Mixing
Extending ${u_1, u_2}$ to an orthonormal basis when finding an SVD
$begingroup$
I've been working through my linear algebra textbook, and when finding an SVD there's just one thing I don't understand.
For example, finding an SVD for the 3x2 matrix A. I will skip the steps of finding the eigenvectors, eigenvalues, singular values... anyway, we find that
$$
V = begin{bmatrix}vec{v}_1 & vec{v}_2end{bmatrix} =
begin{bmatrix}
1/sqrt2 & -1/sqrt2\
1/sqrt2 & 1/sqrt2
end{bmatrix}
$$
and we know that
$$
vec{u}_n = frac{1}{sigma_n}Avec{v}_n
$$
which gives
$$
vec{u}_1 = begin{bmatrix}2/sqrt6\1/sqrt6\1/sqrt6end{bmatrix}, vec{u}_2 = begin{bmatrix}0\-1/sqrt2\1/sqrt2end{bmatrix}
$$
but we know that $U$ is a $m$ by $m$ matrix, so it must be 3 by 3, and so we have to find $vec{u}_3$. This is where I get stuck; the book says that one method to do this is to use the Gram-Schmidt Process, but I just can't seem to wrap my head around how to do this with the vectors shown above.
linear-algebra orthonormal svd
asked May 17 '16 at 20:31
user5368737user5368737
1477
$endgroup$
add a comment |
$begingroup$
I've been working through my linear algebra textbook, and when finding an SVD there's just one thing I don't understand.
For example, finding an SVD for the 3x2 matrix A. I will skip the steps of finding the eigenvectors, eigenvalues, singular values... anyway, we find that
$$
V = begin{bmatrix}vec{v}_1 & vec{v}_2end{bmatrix} =
begin{bmatrix}
1/sqrt2 & -1/sqrt2\
1/sqrt2 & 1/sqrt2
end{bmatrix}
$$
and we know that
$$
vec{u}_n = frac{1}{sigma_n}Avec{v}_n
$$
which gives
$$
vec{u}_1 = begin{bmatrix}2/sqrt6\1/sqrt6\1/sqrt6end{bmatrix}, vec{u}_2 = begin{bmatrix}0\-1/sqrt2\1/sqrt2end{bmatrix}
$$
but we know that $U$ is a $m$ by $m$ matrix, so it must be 3 by 3, and so we have to find $vec{u}_3$. This is where I get stuck; the book says that one method to do this is to use the Gram-Schmidt Process, but I just can't seem to wrap my head around how to do this with the vectors shown above.
linear-algebra orthonormal svd
asked May 17 '16 at 20:31
user5368737user5368737
1477
$endgroup$
$begingroup$
Choose a vector $x$ linearly independent of $u_1,u_2$ basically at random; one possible choice is $begin{bmatrix} 0 \ 1 \ 0 end{bmatrix}$. Then run Gram-Schmidt on ${ u_1,u_2,x }$.
$endgroup$
– Ian
May 17 '16 at 20:36
add a comment |
$begingroup$
I've been working through my linear algebra textbook, and when finding an SVD there's just one thing I don't understand.
For example, finding an SVD for the 3x2 matrix A. I will skip the steps of finding the eigenvectors, eigenvalues, singular values... anyway, we find that
$$
V = begin{bmatrix}vec{v}_1 & vec{v}_2end{bmatrix} =
begin{bmatrix}
1/sqrt2 & -1/sqrt2\
1/sqrt2 & 1/sqrt2
end{bmatrix}
$$
and we know that
$$
vec{u}_n = frac{1}{sigma_n}Avec{v}_n
$$
which gives
$$
vec{u}_1 = begin{bmatrix}2/sqrt6\1/sqrt6\1/sqrt6end{bmatrix}, vec{u}_2 = begin{bmatrix}0\-1/sqrt2\1/sqrt2end{bmatrix}
$$
but we know that $U$ is a $m$ by $m$ matrix, so it must be 3 by 3, and so we have to find $vec{u}_3$. This is where I get stuck; the book says that one method to do this is to use the Gram-Schmidt Process, but I just can't seem to wrap my head around how to do this with the vectors shown above.
linear-algebra orthonormal svd
asked May 17 '16 at 20:31
user5368737user5368737
1477
$endgroup$
I've been working through my linear algebra textbook, and when finding an SVD there's just one thing I don't understand.
For example, finding an SVD for the 3x2 matrix A. I will skip the steps of finding the eigenvectors, eigenvalues, singular values... anyway, we find that
$$
V = begin{bmatrix}vec{v}_1 & vec{v}_2end{bmatrix} =
begin{bmatrix}
1/sqrt2 & -1/sqrt2\
1/sqrt2 & 1/sqrt2
end{bmatrix}
$$
and we know that
$$
vec{u}_n = frac{1}{sigma_n}Avec{v}_n
$$
which gives
$$
vec{u}_1 = begin{bmatrix}2/sqrt6\1/sqrt6\1/sqrt6end{bmatrix}, vec{u}_2 = begin{bmatrix}0\-1/sqrt2\1/sqrt2end{bmatrix}
$$
but we know that $U$ is a $m$ by $m$ matrix, so it must be 3 by 3, and so we have to find $vec{u}_3$. This is where I get stuck; the book says that one method to do this is to use the Gram-Schmidt Process, but I just can't seem to wrap my head around how to do this with the vectors shown above.
linear-algebra orthonormal svd
linear-algebra orthonormal svd
asked May 17 '16 at 20:31
user5368737user5368737
1477
asked May 17 '16 at 20:31
user5368737user5368737
1477
asked May 17 '16 at 20:31
user5368737user5368737
1477
asked May 17 '16 at 20:31
user5368737user5368737
1477
asked May 17 '16 at 20:31
user5368737user5368737
1477
1477
$begingroup$
Choose a vector $x$ linearly independent of $u_1,u_2$ basically at random; one possible choice is $begin{bmatrix} 0 \ 1 \ 0 end{bmatrix}$. Then run Gram-Schmidt on ${ u_1,u_2,x }$.
$endgroup$
– Ian
May 17 '16 at 20:36
add a comment |
$begingroup$
Choose a vector $x$ linearly independent of $u_1,u_2$ basically at random; one possible choice is $begin{bmatrix} 0 \ 1 \ 0 end{bmatrix}$. Then run Gram-Schmidt on ${ u_1,u_2,x }$.
$endgroup$
– Ian
May 17 '16 at 20:36
$begingroup$
Choose a vector $x$ linearly independent of $u_1,u_2$ basically at random; one possible choice is $begin{bmatrix} 0 \ 1 \ 0 end{bmatrix}$. Then run Gram-Schmidt on ${ u_1,u_2,x }$.
$endgroup$
– Ian
May 17 '16 at 20:36
$begingroup$
Choose a vector $x$ linearly independent of $u_1,u_2$ basically at random; one possible choice is $begin{bmatrix} 0 \ 1 \ 0 end{bmatrix}$. Then run Gram-Schmidt on ${ u_1,u_2,x }$.
$endgroup$
– Ian
May 17 '16 at 20:36
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There are a few ways to approach this problem.
Eyeball method
Scrape off the distractions of normalization. The column vectors are
$$
tilde{v}_{1} =
%
left[ begin{array}{c}
2 \ 1 \ 1
end{array} right], qquad
%
tilde{v}_{2} =
%
left[ begin{array}{r}
0 \ -1 \ 1
end{array} right].
%
$$
Find a vector perpendicular to both. One such solution is
$$
tilde{v}_{3} =
%
left[ begin{array}{r}
-1 \ 1 \ 1
end{array} right].
$$
Systematic approach
Start with
$$
mathbf{A} =
left[
begin{array}{cr}
2 & 0 \
1 & -1 \
1 & 1 \
end{array}
right].
$$
Find the nullspace $mathcal{N}left(mathbf{A}^{*} right)$. The row reduced form is
$$
begin{align}
%
mathbf{A}^{T} &mapsto mathbf{E}_{mathbf{A}^{T}} \
%
left[
begin{array}{crc}
2 & 1 & 1 \
0 & -1 & 1 \
end{array}
right]
%
&mapsto
%
left[
begin{array}{ccr}
1 & 0 & 1 \
0 & 1 & -1 \
end{array}
right]
end{align}
$$
In terms of basic variables,
$$
begin{align}
x_{1} &= -x_{3}, \
x_{2} &= x_{3}.
end{align}
$$
Making the natural choice that $x_{3}=1$ produces the column vector
$$
%
left[
begin{array}{r}
x_{1} \
x_{2} \
x_{3}
end{array}
right]
%
=
%
left[
begin{array}{r}
-1 \
1 \
1
end{array}
right]
$$
Gram-Schmidt
Make any choice for the third vector and use the process of Gram and Schmidt to make it an orthogonal vector. A wise choice to begin is
$$
tilde{v}_{3} =
left[ begin{array}{c}
0 \ 0 \ 1
end{array} right]
$$
Why is this a wise choice? It is rich in $0$s, which make manipulation easy.
Define the operator which projects the vector $u$ onto the vector $v$ as
$$
p_{uto v} =
frac{vcdot u}{u cdot u}
u
$$
The Gram-Schmidt process fixes $v_{3}$ using the prescription
$$
v_{GS} = v_{3} -
frac{v_{3} cdot v_{1}} {v_{1} cdot v_{1}} v_{1} -
frac{v_{3} cdot v_{2}} {v_{2} cdot v_{2}} v_{2}
$$
$$
frac{1}{3}
left[ begin{array}{r}
-1 \ 1 \ 1
end{array} right]
%
=
%
left[ begin{array}{c}
0 \ 0 \ 1
end{array} right]
%
-
%
frac{1}{6}
left[ begin{array}{c}
2 \ 1 \ 1
end{array} right]
%
-
%
frac{1}{2}
left[ begin{array}{r}
0 \ -1 \ 1
end{array} right]
%
$$
The normalized form is the column vector you want
$$
frac{1}{sqrt{3}}
left[ begin{array}{r}
-1 \ 1 \ 1
end{array} right]
$$
answered Apr 15 '17 at 20:51
dantopadantopa
6,64942245
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1789474%2fextending-u-1-u-2-to-an-orthonormal-basis-when-finding-an-svd%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There are a few ways to approach this problem.
Eyeball method
Scrape off the distractions of normalization. The column vectors are
$$
tilde{v}_{1} =
%
left[ begin{array}{c}
2 \ 1 \ 1
end{array} right], qquad
%
tilde{v}_{2} =
%
left[ begin{array}{r}
0 \ -1 \ 1
end{array} right].
%
$$
Find a vector perpendicular to both. One such solution is
$$
tilde{v}_{3} =
%
left[ begin{array}{r}
-1 \ 1 \ 1
end{array} right].
$$
Systematic approach
Start with
$$
mathbf{A} =
left[
begin{array}{cr}
2 & 0 \
1 & -1 \
1 & 1 \
end{array}
right].
$$
Find the nullspace $mathcal{N}left(mathbf{A}^{*} right)$. The row reduced form is
$$
begin{align}
%
mathbf{A}^{T} &mapsto mathbf{E}_{mathbf{A}^{T}} \
%
left[
begin{array}{crc}
2 & 1 & 1 \
0 & -1 & 1 \
end{array}
right]
%
&mapsto
%
left[
begin{array}{ccr}
1 & 0 & 1 \
0 & 1 & -1 \
end{array}
right]
end{align}
$$
In terms of basic variables,
$$
begin{align}
x_{1} &= -x_{3}, \
x_{2} &= x_{3}.
end{align}
$$
Making the natural choice that $x_{3}=1$ produces the column vector
$$
%
left[
begin{array}{r}
x_{1} \
x_{2} \
x_{3}
end{array}
right]
%
=
%
left[
begin{array}{r}
-1 \
1 \
1
end{array}
right]
$$
Gram-Schmidt
Make any choice for the third vector and use the process of Gram and Schmidt to make it an orthogonal vector. A wise choice to begin is
$$
tilde{v}_{3} =
left[ begin{array}{c}
0 \ 0 \ 1
end{array} right]
$$
Why is this a wise choice? It is rich in $0$s, which make manipulation easy.
Define the operator which projects the vector $u$ onto the vector $v$ as
$$
p_{uto v} =
frac{vcdot u}{u cdot u}
u
$$
The Gram-Schmidt process fixes $v_{3}$ using the prescription
$$
v_{GS} = v_{3} -
frac{v_{3} cdot v_{1}} {v_{1} cdot v_{1}} v_{1} -
frac{v_{3} cdot v_{2}} {v_{2} cdot v_{2}} v_{2}
$$
$$
frac{1}{3}
left[ begin{array}{r}
-1 \ 1 \ 1
end{array} right]
%
=
%
left[ begin{array}{c}
0 \ 0 \ 1
end{array} right]
%
-
%
frac{1}{6}
left[ begin{array}{c}
2 \ 1 \ 1
end{array} right]
%
-
%
frac{1}{2}
left[ begin{array}{r}
0 \ -1 \ 1
end{array} right]
%
$$
The normalized form is the column vector you want
$$
frac{1}{sqrt{3}}
left[ begin{array}{r}
-1 \ 1 \ 1
end{array} right]
$$
answered Apr 15 '17 at 20:51
dantopadantopa
6,64942245
$endgroup$
add a comment |
$begingroup$
There are a few ways to approach this problem.
Eyeball method
Scrape off the distractions of normalization. The column vectors are
$$
tilde{v}_{1} =
%
left[ begin{array}{c}
2 \ 1 \ 1
end{array} right], qquad
%
tilde{v}_{2} =
%
left[ begin{array}{r}
0 \ -1 \ 1
end{array} right].
%
$$
Find a vector perpendicular to both. One such solution is
$$
tilde{v}_{3} =
%
left[ begin{array}{r}
-1 \ 1 \ 1
end{array} right].
$$
Systematic approach
Start with
$$
mathbf{A} =
left[
begin{array}{cr}
2 & 0 \
1 & -1 \
1 & 1 \
end{array}
right].
$$
Find the nullspace $mathcal{N}left(mathbf{A}^{*} right)$. The row reduced form is
$$
begin{align}
%
mathbf{A}^{T} &mapsto mathbf{E}_{mathbf{A}^{T}} \
%
left[
begin{array}{crc}
2 & 1 & 1 \
0 & -1 & 1 \
end{array}
right]
%
&mapsto
%
left[
begin{array}{ccr}
1 & 0 & 1 \
0 & 1 & -1 \
end{array}
right]
end{align}
$$
In terms of basic variables,
$$
begin{align}
x_{1} &= -x_{3}, \
x_{2} &= x_{3}.
end{align}
$$
Making the natural choice that $x_{3}=1$ produces the column vector
$$
%
left[
begin{array}{r}
x_{1} \
x_{2} \
x_{3}
end{array}
right]
%
=
%
left[
begin{array}{r}
-1 \
1 \
1
end{array}
right]
$$
Gram-Schmidt
Make any choice for the third vector and use the process of Gram and Schmidt to make it an orthogonal vector. A wise choice to begin is
$$
tilde{v}_{3} =
left[ begin{array}{c}
0 \ 0 \ 1
end{array} right]
$$
Why is this a wise choice? It is rich in $0$s, which make manipulation easy.
Define the operator which projects the vector $u$ onto the vector $v$ as
$$
p_{uto v} =
frac{vcdot u}{u cdot u}
u
$$
The Gram-Schmidt process fixes $v_{3}$ using the prescription
$$
v_{GS} = v_{3} -
frac{v_{3} cdot v_{1}} {v_{1} cdot v_{1}} v_{1} -
frac{v_{3} cdot v_{2}} {v_{2} cdot v_{2}} v_{2}
$$
$$
frac{1}{3}
left[ begin{array}{r}
-1 \ 1 \ 1
end{array} right]
%
=
%
left[ begin{array}{c}
0 \ 0 \ 1
end{array} right]
%
-
%
frac{1}{6}
left[ begin{array}{c}
2 \ 1 \ 1
end{array} right]
%
-
%
frac{1}{2}
left[ begin{array}{r}
0 \ -1 \ 1
end{array} right]
%
$$
The normalized form is the column vector you want
$$
frac{1}{sqrt{3}}
left[ begin{array}{r}
-1 \ 1 \ 1
end{array} right]
$$
answered Apr 15 '17 at 20:51
dantopadantopa
6,64942245
$endgroup$
add a comment |
$begingroup$
There are a few ways to approach this problem.
Eyeball method
Scrape off the distractions of normalization. The column vectors are
$$
tilde{v}_{1} =
%
left[ begin{array}{c}
2 \ 1 \ 1
end{array} right], qquad
%
tilde{v}_{2} =
%
left[ begin{array}{r}
0 \ -1 \ 1
end{array} right].
%
$$
Find a vector perpendicular to both. One such solution is
$$
tilde{v}_{3} =
%
left[ begin{array}{r}
-1 \ 1 \ 1
end{array} right].
$$
Systematic approach
Start with
$$
mathbf{A} =
left[
begin{array}{cr}
2 & 0 \
1 & -1 \
1 & 1 \
end{array}
right].
$$
Find the nullspace $mathcal{N}left(mathbf{A}^{*} right)$. The row reduced form is
$$
begin{align}
%
mathbf{A}^{T} &mapsto mathbf{E}_{mathbf{A}^{T}} \
%
left[
begin{array}{crc}
2 & 1 & 1 \
0 & -1 & 1 \
end{array}
right]
%
&mapsto
%
left[
begin{array}{ccr}
1 & 0 & 1 \
0 & 1 & -1 \
end{array}
right]
end{align}
$$
In terms of basic variables,
$$
begin{align}
x_{1} &= -x_{3}, \
x_{2} &= x_{3}.
end{align}
$$
Making the natural choice that $x_{3}=1$ produces the column vector
$$
%
left[
begin{array}{r}
x_{1} \
x_{2} \
x_{3}
end{array}
right]
%
=
%
left[
begin{array}{r}
-1 \
1 \
1
end{array}
right]
$$
Gram-Schmidt
Make any choice for the third vector and use the process of Gram and Schmidt to make it an orthogonal vector. A wise choice to begin is
$$
tilde{v}_{3} =
left[ begin{array}{c}
0 \ 0 \ 1
end{array} right]
$$
Why is this a wise choice? It is rich in $0$s, which make manipulation easy.
Define the operator which projects the vector $u$ onto the vector $v$ as
$$
p_{uto v} =
frac{vcdot u}{u cdot u}
u
$$
The Gram-Schmidt process fixes $v_{3}$ using the prescription
$$
v_{GS} = v_{3} -
frac{v_{3} cdot v_{1}} {v_{1} cdot v_{1}} v_{1} -
frac{v_{3} cdot v_{2}} {v_{2} cdot v_{2}} v_{2}
$$
$$
frac{1}{3}
left[ begin{array}{r}
-1 \ 1 \ 1
end{array} right]
%
=
%
left[ begin{array}{c}
0 \ 0 \ 1
end{array} right]
%
-
%
frac{1}{6}
left[ begin{array}{c}
2 \ 1 \ 1
end{array} right]
%
-
%
frac{1}{2}
left[ begin{array}{r}
0 \ -1 \ 1
end{array} right]
%
$$
The normalized form is the column vector you want
$$
frac{1}{sqrt{3}}
left[ begin{array}{r}
-1 \ 1 \ 1
end{array} right]
$$
answered Apr 15 '17 at 20:51
dantopadantopa
6,64942245
$endgroup$
There are a few ways to approach this problem.
Eyeball method
Scrape off the distractions of normalization. The column vectors are
$$
tilde{v}_{1} =
%
left[ begin{array}{c}
2 \ 1 \ 1
end{array} right], qquad
%
tilde{v}_{2} =
%
left[ begin{array}{r}
0 \ -1 \ 1
end{array} right].
%
$$
Find a vector perpendicular to both. One such solution is
$$
tilde{v}_{3} =
%
left[ begin{array}{r}
-1 \ 1 \ 1
end{array} right].
$$
Systematic approach
Start with
$$
mathbf{A} =
left[
begin{array}{cr}
2 & 0 \
1 & -1 \
1 & 1 \
end{array}
right].
$$
Find the nullspace $mathcal{N}left(mathbf{A}^{*} right)$. The row reduced form is
$$
begin{align}
%
mathbf{A}^{T} &mapsto mathbf{E}_{mathbf{A}^{T}} \
%
left[
begin{array}{crc}
2 & 1 & 1 \
0 & -1 & 1 \
end{array}
right]
%
&mapsto
%
left[
begin{array}{ccr}
1 & 0 & 1 \
0 & 1 & -1 \
end{array}
right]
end{align}
$$
In terms of basic variables,
$$
begin{align}
x_{1} &= -x_{3}, \
x_{2} &= x_{3}.
end{align}
$$
Making the natural choice that $x_{3}=1$ produces the column vector
$$
%
left[
begin{array}{r}
x_{1} \
x_{2} \
x_{3}
end{array}
right]
%
=
%
left[
begin{array}{r}
-1 \
1 \
1
end{array}
right]
$$
Gram-Schmidt
Make any choice for the third vector and use the process of Gram and Schmidt to make it an orthogonal vector. A wise choice to begin is
$$
tilde{v}_{3} =
left[ begin{array}{c}
0 \ 0 \ 1
end{array} right]
$$
Why is this a wise choice? It is rich in $0$s, which make manipulation easy.
Define the operator which projects the vector $u$ onto the vector $v$ as
$$
p_{uto v} =
frac{vcdot u}{u cdot u}
u
$$
The Gram-Schmidt process fixes $v_{3}$ using the prescription
$$
v_{GS} = v_{3} -
frac{v_{3} cdot v_{1}} {v_{1} cdot v_{1}} v_{1} -
frac{v_{3} cdot v_{2}} {v_{2} cdot v_{2}} v_{2}
$$
$$
frac{1}{3}
left[ begin{array}{r}
-1 \ 1 \ 1
end{array} right]
%
=
%
left[ begin{array}{c}
0 \ 0 \ 1
end{array} right]
%
-
%
frac{1}{6}
left[ begin{array}{c}
2 \ 1 \ 1
end{array} right]
%
-
%
frac{1}{2}
left[ begin{array}{r}
0 \ -1 \ 1
end{array} right]
%
$$
The normalized form is the column vector you want
$$
frac{1}{sqrt{3}}
left[ begin{array}{r}
-1 \ 1 \ 1
end{array} right]
$$
answered Apr 15 '17 at 20:51
dantopadantopa
6,64942245
edited Apr 17 '17 at 3:33
answered Apr 15 '17 at 20:51
dantopadantopa
6,64942245
answered Apr 15 '17 at 20:51
dantopadantopa
6,64942245
answered Apr 15 '17 at 20:51
dantopadantopa
6,64942245
6,64942245
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1789474%2fextending-u-1-u-2-to-an-orthonormal-basis-when-finding-an-svd%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Choose a vector $x$ linearly independent of $u_1,u_2$ basically at random; one possible choice is $begin{bmatrix} 0 \ 1 \ 0 end{bmatrix}$. Then run Gram-Schmidt on ${ u_1,u_2,x }$.
$endgroup$
– Ian
May 17 '16 at 20:36