Fdtxjr

ELI5: Why do they say that Israel would have been the fourth country to land a spacecraft on the Moon and why do they call it low cost?

Am I ethically obligated to go into work on an off day if the reason is sudden?

Why did Peik Lin say, "I'm not an animal"?

"... to apply for a visa" or "... and applied for a visa"?

How to read αἱμύλιος or when to aspirate

Visa regaring travelling European country

Didn't get enough time to take a Coding Test - what to do now?

Match Roman Numerals

how can a perfect fourth interval be considered either consonant or dissonant?

Loose spokes after only a few rides

How to support a colleague who finds meetings extremely tiring?

Circular reasoning in L'Hopital's rule

Can withdrawing asylum be illegal?

Why are PDP-7-style microprogrammed instructions out of vogue?

First use of “packing” as in carrying a gun

How to type a long/em dash `—`

How did the audience guess the pentatonic scale in Bobby McFerrin's presentation?

Is every episode of "Where are my Pants?" identical?

"is" operation returns false even though two objects have same id

Sort list of array linked objects by keys and values

Homework question about an engine pulling a train

Do I have Disadvantage attacking with an off-hand weapon?

60's-70's movie: home appliances revolting against the owners

What do I do when my TA workload is more than expected?

Linear independence of gradient vectors

0

$begingroup$

I have to find all feasible points that are regular for a function with constraints:

$h_1(x_1,x_2,x_3) = 2x_1x_2+x_3^2=0$

$h_2(x_1,x_2,x_3) = x_1^2+x_2^2+x_3^2=4$

The definition says that if a point $x^*$ satisfies the contraints $h_i(x^*)=0, i = 1,...,m$ and if the gradient vectors $∇h_i(x^*), i=1,...,m$ are linearly independent then the point $x^*$ is a regular point.

So the gradient vectors I get in this case are

$∇h_1(x)=[2x_2,2x_1,2x_3]^T$

$∇h_2(x)=[2x_1,2x_2,2x_3]^T$

After messing around a bit, I found 6 points that satisfy the constraints.

Those points are $(-1,1,sqrt2),(1,-1,sqrt2),(0,2,0),(2,0,0),(0-2,0)$ and $(-2,0,0)$

I didn't bother to check all the points because surely there are more that satisfy the constraints. Instead I tried looking at the general case.

If $a(2x_2,2x_1,2x_3)+b(2x_1,2x_2,2x_3)=0$ is true such that $a,b$ are not $0$ then they are not linearly independent. But this is where I'm stuck. I can't tell for which $x$-s the vectors are linearly independent.

linear-algebra

share|cite|improve this question

asked Mar 24 at 19:06

Nick202Nick202

605

$endgroup$

add a comment | 

0

$begingroup$

I have to find all feasible points that are regular for a function with constraints:

$h_1(x_1,x_2,x_3) = 2x_1x_2+x_3^2=0$

$h_2(x_1,x_2,x_3) = x_1^2+x_2^2+x_3^2=4$

The definition says that if a point $x^*$ satisfies the contraints $h_i(x^*)=0, i = 1,...,m$ and if the gradient vectors $∇h_i(x^*), i=1,...,m$ are linearly independent then the point $x^*$ is a regular point.

So the gradient vectors I get in this case are

$∇h_1(x)=[2x_2,2x_1,2x_3]^T$

$∇h_2(x)=[2x_1,2x_2,2x_3]^T$

After messing around a bit, I found 6 points that satisfy the constraints.

Those points are $(-1,1,sqrt2),(1,-1,sqrt2),(0,2,0),(2,0,0),(0-2,0)$ and $(-2,0,0)$

I didn't bother to check all the points because surely there are more that satisfy the constraints. Instead I tried looking at the general case.

If $a(2x_2,2x_1,2x_3)+b(2x_1,2x_2,2x_3)=0$ is true such that $a,b$ are not $0$ then they are not linearly independent. But this is where I'm stuck. I can't tell for which $x$-s the vectors are linearly independent.

linear-algebra

share|cite|improve this question

asked Mar 24 at 19:06

Nick202Nick202

605

$endgroup$

add a comment | 

$begingroup$

I have to find all feasible points that are regular for a function with constraints:

$h_1(x_1,x_2,x_3) = 2x_1x_2+x_3^2=0$

$h_2(x_1,x_2,x_3) = x_1^2+x_2^2+x_3^2=4$

The definition says that if a point $x^*$ satisfies the contraints $h_i(x^*)=0, i = 1,...,m$ and if the gradient vectors $∇h_i(x^*), i=1,...,m$ are linearly independent then the point $x^*$ is a regular point.

So the gradient vectors I get in this case are

$∇h_1(x)=[2x_2,2x_1,2x_3]^T$

$∇h_2(x)=[2x_1,2x_2,2x_3]^T$

After messing around a bit, I found 6 points that satisfy the constraints.

Those points are $(-1,1,sqrt2),(1,-1,sqrt2),(0,2,0),(2,0,0),(0-2,0)$ and $(-2,0,0)$

I didn't bother to check all the points because surely there are more that satisfy the constraints. Instead I tried looking at the general case.

If $a(2x_2,2x_1,2x_3)+b(2x_1,2x_2,2x_3)=0$ is true such that $a,b$ are not $0$ then they are not linearly independent. But this is where I'm stuck. I can't tell for which $x$-s the vectors are linearly independent.

linear-algebra

share|cite|improve this question

asked Mar 24 at 19:06

Nick202Nick202

605

$endgroup$

share|cite|improve this question

asked Mar 24 at 19:06

Nick202Nick202

605

share|cite|improve this question

asked Mar 24 at 19:06

Nick202Nick202

605

asked Mar 24 at 19:06

Nick202Nick202

605

asked Mar 24 at 19:06

Nick202Nick202

605

1 Answer
1

active

oldest

votes

0

$begingroup$

Find the rank of the matrix
$$
pmatrix2x_2 &2x_1 & 2x_3\
2x_1 &2x_2 & 2x_3.
$$
I.e., calculate the determinants of the three $2times 2$ minors. Rank 2 (some minor $ne 0$) means that the vectors are l.i.

share|cite|improve this answer

answered Mar 24 at 20:28

Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

35.5k42972

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But those determinants could be 0, for certain values, no?
  $endgroup$
  – Nick202
  Mar 24 at 21:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Nick202, Yes...
  $endgroup$
  – Martín-Blas Pérez Pinilla
  Mar 25 at 3:42
  

  
  
  
  

add a comment | 

Your Answer

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

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

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3160900%2flinear-independence-of-gradient-vectors%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

0

$begingroup$

Find the rank of the matrix
$$
pmatrix2x_2 &2x_1 & 2x_3\
2x_1 &2x_2 & 2x_3.
$$
I.e., calculate the determinants of the three $2times 2$ minors. Rank 2 (some minor $ne 0$) means that the vectors are l.i.

share|cite|improve this answer

answered Mar 24 at 20:28

Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

35.5k42972

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But those determinants could be 0, for certain values, no?
  $endgroup$
  – Nick202
  Mar 24 at 21:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Nick202, Yes...
  $endgroup$
  – Martín-Blas Pérez Pinilla
  Mar 25 at 3:42
  

  
  
  
  

add a comment | 

0

$begingroup$

Find the rank of the matrix
$$
pmatrix2x_2 &2x_1 & 2x_3\
2x_1 &2x_2 & 2x_3.
$$
I.e., calculate the determinants of the three $2times 2$ minors. Rank 2 (some minor $ne 0$) means that the vectors are l.i.

share|cite|improve this answer

answered Mar 24 at 20:28

Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

35.5k42972

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  But those determinants could be 0, for certain values, no?
  $endgroup$
  – Nick202
  Mar 24 at 21:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Nick202, Yes...
  $endgroup$
  – Martín-Blas Pérez Pinilla
  Mar 25 at 3:42
  

  
  
  
  

add a comment | 

$begingroup$

Find the rank of the matrix
$$
pmatrix2x_2 &2x_1 & 2x_3\
2x_1 &2x_2 & 2x_3.
$$
I.e., calculate the determinants of the three $2times 2$ minors. Rank 2 (some minor $ne 0$) means that the vectors are l.i.

share|cite|improve this answer

answered Mar 24 at 20:28

Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

35.5k42972

$endgroup$

share|cite|improve this answer

answered Mar 24 at 20:28

Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

35.5k42972

answered Mar 24 at 20:28

Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

35.5k42972

answered Mar 24 at 20:28

Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla

35.5k42972