 '18 at 13:09
KristyKristy
204
asked Dec 17 '18 at 13:09
KristyKristy
204
204
closed as off-topic by Arnaud D., GNUSupporter 8964民主女神 地下教會, Adrian Keister, Paul Frost, José Carlos Santos Dec 17 '18 at 17:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Arnaud D., GNUSupporter 8964民主女神 地下教會, Adrian Keister, Paul Frost, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Arnaud D., GNUSupporter 8964民主女神 地下教會, Adrian Keister, Paul Frost, José Carlos Santos Dec 17 '18 at 17:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Arnaud D., GNUSupporter 8964民主女神 地下教會, Adrian Keister, Paul Frost, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
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1 Answer
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$begingroup$
Since $H = I- 2ww^*$ is hemitian, the singular values of $H$are just the absolute values of the eigenvalues of $H$.
Notice that if $uperp w$, then $$Hu = u - 2ww^*u = u-2langle u,wrangle w = u$$
so $u$ is an eigenvector for $H$ with eigenvalue $1$. If you are working in $mathbb{C}^n$, you can pick $n-1$ linearly independent vectors orthogonal to $v$ so the multiplicity of the eigenvalue $1$ is at least $n-1$.
On the other hand, we have
$$Hw = w - 2ww^*w = w - 2|w|^2w = w-2w = -w$$
so $w$ is the eigenvector for $H$ with eigenvalue $-1$. Therefore, the only eigenvalues are $pm 1$ so the desired singular value is $1$.
answered Dec 17 '18 at 13:31
mechanodroidmechanodroid
27.8k62447
$endgroup$
$begingroup$
If I calculate the eigenvalue like this:$|H-lambda I|=|（1-lambda ）I-2ww^*|=begin{vmatrix} 1-lambda & w^*\ 2w & I end{vmatrix}=begin{vmatrix} 1-lambda+2w^*w & w^*\ 0 & I end{vmatrix}=0$，implying that $lambda=3$, where did I go wrong?
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– Kristy
Dec 17 '18 at 13:48
$begingroup$
@Kristy How did you get $$det((1-lambda) I - 2ww^*) = begin{vmatrix} 1-lambda & w^* \ 2w & Iend{vmatrix}$$ It doesn't seem correct.
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– mechanodroid
Dec 17 '18 at 13:56
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since $H = I- 2ww^*$ is hemitian, the singular values of $H$are just the absolute values of the eigenvalues of $H$.
Notice that if $uperp w$, then $$Hu = u - 2ww^*u = u-2langle u,wrangle w = u$$
so $u$ is an eigenvector for $H$ with eigenvalue $1$. If you are working in $mathbb{C}^n$, you can pick $n-1$ linearly independent vectors orthogonal to $v$ so the multiplicity of the eigenvalue $1$ is at least $n-1$.
On the other hand, we have
$$Hw = w - 2ww^*w = w - 2|w|^2w = w-2w = -w$$
so $w$ is the eigenvector for $H$ with eigenvalue $-1$. Therefore, the only eigenvalues are $pm 1$ so the desired singular value is $1$.
answered Dec 17 '18 at 13:31
mechanodroidmechanodroid
27.8k62447
$endgroup$
$begingroup$
If I calculate the eigenvalue like this:$|H-lambda I|=|（1-lambda ）I-2ww^*|=begin{vmatrix} 1-lambda & w^*\ 2w & I end{vmatrix}=begin{vmatrix} 1-lambda+2w^*w & w^*\ 0 & I end{vmatrix}=0$，implying that $lambda=3$, where did I go wrong?
$endgroup$
– Kristy
Dec 17 '18 at 13:48
$begingroup$
@Kristy How did you get $$det((1-lambda) I - 2ww^*) = begin{vmatrix} 1-lambda & w^* \ 2w & Iend{vmatrix}$$ It doesn't seem correct.
$endgroup$
– mechanodroid
Dec 17 '18 at 13:56
add a comment |
$begingroup$
Since $H = I- 2ww^*$ is hemitian, the singular values of $H$are just the absolute values of the eigenvalues of $H$.
Notice that if $uperp w$, then $$Hu = u - 2ww^*u = u-2langle u,wrangle w = u$$
so $u$ is an eigenvector for $H$ with eigenvalue $1$. If you are working in $mathbb{C}^n$, you can pick $n-1$ linearly independent vectors orthogonal to $v$ so the multiplicity of the eigenvalue $1$ is at least $n-1$.
On the other hand, we have
$$Hw = w - 2ww^*w = w - 2|w|^2w = w-2w = -w$$
so $w$ is the eigenvector for $H$ with eigenvalue $-1$. Therefore, the only eigenvalues are $pm 1$ so the desired singular value is $1$.
answered Dec 17 '18 at 13:31
mechanodroidmechanodroid
27.8k62447
$endgroup$
$begingroup$
If I calculate the eigenvalue like this:$|H-lambda I|=|（1-lambda ）I-2ww^*|=begin{vmatrix} 1-lambda & w^*\ 2w & I end{vmatrix}=begin{vmatrix} 1-lambda+2w^*w & w^*\ 0 & I end{vmatrix}=0$，implying that $lambda=3$, where did I go wrong?
$endgroup$
– Kristy
Dec 17 '18 at 13:48
$begingroup$
@Kristy How did you get $$det((1-lambda) I - 2ww^*) = begin{vmatrix} 1-lambda & w^* \ 2w & Iend{vmatrix}$$ It doesn't seem correct.
$endgroup$
– mechanodroid
Dec 17 '18 at 13:56
add a comment |
$begingroup$
Since $H = I- 2ww^*$ is hemitian, the singular values of $H$are just the absolute values of the eigenvalues of $H$.
Notice that if $uperp w$, then $$Hu = u - 2ww^*u = u-2langle u,wrangle w = u$$
so $u$ is an eigenvector for $H$ with eigenvalue $1$. If you are working in $mathbb{C}^n$, you can pick $n-1$ linearly independent vectors orthogonal to $v$ so the multiplicity of the eigenvalue $1$ is at least $n-1$.
On the other hand, we have
$$Hw = w - 2ww^*w = w - 2|w|^2w = w-2w = -w$$
so $w$ is the eigenvector for $H$ with eigenvalue $-1$. Therefore, the only eigenvalues are $pm 1$ so the desired singular value is $1$.
answered Dec 17 '18 at 13:31
mechanodroidmechanodroid
27.8k62447
$endgroup$
Since $H = I- 2ww^*$ is hemitian, the singular values of $H$are just the absolute values of the eigenvalues of $H$.
Notice that if $uperp w$, then $$Hu = u - 2ww^*u = u-2langle u,wrangle w = u$$
so $u$ is an eigenvector for $H$ with eigenvalue $1$. If you are working in $mathbb{C}^n$, you can pick $n-1$ linearly independent vectors orthogonal to $v$ so the multiplicity of the eigenvalue $1$ is at least $n-1$.
On the other hand, we have
$$Hw = w - 2ww^*w = w - 2|w|^2w = w-2w = -w$$
so $w$ is the eigenvector for $H$ with eigenvalue $-1$. Therefore, the only eigenvalues are $pm 1$ so the desired singular value is $1$.
answered Dec 17 '18 at 13:31
mechanodroidmechanodroid
27.8k62447
answered Dec 17 '18 at 13:31
mechanodroidmechanodroid
27.8k62447
answered Dec 17 '18 at 13:31
mechanodroidmechanodroid
27.8k62447
answered Dec 17 '18 at 13:31
mechanodroidmechanodroid
27.8k62447
27.8k62447
$begingroup$
If I calculate the eigenvalue like this:$|H-lambda I|=|（1-lambda ）I-2ww^*|=begin{vmatrix} 1-lambda & w^*\ 2w & I end{vmatrix}=begin{vmatrix} 1-lambda+2w^*w & w^*\ 0 & I end{vmatrix}=0$，implying that $lambda=3$, where did I go wrong?
$endgroup$
– Kristy
Dec 17 '18 at 13:48
$begingroup$
@Kristy How did you get $$det((1-lambda) I - 2ww^*) = begin{vmatrix} 1-lambda & w^* \ 2w & Iend{vmatrix}$$ It doesn't seem correct.
$endgroup$
– mechanodroid
Dec 17 '18 at 13:56
add a comment |
$begingroup$
If I calculate the eigenvalue like this:$|H-lambda I|=|（1-lambda ）I-2ww^*|=begin{vmatrix} 1-lambda & w^*\ 2w & I end{vmatrix}=begin{vmatrix} 1-lambda+2w^*w & w^*\ 0 & I end{vmatrix}=0$，implying that $lambda=3$, where did I go wrong?
$endgroup$
– Kristy
Dec 17 '18 at 13:48
$begingroup$
@Kristy How did you get $$det((1-lambda) I - 2ww^*) = begin{vmatrix} 1-lambda & w^* \ 2w & Iend{vmatrix}$$ It doesn't seem correct.
$endgroup$
– mechanodroid
Dec 17 '18 at 13:56
$begingroup$
If I calculate the eigenvalue like this:$|H-lambda I|=|（1-lambda ）I-2ww^*|=begin{vmatrix} 1-lambda & w^*\ 2w & I end{vmatrix}=begin{vmatrix} 1-lambda+2w^*w & w^*\ 0 & I end{vmatrix}=0$，implying that $lambda=3$, where did I go wrong?
$endgroup$
– Kristy
Dec 17 '18 at 13:48
$begingroup$
If I calculate the eigenvalue like this:$|H-lambda I|=|（1-lambda ）I-2ww^*|=begin{vmatrix} 1-lambda & w^*\ 2w & I end{vmatrix}=begin{vmatrix} 1-lambda+2w^*w & w^*\ 0 & I end{vmatrix}=0$，implying that $lambda=3$, where did I go wrong?
$endgroup$
– Kristy
Dec 17 '18 at 13:48
$begingroup$
@Kristy How did you get $$det((1-lambda) I - 2ww^*) = begin{vmatrix} 1-lambda & w^* \ 2w & Iend{vmatrix}$$ It doesn't seem correct.
$endgroup$
– mechanodroid
Dec 17 '18 at 13:56
$begingroup$
@Kristy How did you get $$det((1-lambda) I - 2ww^*) = begin{vmatrix} 1-lambda & w^* \ 2w & Iend{vmatrix}$$ It doesn't seem correct.
$endgroup$
– mechanodroid
Dec 17 '18 at 13:56
add a comment |
