Is $Av_1,Av_2,Av_3$ orthogonal if you have eigenvector of $A^TA$

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
3
down vote

favorite
1












Let $Ain M_3(mathbb R)$ and if $v_1,v_2,v_3$ orthonormed eigenvectors of matrix $A^TA$ and which eigenvalues is $1,2,3$ then vectors $Av_1,Av_2,Av_3$ is orthogonal?



I only know that we need to prove that $(Av_1,Av_2)=0$ and $(Av_1,Av_3)=0$ but I write that $(Av_1,Av_2)=(Av_1)^TAv_2=v_1^TA^TAv_2=2v_1^Tv_2=2(v_1,v_2)=0$ but I do not know is this prove ok?






linear-algebra orthogonality orthonormal orthogonal-matrices







share|cite|improve this question

















  • 2




    Yes, your proof is fine.
    – Kavi Rama Murthy
    Sep 4 at 7:55














up vote
3
down vote

favorite
1












Let $Ain M_3(mathbb R)$ and if $v_1,v_2,v_3$ orthonormed eigenvectors of matrix $A^TA$ and which eigenvalues is $1,2,3$ then vectors $Av_1,Av_2,Av_3$ is orthogonal?



I only know that we need to prove that $(Av_1,Av_2)=0$ and $(Av_1,Av_3)=0$ but I write that $(Av_1,Av_2)=(Av_1)^TAv_2=v_1^TA^TAv_2=2v_1^Tv_2=2(v_1,v_2)=0$ but I do not know is this prove ok?






linear-algebra orthogonality orthonormal orthogonal-matrices







share|cite|improve this question

















  • 2




    Yes, your proof is fine.
    – Kavi Rama Murthy
    Sep 4 at 7:55












up vote
3
down vote

favorite
1









up vote
3
down vote

favorite
1






1





Let $Ain M_3(mathbb R)$ and if $v_1,v_2,v_3$ orthonormed eigenvectors of matrix $A^TA$ and which eigenvalues is $1,2,3$ then vectors $Av_1,Av_2,Av_3$ is orthogonal?



I only know that we need to prove that $(Av_1,Av_2)=0$ and $(Av_1,Av_3)=0$ but I write that $(Av_1,Av_2)=(Av_1)^TAv_2=v_1^TA^TAv_2=2v_1^Tv_2=2(v_1,v_2)=0$ but I do not know is this prove ok?






linear-algebra orthogonality orthonormal orthogonal-matrices







share|cite|improve this question













Let $Ain M_3(mathbb R)$ and if $v_1,v_2,v_3$ orthonormed eigenvectors of matrix $A^TA$ and which eigenvalues is $1,2,3$ then vectors $Av_1,Av_2,Av_3$ is orthogonal?



I only know that we need to prove that $(Av_1,Av_2)=0$ and $(Av_1,Av_3)=0$ but I write that $(Av_1,Av_2)=(Av_1)^TAv_2=v_1^TA^TAv_2=2v_1^Tv_2=2(v_1,v_2)=0$ but I do not know is this prove ok?






linear-algebra orthogonality orthonormal orthogonal-matrices




linear-algebra orthogonality orthonormal orthogonal-matrices






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Sep 4 at 7:50









Marko Škorić

3908




3908







  • 2




    Yes, your proof is fine.
    – Kavi Rama Murthy
    Sep 4 at 7:55












  • 2




    Yes, your proof is fine.
    – Kavi Rama Murthy
    Sep 4 at 7:55







2




2




Yes, your proof is fine.
– Kavi Rama Murthy
Sep 4 at 7:55




Yes, your proof is fine.
– Kavi Rama Murthy
Sep 4 at 7:55















active

oldest

votes











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',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);













 

draft saved


draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2904759%2fis-av-1-av-2-av-3-orthogonal-if-you-have-eigenvector-of-ata%23new-answer', 'question_page');

);

Post as a guest






















active

oldest

votes













active

oldest

votes









active

oldest

votes






active

oldest

votes















 

draft saved


draft discarded















































 


draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2904759%2fis-av-1-av-2-av-3-orthogonal-if-you-have-eigenvector-of-ata%23new-answer', 'question_page');

);

Post as a guest
































































-

這個網誌中的熱門文章

Is there any way to eliminate the singular point to solve this integral by hand or by approximations?

圖片
Clash Royale CLAN TAG #URR8PPP up vote 0 down vote favorite 1 $$int_0^1 (1+(x^-p-1)^1-p)^frac1pdx, pin mathbbR, pge 1$$ I find that $x=0,1$ are singular points(discontinuous points of this integral), is there a way to eliminate those singular points to make sure that this integral can be calculated by hand or by Taylor approximations. In either cases, what are those steps of calculating the integral? calculus real-analysis analysis approximation share | cite | improve this question edited Sep 11 at 5:07 asked Sep 11 at 4:48 Mclalalala 126 8 1 What is $p$? Please give more information to make it complete. – xbh Sep 11 at 5:02 @xbh already fixed it, thanks for reminding. – Mclalalala Sep 11 at 5:10 Seems that $0$ is not singular here… [I might compute wrong] – xbh Sep 11 at 5:30 @xbh can you show me how to get the close formula for the integral above? – Mclalalala Sep 11 at 5:33
閱讀完整內容

Why am i infinitely getting the same tweet with the Twitter Search API?

圖片
up vote 0 down vote favorite My code workes perfectly for other queries, but for one query I am infinitely getting the same tweet. I can't understand what the problem is. API call: query='Allergic asthma OR Nonallergic asthma OR Occupational asthma OR EIB OR Exercise-induced bronchoconstriction OR Nocturnal asthma OR Cough-variant asthma' new_tweets = api.search(q=searchQuery, count=100, since_id=since_id, lang='en',tweet_mode='extended') if not new_tweets: print("No more tweets found") break for tweet in new_tweets: if True: print(jsonpickle.encode(tweet._json, unpicklable=False)) my_file = open('searchTweets.txt','a') my_file.write(jsonpickle.encode(tweet._json, unpicklable=False)+'n') my_file.close() else: continue Output in file: https://pastebin.com/JmRCsTeE These are the JSONs of the same Tweet. Other queries that work: Breast cancer OR Prostate cancer OR Colon cancer OR Lung cancer Type 2 dia
閱讀完整內容

Carbon dioxide

圖片
"CO2" redirects here. For other uses, see CO2 (disambiguation). Carbon dioxide Names Other names Carbonic acid gas Carbonic anhydride Carbonic oxide Carbon oxide Carbon(IV) oxide Dry ice (solid phase) Identifiers CAS Number 124-38-9   Y 3D model (JSmol) Interactive image Interactive image 3DMet B01131 Beilstein Reference 1900390 ChEBI CHEBI:16526   Y ChEMBL ChEMBL1231871   N ChemSpider 274   Y ECHA InfoCard 100.004.271 EC Number 204-696-9 E number E290 (preservatives) Gmelin Reference 989 KEGG D00004   Y MeSH Carbon+dioxide PubChem CID 280 RTECS number FF6400000 UNII 142M471B3J   Y UN number 1013 (gas), 1845 (solid) InChI InChI=1S/CO2/c2-1-3   Y Key: CURLTUGMZLYLDI-UHFFFAOYSA-N   Y InChI=1/CO2/c2-1-3 Key: CURLTUGMZLYLDI-UHFFFAOYAO SMILES O=C=O C(=O)=O Properties Chemical formula C O 2 Molar mass 44.01 g·mol −1 Appearance Colorless gas Odor Low concentrations: none High concentrations: sharp; acidic [1] Density 1562 kg/m 3 (solid at 1 atm and −78.5 °C) 1101 kg/m 3 (liquid at
閱讀完整內容