Gram-Schmidt Procedure from “Linear Algebra Done Right” The Next CEO of Stack OverflowUnderstanding the Gram-Schmidt processExtension to the complex numbers for ex. 12 in ch. 6 of Axler's “Linear Algebra Done Right”Help to understand Gram-Schmidt ProofPrerequisites for Linear Algebra Done Right by Sheldon Axler.Using the Gram -schmidt procedure to find the orthonormal set (Linear Algebra)Axioms for vector space in Axler's “Linear Algebra Done Right” - distributivity of scalar multiplication missingGram Schmidt and Inner ProductA Proof for Gram-Schmidt Procedure in Linear Algebra Done RightFormula for the Ratio of Gram DeterminantsConfusing Definition of field in Sheldon Axler's Linear Algebra done right. What does “1+1 is defined to equal 0” in the definition of field mean?
Can you teleport closer to a creature you are Frightened of?
Airship steam engine room - problems and conflict
Are British MPs missing the point, with these 'Indicative Votes'?
Upgrading From a 9 Speed Sora Derailleur?
Does int main() need a declaration on C++?
How to show a landlord what we have in savings?
Raspberry pi 3 B with Ubuntu 18.04 server arm64: what pi version
Can Sri Krishna be called 'a person'?
Can I cast Thunderwave and be at the center of its bottom face, but not be affected by it?
Why does freezing point matter when picking cooler ice packs?
How to find if SQL server backup is encrypted with TDE without restoring the backup
pgfplots: How to draw a tangent graph below two others?
How to compactly explain secondary and tertiary characters without resorting to stereotypes?
Would a grinding machine be a simple and workable propulsion system for an interplanetary spacecraft?
Read/write a pipe-delimited file line by line with some simple text manipulation
Ising model simulation
MT "will strike" & LXX "will watch carefully" (Gen 3:15)?
Can a PhD from a non-TU9 German university become a professor in a TU9 university?
My boss doesn't want me to have a side project
What difference does it make matching a word with/without a trailing whitespace?
Traveling with my 5 year old daughter (as the father) without the mother from Germany to Mexico
Early programmable calculators with RS-232
Car headlights in a world without electricity
Do I need to write [sic] when including a quotation with a number less than 10 that isn't written out?
Gram-Schmidt Procedure from “Linear Algebra Done Right”
The Next CEO of Stack OverflowUnderstanding the Gram-Schmidt processExtension to the complex numbers for ex. 12 in ch. 6 of Axler's “Linear Algebra Done Right”Help to understand Gram-Schmidt ProofPrerequisites for Linear Algebra Done Right by Sheldon Axler.Using the Gram -schmidt procedure to find the orthonormal set (Linear Algebra)Axioms for vector space in Axler's “Linear Algebra Done Right” - distributivity of scalar multiplication missingGram Schmidt and Inner ProductA Proof for Gram-Schmidt Procedure in Linear Algebra Done RightFormula for the Ratio of Gram DeterminantsConfusing Definition of field in Sheldon Axler's Linear Algebra done right. What does “1+1 is defined to equal 0” in the definition of field mean?
$begingroup$
The following content is from "Linear Algebra Done Right" book by Sheldon Axler, 6.31.
There was a part of the proof what I don't understand is that
$beginalign*
leftlangle e_j, e_krightrangle &= leftlanglefracv_j - leftlangle v_j, e_1rightrangle e_1 - ...... - leftlangle v_j, e_j-1rightrangle e_j-1, e_krightrangle\ &= fracleftlangle v_j, e_krightrangle - leftlangle v_j, e_krightrangle \
&= 0
endalign*$
How did he got from the first equation into the second equation?
linear-algebra abstract-algebra proof-verification gram-schmidt
edited Mar 20 at 10:02
Dan Uznanski
7,15821528
asked Mar 18 at 23:35
AlgorisumAlgorisum
457
$endgroup$
add a comment |
$begingroup$
The following content is from "Linear Algebra Done Right" book by Sheldon Axler, 6.31.
There was a part of the proof what I don't understand is that
$beginalign*
leftlangle e_j, e_krightrangle &= leftlanglefracv_j - leftlangle v_j, e_1rightrangle e_1 - ...... - leftlangle v_j, e_j-1rightrangle e_j-1, e_krightrangle\ &= fracleftlangle v_j, e_krightrangle - leftlangle v_j, e_krightrangle \
&= 0
endalign*$
How did he got from the first equation into the second equation?
linear-algebra abstract-algebra proof-verification gram-schmidt
edited Mar 20 at 10:02
Dan Uznanski
7,15821528
asked Mar 18 at 23:35
AlgorisumAlgorisum
457
$endgroup$
$begingroup$
I'm guessing the basis $e_i_i$ is orthonormal? First use the fact that the inner product $langle cdot, cdot rangle$ is bilinear to distribute it to each term in that sum. Then use that $langle e_i, e_j rangle = 0$ if $i neq j$ and $1$ if $i=j$.
$endgroup$
– André 3000
Mar 18 at 23:47
add a comment |
$begingroup$
The following content is from "Linear Algebra Done Right" book by Sheldon Axler, 6.31.
There was a part of the proof what I don't understand is that
$beginalign*
leftlangle e_j, e_krightrangle &= leftlanglefracv_j - leftlangle v_j, e_1rightrangle e_1 - ...... - leftlangle v_j, e_j-1rightrangle e_j-1, e_krightrangle\ &= fracleftlangle v_j, e_krightrangle - leftlangle v_j, e_krightrangle \
&= 0
endalign*$
How did he got from the first equation into the second equation?
linear-algebra abstract-algebra proof-verification gram-schmidt
edited Mar 20 at 10:02
Dan Uznanski
7,15821528
asked Mar 18 at 23:35
AlgorisumAlgorisum
457
$endgroup$
The following content is from "Linear Algebra Done Right" book by Sheldon Axler, 6.31.
There was a part of the proof what I don't understand is that
$beginalign*
leftlangle e_j, e_krightrangle &= leftlanglefracv_j - leftlangle v_j, e_1rightrangle e_1 - ...... - leftlangle v_j, e_j-1rightrangle e_j-1, e_krightrangle\ &= fracleftlangle v_j, e_krightrangle - leftlangle v_j, e_krightrangle \
&= 0
endalign*$
How did he got from the first equation into the second equation?
linear-algebra abstract-algebra proof-verification gram-schmidt
linear-algebra abstract-algebra proof-verification gram-schmidt
edited Mar 20 at 10:02
Dan Uznanski
7,15821528
asked Mar 18 at 23:35
AlgorisumAlgorisum
457
edited Mar 20 at 10:02
Dan Uznanski
7,15821528
asked Mar 18 at 23:35
AlgorisumAlgorisum
457
edited Mar 20 at 10:02
Dan Uznanski
7,15821528
edited Mar 20 at 10:02
Dan Uznanski
7,15821528
edited Mar 20 at 10:02
Dan Uznanski
7,15821528
7,15821528
asked Mar 18 at 23:35
AlgorisumAlgorisum
457
asked Mar 18 at 23:35
AlgorisumAlgorisum
457
asked Mar 18 at 23:35
AlgorisumAlgorisum
457
457
$begingroup$
I'm guessing the basis $e_i_i$ is orthonormal? First use the fact that the inner product $langle cdot, cdot rangle$ is bilinear to distribute it to each term in that sum. Then use that $langle e_i, e_j rangle = 0$ if $i neq j$ and $1$ if $i=j$.
$endgroup$
– André 3000
Mar 18 at 23:47
add a comment |
$begingroup$
I'm guessing the basis $e_i_i$ is orthonormal? First use the fact that the inner product $langle cdot, cdot rangle$ is bilinear to distribute it to each term in that sum. Then use that $langle e_i, e_j rangle = 0$ if $i neq j$ and $1$ if $i=j$.
$endgroup$
– André 3000
Mar 18 at 23:47
$begingroup$
I'm guessing the basis $e_i_i$ is orthonormal? First use the fact that the inner product $langle cdot, cdot rangle$ is bilinear to distribute it to each term in that sum. Then use that $langle e_i, e_j rangle = 0$ if $i neq j$ and $1$ if $i=j$.
$endgroup$
– André 3000
Mar 18 at 23:47
$begingroup$
I'm guessing the basis $e_i_i$ is orthonormal? First use the fact that the inner product $langle cdot, cdot rangle$ is bilinear to distribute it to each term in that sum. Then use that $langle e_i, e_j rangle = 0$ if $i neq j$ and $1$ if $i=j$.
$endgroup$
– André 3000
Mar 18 at 23:47
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The inner product is bi-linear. This basically means that it can distribute across its inputs. For the purpose of my answer I am going to assume you are working in a real vector space; if the space is complex then the modifications necessary are very minor.
The bi-linearity of the inner product means that the following holds.
$$ langle a V_1 + b V_2, c U_1 + d U_2 rangle = ac langle V_1, U_2rangle + ad langle V_2, U_1 rangle + bc langle V_2, U_1 rangle + bd langle V_2, U_2 rangle$$
If your vector space is complex, then some of these coefficients will be complex conjugated.
Lets apply this identity to your problem.
$$langle e_j, e_k rangle = Biglangle fracv_j - langle v_j, e_1 rangle e_1 - dots - langle v_j, e_j-1rangle e_j-1, e_kBigrangle$$
$$= fracBiglangle v_j - langle v_j, e_1 rangle e_1 - dots - langle v_j, e_j-1rangle e_j-1, e_kBigrangle$$
$$= fracBiglangle v_j, e_kBigrangle - langle v_j, e_1 rangle Biglangle e_1, e_kBigrangle - dots - langle v_j, e_krangle Biglangle e_k, e_kBigrangle dots- langle v_j, e_j-1rangle Biglangle e_j-1, e_kBigrangle$$
By construction the basis vectors $e_1 dots e_k dots e_j-1$ are ortho-normal. This means that $langle e_i, e_k rangle = 0$ when $i neq k$ and $langle e_k, e_k rangle = 1$ for $1leq i leq j-1$.
$$= fracBiglangle v_j, e_kBigrangle - langle v_j, e_1 rangle 0 - dots - langle v_j, e_krangle 1 dots- langle v_j, e_j-1rangle 0$$
$$= fracBiglangle v_j, e_kBigrangle - 0 - dots - langle v_j, e_krangle dots- 0$$
$$= fracBiglangle v_j, e_kBigrangle - langle v_j, e_krangle $$
$$= frac0 $$
$$= 0$$
answered Mar 19 at 0:31
SpencerSpencer
8,67012056
$endgroup$
add a comment |
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%2f3153458%2fgram-schmidt-procedure-from-linear-algebra-done-right%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
$begingroup$
The inner product is bi-linear. This basically means that it can distribute across its inputs. For the purpose of my answer I am going to assume you are working in a real vector space; if the space is complex then the modifications necessary are very minor.
The bi-linearity of the inner product means that the following holds.
$$ langle a V_1 + b V_2, c U_1 + d U_2 rangle = ac langle V_1, U_2rangle + ad langle V_2, U_1 rangle + bc langle V_2, U_1 rangle + bd langle V_2, U_2 rangle$$
If your vector space is complex, then some of these coefficients will be complex conjugated.
Lets apply this identity to your problem.
$$langle e_j, e_k rangle = Biglangle fracv_j - langle v_j, e_1 rangle e_1 - dots - langle v_j, e_j-1rangle e_j-1, e_kBigrangle$$
$$= fracBiglangle v_j - langle v_j, e_1 rangle e_1 - dots - langle v_j, e_j-1rangle e_j-1, e_kBigrangle$$
$$= fracBiglangle v_j, e_kBigrangle - langle v_j, e_1 rangle Biglangle e_1, e_kBigrangle - dots - langle v_j, e_krangle Biglangle e_k, e_kBigrangle dots- langle v_j, e_j-1rangle Biglangle e_j-1, e_kBigrangle$$
By construction the basis vectors $e_1 dots e_k dots e_j-1$ are ortho-normal. This means that $langle e_i, e_k rangle = 0$ when $i neq k$ and $langle e_k, e_k rangle = 1$ for $1leq i leq j-1$.
$$= fracBiglangle v_j, e_kBigrangle - langle v_j, e_1 rangle 0 - dots - langle v_j, e_krangle 1 dots- langle v_j, e_j-1rangle 0$$
$$= fracBiglangle v_j, e_kBigrangle - 0 - dots - langle v_j, e_krangle dots- 0$$
$$= fracBiglangle v_j, e_kBigrangle - langle v_j, e_krangle $$
$$= frac0 $$
$$= 0$$
answered Mar 19 at 0:31
SpencerSpencer
8,67012056
$endgroup$
add a comment |
$begingroup$
The inner product is bi-linear. This basically means that it can distribute across its inputs. For the purpose of my answer I am going to assume you are working in a real vector space; if the space is complex then the modifications necessary are very minor.
The bi-linearity of the inner product means that the following holds.
$$ langle a V_1 + b V_2, c U_1 + d U_2 rangle = ac langle V_1, U_2rangle + ad langle V_2, U_1 rangle + bc langle V_2, U_1 rangle + bd langle V_2, U_2 rangle$$
If your vector space is complex, then some of these coefficients will be complex conjugated.
Lets apply this identity to your problem.
$$langle e_j, e_k rangle = Biglangle fracv_j - langle v_j, e_1 rangle e_1 - dots - langle v_j, e_j-1rangle e_j-1, e_kBigrangle$$
$$= fracBiglangle v_j - langle v_j, e_1 rangle e_1 - dots - langle v_j, e_j-1rangle e_j-1, e_kBigrangle$$
$$= fracBiglangle v_j, e_kBigrangle - langle v_j, e_1 rangle Biglangle e_1, e_kBigrangle - dots - langle v_j, e_krangle Biglangle e_k, e_kBigrangle dots- langle v_j, e_j-1rangle Biglangle e_j-1, e_kBigrangle$$
By construction the basis vectors $e_1 dots e_k dots e_j-1$ are ortho-normal. This means that $langle e_i, e_k rangle = 0$ when $i neq k$ and $langle e_k, e_k rangle = 1$ for $1leq i leq j-1$.
$$= fracBiglangle v_j, e_kBigrangle - langle v_j, e_1 rangle 0 - dots - langle v_j, e_krangle 1 dots- langle v_j, e_j-1rangle 0$$
$$= fracBiglangle v_j, e_kBigrangle - 0 - dots - langle v_j, e_krangle dots- 0$$
$$= fracBiglangle v_j, e_kBigrangle - langle v_j, e_krangle $$
$$= frac0 $$
$$= 0$$
answered Mar 19 at 0:31
SpencerSpencer
8,67012056
$endgroup$
add a comment |
$begingroup$
The inner product is bi-linear. This basically means that it can distribute across its inputs. For the purpose of my answer I am going to assume you are working in a real vector space; if the space is complex then the modifications necessary are very minor.
The bi-linearity of the inner product means that the following holds.
$$ langle a V_1 + b V_2, c U_1 + d U_2 rangle = ac langle V_1, U_2rangle + ad langle V_2, U_1 rangle + bc langle V_2, U_1 rangle + bd langle V_2, U_2 rangle$$
If your vector space is complex, then some of these coefficients will be complex conjugated.
Lets apply this identity to your problem.
$$langle e_j, e_k rangle = Biglangle fracv_j - langle v_j, e_1 rangle e_1 - dots - langle v_j, e_j-1rangle e_j-1, e_kBigrangle$$
$$= fracBiglangle v_j - langle v_j, e_1 rangle e_1 - dots - langle v_j, e_j-1rangle e_j-1, e_kBigrangle$$
$$= fracBiglangle v_j, e_kBigrangle - langle v_j, e_1 rangle Biglangle e_1, e_kBigrangle - dots - langle v_j, e_krangle Biglangle e_k, e_kBigrangle dots- langle v_j, e_j-1rangle Biglangle e_j-1, e_kBigrangle$$
By construction the basis vectors $e_1 dots e_k dots e_j-1$ are ortho-normal. This means that $langle e_i, e_k rangle = 0$ when $i neq k$ and $langle e_k, e_k rangle = 1$ for $1leq i leq j-1$.
$$= fracBiglangle v_j, e_kBigrangle - langle v_j, e_1 rangle 0 - dots - langle v_j, e_krangle 1 dots- langle v_j, e_j-1rangle 0$$
$$= fracBiglangle v_j, e_kBigrangle - 0 - dots - langle v_j, e_krangle dots- 0$$
$$= fracBiglangle v_j, e_kBigrangle - langle v_j, e_krangle $$
$$= frac0 $$
$$= 0$$
answered Mar 19 at 0:31
SpencerSpencer
8,67012056
$endgroup$
The inner product is bi-linear. This basically means that it can distribute across its inputs. For the purpose of my answer I am going to assume you are working in a real vector space; if the space is complex then the modifications necessary are very minor.
The bi-linearity of the inner product means that the following holds.
$$ langle a V_1 + b V_2, c U_1 + d U_2 rangle = ac langle V_1, U_2rangle + ad langle V_2, U_1 rangle + bc langle V_2, U_1 rangle + bd langle V_2, U_2 rangle$$
If your vector space is complex, then some of these coefficients will be complex conjugated.
Lets apply this identity to your problem.
$$langle e_j, e_k rangle = Biglangle fracv_j - langle v_j, e_1 rangle e_1 - dots - langle v_j, e_j-1rangle e_j-1, e_kBigrangle$$
$$= fracBiglangle v_j - langle v_j, e_1 rangle e_1 - dots - langle v_j, e_j-1rangle e_j-1, e_kBigrangle$$
$$= fracBiglangle v_j, e_kBigrangle - langle v_j, e_1 rangle Biglangle e_1, e_kBigrangle - dots - langle v_j, e_krangle Biglangle e_k, e_kBigrangle dots- langle v_j, e_j-1rangle Biglangle e_j-1, e_kBigrangle$$
By construction the basis vectors $e_1 dots e_k dots e_j-1$ are ortho-normal. This means that $langle e_i, e_k rangle = 0$ when $i neq k$ and $langle e_k, e_k rangle = 1$ for $1leq i leq j-1$.
$$= fracBiglangle v_j, e_kBigrangle - langle v_j, e_1 rangle 0 - dots - langle v_j, e_krangle 1 dots- langle v_j, e_j-1rangle 0$$
$$= fracBiglangle v_j, e_kBigrangle - 0 - dots - langle v_j, e_krangle dots- 0$$
$$= fracBiglangle v_j, e_kBigrangle - langle v_j, e_krangle $$
$$= frac0 $$
$$= 0$$
answered Mar 19 at 0:31
SpencerSpencer
8,67012056
edited Mar 19 at 2:52
answered Mar 19 at 0:31
SpencerSpencer
8,67012056
answered Mar 19 at 0:31
SpencerSpencer
8,67012056
answered Mar 19 at 0:31
SpencerSpencer
8,67012056
8,67012056
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%2f3153458%2fgram-schmidt-procedure-from-linear-algebra-done-right%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
$begingroup$
I'm guessing the basis $e_i_i$ is orthonormal? First use the fact that the inner product $langle cdot, cdot rangle$ is bilinear to distribute it to each term in that sum. Then use that $langle e_i, e_j rangle = 0$ if $i neq j$ and $1$ if $i=j$.
$endgroup$
– André 3000
Mar 18 at 23:47