Find orthogonal projection of $ [n,0,0,…,0]^T$ on subspace $V$












0














$n>1$ Given is $$V = left{ vec{x} in mathbb R^n : x_1+x_2 + ... + x_n = 0 right} $$
a) Find orthogonal basis of $V^{perp} $

b) Find orthogonal projection $vec{x} = [n,0,0,...,0]^T$ on subspace $V$



If it comes to a)
$$dim V^{perp} = n - dim V = dim V^{perp} = n - n + 1 = 1$$ So $V^{perp} = span$ one_vector_perpendicular_to_v


Put $[1,1,1,...,1,1]^T$ - it is perpendicular to $V$

Let's start Gram–Schmidt process - but we have $1$ vector so $u_1 = [1,1,1,...,1,1]^T$ = orthogonal basis of $V^{perp}$



b) It seems to be very interesting and hard. I found basis of $V$:
$$[-1,1,0,0,...,0] = vec{v_1}$$
$$[-1,0,1,0,...,0] = vec{v_2}$$
$$[-1,0,0,1,...,0] = vec{v_3}$$
$$...$$
$$[-1,0,0,0,...,1] = vec{v_{n-1}}$$



Now I start Gram–Schmidt process
$$u_1 = v_1 $$
$$u_2 = v_2 - frac{1}{2} cdot v_1$$
$$u_ 3 = v_3 - frac{1}{4} cdot v_2 + frac{1}{8} cdot v_1 $$
$$u_4 = v_4 - frac{7}{16} cdot v_3 + frac{7}{64} cdot v_2 - frac{7}{64} cdot v_1$$
I don't even know if I don't take mistake.

Moreover the calculations getting harder and harder and I still don't see any regular sequence in it. Can somebody help me with this task?






linear-algebra orthogonality







share|cite|improve this question






















  • Why do you need an orthogonal basis of $V$? One can find the projection if one follows the vector $(1 1 ldots 1)$ until the line meets $V$.
    – A.Γ.
    yesterday










  • I need orthogonal basis because I have that formula for orthogonal projection: $ P_z(x) = sum_{j=1}^k <z_j,x>z_j $ where $z_1,...,z_k $ is orthogonal basis and I don't have any other idea how to do this task @A.Γ.
    – VirtualUser
    yesterday












  • You can save yourself a lot of work by taking advantage of the fact that the orthogonal projection onto a subspace is what’s left after subtracting the orthogonal projection onto its complement.
    – amd
    yesterday
















0














$n>1$ Given is $$V = left{ vec{x} in mathbb R^n : x_1+x_2 + ... + x_n = 0 right} $$
a) Find orthogonal basis of $V^{perp} $

b) Find orthogonal projection $vec{x} = [n,0,0,...,0]^T$ on subspace $V$



If it comes to a)
$$dim V^{perp} = n - dim V = dim V^{perp} = n - n + 1 = 1$$ So $V^{perp} = span$ one_vector_perpendicular_to_v


Put $[1,1,1,...,1,1]^T$ - it is perpendicular to $V$

Let's start Gram–Schmidt process - but we have $1$ vector so $u_1 = [1,1,1,...,1,1]^T$ = orthogonal basis of $V^{perp}$



b) It seems to be very interesting and hard. I found basis of $V$:
$$[-1,1,0,0,...,0] = vec{v_1}$$
$$[-1,0,1,0,...,0] = vec{v_2}$$
$$[-1,0,0,1,...,0] = vec{v_3}$$
$$...$$
$$[-1,0,0,0,...,1] = vec{v_{n-1}}$$



Now I start Gram–Schmidt process
$$u_1 = v_1 $$
$$u_2 = v_2 - frac{1}{2} cdot v_1$$
$$u_ 3 = v_3 - frac{1}{4} cdot v_2 + frac{1}{8} cdot v_1 $$
$$u_4 = v_4 - frac{7}{16} cdot v_3 + frac{7}{64} cdot v_2 - frac{7}{64} cdot v_1$$
I don't even know if I don't take mistake.

Moreover the calculations getting harder and harder and I still don't see any regular sequence in it. Can somebody help me with this task?






linear-algebra orthogonality







share|cite|improve this question






















  • Why do you need an orthogonal basis of $V$? One can find the projection if one follows the vector $(1 1 ldots 1)$ until the line meets $V$.
    – A.Γ.
    yesterday










  • I need orthogonal basis because I have that formula for orthogonal projection: $ P_z(x) = sum_{j=1}^k <z_j,x>z_j $ where $z_1,...,z_k $ is orthogonal basis and I don't have any other idea how to do this task @A.Γ.
    – VirtualUser
    yesterday












  • You can save yourself a lot of work by taking advantage of the fact that the orthogonal projection onto a subspace is what’s left after subtracting the orthogonal projection onto its complement.
    – amd
    yesterday














0












0








0







$n>1$ Given is $$V = left{ vec{x} in mathbb R^n : x_1+x_2 + ... + x_n = 0 right} $$
a) Find orthogonal basis of $V^{perp} $

b) Find orthogonal projection $vec{x} = [n,0,0,...,0]^T$ on subspace $V$



If it comes to a)
$$dim V^{perp} = n - dim V = dim V^{perp} = n - n + 1 = 1$$ So $V^{perp} = span$ one_vector_perpendicular_to_v


Put $[1,1,1,...,1,1]^T$ - it is perpendicular to $V$

Let's start Gram–Schmidt process - but we have $1$ vector so $u_1 = [1,1,1,...,1,1]^T$ = orthogonal basis of $V^{perp}$



b) It seems to be very interesting and hard. I found basis of $V$:
$$[-1,1,0,0,...,0] = vec{v_1}$$
$$[-1,0,1,0,...,0] = vec{v_2}$$
$$[-1,0,0,1,...,0] = vec{v_3}$$
$$...$$
$$[-1,0,0,0,...,1] = vec{v_{n-1}}$$



Now I start Gram–Schmidt process
$$u_1 = v_1 $$
$$u_2 = v_2 - frac{1}{2} cdot v_1$$
$$u_ 3 = v_3 - frac{1}{4} cdot v_2 + frac{1}{8} cdot v_1 $$
$$u_4 = v_4 - frac{7}{16} cdot v_3 + frac{7}{64} cdot v_2 - frac{7}{64} cdot v_1$$
I don't even know if I don't take mistake.

Moreover the calculations getting harder and harder and I still don't see any regular sequence in it. Can somebody help me with this task?






linear-algebra orthogonality







share|cite|improve this question













$n>1$ Given is $$V = left{ vec{x} in mathbb R^n : x_1+x_2 + ... + x_n = 0 right} $$
a) Find orthogonal basis of $V^{perp} $

b) Find orthogonal projection $vec{x} = [n,0,0,...,0]^T$ on subspace $V$



If it comes to a)
$$dim V^{perp} = n - dim V = dim V^{perp} = n - n + 1 = 1$$ So $V^{perp} = span$ one_vector_perpendicular_to_v


Put $[1,1,1,...,1,1]^T$ - it is perpendicular to $V$

Let's start Gram–Schmidt process - but we have $1$ vector so $u_1 = [1,1,1,...,1,1]^T$ = orthogonal basis of $V^{perp}$



b) It seems to be very interesting and hard. I found basis of $V$:
$$[-1,1,0,0,...,0] = vec{v_1}$$
$$[-1,0,1,0,...,0] = vec{v_2}$$
$$[-1,0,0,1,...,0] = vec{v_3}$$
$$...$$
$$[-1,0,0,0,...,1] = vec{v_{n-1}}$$



Now I start Gram–Schmidt process
$$u_1 = v_1 $$
$$u_2 = v_2 - frac{1}{2} cdot v_1$$
$$u_ 3 = v_3 - frac{1}{4} cdot v_2 + frac{1}{8} cdot v_1 $$
$$u_4 = v_4 - frac{7}{16} cdot v_3 + frac{7}{64} cdot v_2 - frac{7}{64} cdot v_1$$
I don't even know if I don't take mistake.

Moreover the calculations getting harder and harder and I still don't see any regular sequence in it. Can somebody help me with this task?






linear-algebra orthogonality




linear-algebra orthogonality






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked yesterday









VirtualUser

44711




44711












  • Why do you need an orthogonal basis of $V$? One can find the projection if one follows the vector $(1 1 ldots 1)$ until the line meets $V$.
    – A.Γ.
    yesterday










  • I need orthogonal basis because I have that formula for orthogonal projection: $ P_z(x) = sum_{j=1}^k <z_j,x>z_j $ where $z_1,...,z_k $ is orthogonal basis and I don't have any other idea how to do this task @A.Γ.
    – VirtualUser
    yesterday












  • You can save yourself a lot of work by taking advantage of the fact that the orthogonal projection onto a subspace is what’s left after subtracting the orthogonal projection onto its complement.
    – amd
    yesterday


















  • Why do you need an orthogonal basis of $V$? One can find the projection if one follows the vector $(1 1 ldots 1)$ until the line meets $V$.
    – A.Γ.
    yesterday










  • I need orthogonal basis because I have that formula for orthogonal projection: $ P_z(x) = sum_{j=1}^k <z_j,x>z_j $ where $z_1,...,z_k $ is orthogonal basis and I don't have any other idea how to do this task @A.Γ.
    – VirtualUser
    yesterday












  • You can save yourself a lot of work by taking advantage of the fact that the orthogonal projection onto a subspace is what’s left after subtracting the orthogonal projection onto its complement.
    – amd
    yesterday
















Why do you need an orthogonal basis of $V$? One can find the projection if one follows the vector $(1 1 ldots 1)$ until the line meets $V$.
– A.Γ.
yesterday




Why do you need an orthogonal basis of $V$? One can find the projection if one follows the vector $(1 1 ldots 1)$ until the line meets $V$.
– A.Γ.
yesterday












I need orthogonal basis because I have that formula for orthogonal projection: $ P_z(x) = sum_{j=1}^k <z_j,x>z_j $ where $z_1,...,z_k $ is orthogonal basis and I don't have any other idea how to do this task @A.Γ.
– VirtualUser
yesterday






I need orthogonal basis because I have that formula for orthogonal projection: $ P_z(x) = sum_{j=1}^k <z_j,x>z_j $ where $z_1,...,z_k $ is orthogonal basis and I don't have any other idea how to do this task @A.Γ.
– VirtualUser
yesterday














You can save yourself a lot of work by taking advantage of the fact that the orthogonal projection onto a subspace is what’s left after subtracting the orthogonal projection onto its complement.
– amd
yesterday




You can save yourself a lot of work by taking advantage of the fact that the orthogonal projection onto a subspace is what’s left after subtracting the orthogonal projection onto its complement.
– amd
yesterday










1 Answer
1






active

oldest

votes


















1














Let us call $vec u$ the projection of $vec{x} = [n,0,0,...,0]^T$ on $V$.
$vec x - vec u$ is orthogonal to $V$, i.e. $vec x - vec u = a vec v$ with $v = (1, dots, 1)^T$ as you already showed



Therefore $vec x = a vec v + vec u$ with $vec u$ satisfying $sum_i u_i = 0$



Then,
$$ sum_i (x_i - a) = 0 $$



And finally $a = 1$ and



$$vec u = (n-1, -1, dots, -1) ^T $$






share|cite|improve this answer























  • the answer should be $vec u = (n-1, 1, dots, 1) ^T $ or $ vec u = (n-1, -1, dots, -1) ^T$?
    – VirtualUser
    yesterday










  • Great, it is a loooot of simpler way than my idea, thanks
    – VirtualUser
    yesterday






  • 1




    @VirtualUser Corrected. You read my answer before I had time to check it!
    – Damien
    yesterday











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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060582%2ffind-orthogonal-projection-of-n-0-0-0t-on-subspace-v%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









1














Let us call $vec u$ the projection of $vec{x} = [n,0,0,...,0]^T$ on $V$.
$vec x - vec u$ is orthogonal to $V$, i.e. $vec x - vec u = a vec v$ with $v = (1, dots, 1)^T$ as you already showed



Therefore $vec x = a vec v + vec u$ with $vec u$ satisfying $sum_i u_i = 0$



Then,
$$ sum_i (x_i - a) = 0 $$



And finally $a = 1$ and



$$vec u = (n-1, -1, dots, -1) ^T $$






share|cite|improve this answer























  • the answer should be $vec u = (n-1, 1, dots, 1) ^T $ or $ vec u = (n-1, -1, dots, -1) ^T$?
    – VirtualUser
    yesterday










  • Great, it is a loooot of simpler way than my idea, thanks
    – VirtualUser
    yesterday






  • 1




    @VirtualUser Corrected. You read my answer before I had time to check it!
    – Damien
    yesterday
















1














Let us call $vec u$ the projection of $vec{x} = [n,0,0,...,0]^T$ on $V$.
$vec x - vec u$ is orthogonal to $V$, i.e. $vec x - vec u = a vec v$ with $v = (1, dots, 1)^T$ as you already showed



Therefore $vec x = a vec v + vec u$ with $vec u$ satisfying $sum_i u_i = 0$



Then,
$$ sum_i (x_i - a) = 0 $$



And finally $a = 1$ and



$$vec u = (n-1, -1, dots, -1) ^T $$






share|cite|improve this answer























  • the answer should be $vec u = (n-1, 1, dots, 1) ^T $ or $ vec u = (n-1, -1, dots, -1) ^T$?
    – VirtualUser
    yesterday










  • Great, it is a loooot of simpler way than my idea, thanks
    – VirtualUser
    yesterday






  • 1




    @VirtualUser Corrected. You read my answer before I had time to check it!
    – Damien
    yesterday














1












1








1






Let us call $vec u$ the projection of $vec{x} = [n,0,0,...,0]^T$ on $V$.
$vec x - vec u$ is orthogonal to $V$, i.e. $vec x - vec u = a vec v$ with $v = (1, dots, 1)^T$ as you already showed



Therefore $vec x = a vec v + vec u$ with $vec u$ satisfying $sum_i u_i = 0$



Then,
$$ sum_i (x_i - a) = 0 $$



And finally $a = 1$ and



$$vec u = (n-1, -1, dots, -1) ^T $$






share|cite|improve this answer














Let us call $vec u$ the projection of $vec{x} = [n,0,0,...,0]^T$ on $V$.
$vec x - vec u$ is orthogonal to $V$, i.e. $vec x - vec u = a vec v$ with $v = (1, dots, 1)^T$ as you already showed



Therefore $vec x = a vec v + vec u$ with $vec u$ satisfying $sum_i u_i = 0$



Then,
$$ sum_i (x_i - a) = 0 $$



And finally $a = 1$ and



$$vec u = (n-1, -1, dots, -1) ^T $$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited yesterday

























answered yesterday









Damien

50714




50714












  • the answer should be $vec u = (n-1, 1, dots, 1) ^T $ or $ vec u = (n-1, -1, dots, -1) ^T$?
    – VirtualUser
    yesterday










  • Great, it is a loooot of simpler way than my idea, thanks
    – VirtualUser
    yesterday






  • 1




    @VirtualUser Corrected. You read my answer before I had time to check it!
    – Damien
    yesterday


















  • the answer should be $vec u = (n-1, 1, dots, 1) ^T $ or $ vec u = (n-1, -1, dots, -1) ^T$?
    – VirtualUser
    yesterday










  • Great, it is a loooot of simpler way than my idea, thanks
    – VirtualUser
    yesterday






  • 1




    @VirtualUser Corrected. You read my answer before I had time to check it!
    – Damien
    yesterday
















the answer should be $vec u = (n-1, 1, dots, 1) ^T $ or $ vec u = (n-1, -1, dots, -1) ^T$?
– VirtualUser
yesterday




the answer should be $vec u = (n-1, 1, dots, 1) ^T $ or $ vec u = (n-1, -1, dots, -1) ^T$?
– VirtualUser
yesterday












Great, it is a loooot of simpler way than my idea, thanks
– VirtualUser
yesterday




Great, it is a loooot of simpler way than my idea, thanks
– VirtualUser
yesterday




1




1




@VirtualUser Corrected. You read my answer before I had time to check it!
– Damien
yesterday




@VirtualUser Corrected. You read my answer before I had time to check it!
– Damien
yesterday


















draft saved

draft discarded




















































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.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • 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.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060582%2ffind-orthogonal-projection-of-n-0-0-0t-on-subspace-v%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

An IMO inspired problem

Image
14 4 This problem from IMO 1988 is said to be one of the most elegant ones in functional equations. Problem : The function $f$ is defined on the set of all positive integers as follows: begin{align} f(1) = 1, f(3) &= 3, f(2n) = f(n), \ f(4n+1) &= 2f(2n+1) - f(n), \ f(4n+3) &= 3 f(2n+1) - 2f(n) end{align} Find the number of $n$ with $f(n) = n, 1 leq n leq 1988$. The main idea towards the solution is realizing that these conditions stand for the fact that $f(n)$ just reverses the digits in the binary representation of the number $n$. So, essentially the solution is finding the number of binary pallindromes $leq 1988_{10}$. My question is the following : How to reformulate the problem so that $f(n)$ reverses the digits of $n$ in its ternary representation. Or even better, can we reformulate it for a
Read more

Management

Image
Management From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Manager" redirects here. For other uses, see Management (disambiguation) and Manager (disambiguation). An organization chart for the United States Coast Guard shows the hierarchy of managerial roles in that organization. Business administration Management of a business Accounting Management accounting Financial accounting Financial audit Business entities Cooperative Corporation Limited liability company Partnership Sole proprietorship State-owned enterprise Corporate governance Annual general meeting Board of directors Supervisory board Advisory board Audit committee Corporate law Commercial law Constitutional documents Contract Corporate crime Corporate liability Insolvency law International trade law Mergers and acquisitions Economics Commodity Pu
Read more

Has there ever been an instance of an active nuclear power plant within or near a war zone?

Image
36 3 The question Has there ever been a military combat going on around an active (running or in outage but not decommissioned) commercial nuclear power plant? Or dangerously near it? If yes, how did the plant and its personnel fare in such situation? Why I ask Let's assume for the scope of this question modern commercial nuclear power plants (CANDU, PWR, BWR of generation II plus all of gen III and newer) are usually reasonably safe to operate given that the engineering has been done right and there is a whole cohort of very well trained personnel on-site at all times, well rested and with considerable resources on their hands. Things hard to get with an armed conflict like the recent Syrian war raging around. I can't seem to find any reference or a comprehensive article about this, hence this question.
Read more