$V=U_1oplus …oplus U_n iff dim(V)=dim(U_1)+…+dim(U_n)$ and $V=U_1+…+U_n$.

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











up vote
1
down vote

favorite












I want to prove that $$textif dim(V)=dim(U_1)+dots+dim(U_n)text and V=U_1+dots+U_ntext, then V=U_1oplusdotsoplus U_n $$



If $n=1$, it's obvious. Set $W=U_1+dots+U_n-1$ and suppose $dim(W)=dim(U_1)+dots+dim(U_n-1)$. By the induction hypothesis, we have that $W=U_1oplusdotsoplus U_n-1$. Let's prove that $V=Woplus U_n$, given that we have that $V=W+U_n$ and $dim(V)=dim(W)+dim(U_n)$. Therefore $$dim(Wcap U_n)=dim(W)+dim(U_n)-dim(W+U_n)=dim(V)-dim(V)=0.$$
Therefore, $$V=Woplus U_n=U_1oplusdotsoplus U_n.$$



Does it work ? My big doubt is about the fact that I implicitly used $$(Aoplus B)oplus C=Aoplus Boplus C,$$
and I'm not really sure if it really holds.




linear-algebra




share|cite|improve this question





















  • Of course it works. The best way to be sure is to prove $(Aoplus B)oplus C=Aoplus Boplus C$ (and it's really not complicated).
    – Surb
    Aug 7 at 17:31











  • In order to proceed by induction you need to be able to prove that $$[dim(V)=dim(U_1)+...+dim(U_n)quad textandquad V=U_1+...+U_n]implies [dim(U_1+...+U_n-1)=dim(U_1)+...+dim(U_n-1)]$$
    – Arnaud Mortier
    Aug 7 at 17:37














up vote
1
down vote

favorite












I want to prove that $$textif dim(V)=dim(U_1)+dots+dim(U_n)text and V=U_1+dots+U_ntext, then V=U_1oplusdotsoplus U_n $$



If $n=1$, it's obvious. Set $W=U_1+dots+U_n-1$ and suppose $dim(W)=dim(U_1)+dots+dim(U_n-1)$. By the induction hypothesis, we have that $W=U_1oplusdotsoplus U_n-1$. Let's prove that $V=Woplus U_n$, given that we have that $V=W+U_n$ and $dim(V)=dim(W)+dim(U_n)$. Therefore $$dim(Wcap U_n)=dim(W)+dim(U_n)-dim(W+U_n)=dim(V)-dim(V)=0.$$
Therefore, $$V=Woplus U_n=U_1oplusdotsoplus U_n.$$



Does it work ? My big doubt is about the fact that I implicitly used $$(Aoplus B)oplus C=Aoplus Boplus C,$$
and I'm not really sure if it really holds.




linear-algebra




share|cite|improve this question





















  • Of course it works. The best way to be sure is to prove $(Aoplus B)oplus C=Aoplus Boplus C$ (and it's really not complicated).
    – Surb
    Aug 7 at 17:31











  • In order to proceed by induction you need to be able to prove that $$[dim(V)=dim(U_1)+...+dim(U_n)quad textandquad V=U_1+...+U_n]implies [dim(U_1+...+U_n-1)=dim(U_1)+...+dim(U_n-1)]$$
    – Arnaud Mortier
    Aug 7 at 17:37












up vote
1
down vote

favorite









up vote
1
down vote

favorite











I want to prove that $$textif dim(V)=dim(U_1)+dots+dim(U_n)text and V=U_1+dots+U_ntext, then V=U_1oplusdotsoplus U_n $$



If $n=1$, it's obvious. Set $W=U_1+dots+U_n-1$ and suppose $dim(W)=dim(U_1)+dots+dim(U_n-1)$. By the induction hypothesis, we have that $W=U_1oplusdotsoplus U_n-1$. Let's prove that $V=Woplus U_n$, given that we have that $V=W+U_n$ and $dim(V)=dim(W)+dim(U_n)$. Therefore $$dim(Wcap U_n)=dim(W)+dim(U_n)-dim(W+U_n)=dim(V)-dim(V)=0.$$
Therefore, $$V=Woplus U_n=U_1oplusdotsoplus U_n.$$



Does it work ? My big doubt is about the fact that I implicitly used $$(Aoplus B)oplus C=Aoplus Boplus C,$$
and I'm not really sure if it really holds.




linear-algebra




share|cite|improve this question













I want to prove that $$textif dim(V)=dim(U_1)+dots+dim(U_n)text and V=U_1+dots+U_ntext, then V=U_1oplusdotsoplus U_n $$



If $n=1$, it's obvious. Set $W=U_1+dots+U_n-1$ and suppose $dim(W)=dim(U_1)+dots+dim(U_n-1)$. By the induction hypothesis, we have that $W=U_1oplusdotsoplus U_n-1$. Let's prove that $V=Woplus U_n$, given that we have that $V=W+U_n$ and $dim(V)=dim(W)+dim(U_n)$. Therefore $$dim(Wcap U_n)=dim(W)+dim(U_n)-dim(W+U_n)=dim(V)-dim(V)=0.$$
Therefore, $$V=Woplus U_n=U_1oplusdotsoplus U_n.$$



Does it work ? My big doubt is about the fact that I implicitly used $$(Aoplus B)oplus C=Aoplus Boplus C,$$
and I'm not really sure if it really holds.





linear-algebra





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Aug 7 at 17:46









zzuussee

1,639420




1,639420









asked Aug 7 at 17:29









Henri

404




404











  • Of course it works. The best way to be sure is to prove $(Aoplus B)oplus C=Aoplus Boplus C$ (and it's really not complicated).
    – Surb
    Aug 7 at 17:31











  • In order to proceed by induction you need to be able to prove that $$[dim(V)=dim(U_1)+...+dim(U_n)quad textandquad V=U_1+...+U_n]implies [dim(U_1+...+U_n-1)=dim(U_1)+...+dim(U_n-1)]$$
    – Arnaud Mortier
    Aug 7 at 17:37
















  • Of course it works. The best way to be sure is to prove $(Aoplus B)oplus C=Aoplus Boplus C$ (and it's really not complicated).
    – Surb
    Aug 7 at 17:31











  • In order to proceed by induction you need to be able to prove that $$[dim(V)=dim(U_1)+...+dim(U_n)quad textandquad V=U_1+...+U_n]implies [dim(U_1+...+U_n-1)=dim(U_1)+...+dim(U_n-1)]$$
    – Arnaud Mortier
    Aug 7 at 17:37















Of course it works. The best way to be sure is to prove $(Aoplus B)oplus C=Aoplus Boplus C$ (and it's really not complicated).
– Surb
Aug 7 at 17:31





Of course it works. The best way to be sure is to prove $(Aoplus B)oplus C=Aoplus Boplus C$ (and it's really not complicated).
– Surb
Aug 7 at 17:31













In order to proceed by induction you need to be able to prove that $$[dim(V)=dim(U_1)+...+dim(U_n)quad textandquad V=U_1+...+U_n]implies [dim(U_1+...+U_n-1)=dim(U_1)+...+dim(U_n-1)]$$
– Arnaud Mortier
Aug 7 at 17:37




In order to proceed by induction you need to be able to prove that $$[dim(V)=dim(U_1)+...+dim(U_n)quad textandquad V=U_1+...+U_n]implies [dim(U_1+...+U_n-1)=dim(U_1)+...+dim(U_n-1)]$$
– Arnaud Mortier
Aug 7 at 17:37















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%2f2875184%2fv-u-1-oplus-oplus-u-n-iff-dimv-dimu-1-dimu-n-and-v-u-1%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%2f2875184%2fv-u-1-oplus-oplus-u-n-iff-dimv-dimu-1-dimu-n-and-v-u-1%23new-answer', 'question_page');

);

Post as a guest
































































-

這個網誌中的熱門文章

How to combine Bézier curves to a surface?

圖片
Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite My aim is to smooth the terrain in a video game. Therefore I contrived an algorithm that makes use of Bézier curves of different orders. But this algorithm is defined in a two dimensional space for now. To shift it into the third dimension I need to somehow combine the Bézier curves. Given two curves in each two dimensions, how can I combine them to build a surface? I thought about something like a curve of a curve. Or maybe I could multiply them somehow. How can I combine them to cause the wanted behavior? Is there a common approach? In the image you can see the input, on the left, and the output, on the right, of the algorithm that works in a two dimensional space. For the real application there is a three dimensional input. The algorithm relays only on the adjacent tiles. Given these it will define the control points of the mentioned Bézier cur...
閱讀完整內容

Mutual Information Always Non-negative

圖片
Clash Royale CLAN TAG #URR8PPP up vote 10 down vote favorite 6 What is the simplest proof that mutual information is always non-negative? i.e., $I(X;Y)ge0$ probability-theory share | cite | improve this question edited Jun 17 '12 at 15:38 Cameron Buie 83.6k 7 71 154 asked Jun 17 '12 at 15:30 Omri 359 3 13 3 Convexity of the function $tmapsto tlog t$. – Did Jun 17 '12 at 15:33 In addition, the convexity properties require the coefficients in the linear combination sum 1. Then, as p(x,y) is a probability distribution, it fullfits such condition. – Francisco Javier Delgado Ceped Aug 10 at 18:44 add a comment  |  up vote 10 down vote favorite 6 What is the simplest proof that mutual information is always non-negative? i.e., $I(X;Y)ge0$ probability-theory share | cite | improve this question edited Jun 17 ...
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