Why can't a matrix of linearly independent functions have determinant vanishing everywhere?
$begingroup$
Say I have an affine algebraic variety set $X$ over an algebraically closed field $k$ and regular functions $f_1,...,f_m:Xrightarrow k$ which are linearly independent over the field $k$. Why do there exist $m$ points $x_1,...,x_min X$ such that the matrix $$(f_i(x_j))_{1leq i,jleq m}$$ is invertible?
linear-algebra
asked Nov 30 '17 at 17:10
Pliny the illPliny the ill
2991316
$endgroup$
add a comment |
$begingroup$
Say I have an affine algebraic variety set $X$ over an algebraically closed field $k$ and regular functions $f_1,...,f_m:Xrightarrow k$ which are linearly independent over the field $k$. Why do there exist $m$ points $x_1,...,x_min X$ such that the matrix $$(f_i(x_j))_{1leq i,jleq m}$$ is invertible?
linear-algebra
asked Nov 30 '17 at 17:10
Pliny the illPliny the ill
2991316
$endgroup$
add a comment |
$begingroup$
Say I have an affine algebraic variety set $X$ over an algebraically closed field $k$ and regular functions $f_1,...,f_m:Xrightarrow k$ which are linearly independent over the field $k$. Why do there exist $m$ points $x_1,...,x_min X$ such that the matrix $$(f_i(x_j))_{1leq i,jleq m}$$ is invertible?
linear-algebra
asked Nov 30 '17 at 17:10
Pliny the illPliny the ill
2991316
$endgroup$
Say I have an affine algebraic variety set $X$ over an algebraically closed field $k$ and regular functions $f_1,...,f_m:Xrightarrow k$ which are linearly independent over the field $k$. Why do there exist $m$ points $x_1,...,x_min X$ such that the matrix $$(f_i(x_j))_{1leq i,jleq m}$$ is invertible?
linear-algebra
linear-algebra
asked Nov 30 '17 at 17:10
Pliny the illPliny the ill
2991316
asked Nov 30 '17 at 17:10
Pliny the illPliny the ill
2991316
edited Nov 30 '17 at 17:40
Pliny the ill
asked Nov 30 '17 at 17:10
Pliny the illPliny the ill
2991316
asked Nov 30 '17 at 17:10
Pliny the illPliny the ill
2991316
asked Nov 30 '17 at 17:10
Pliny the illPliny the ill
2991316
2991316
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Here is an answer to your question:
By hypothesis, span$lbrace left(f_1(x),dots,f_m(x)right)in k^m:xin X rbrace$ cannot have dimension $<m$. Hence this set contains a basis.
answered Dec 5 '18 at 18:14
UnpudUnpud
452419
$endgroup$
add a comment |
$begingroup$
Ok I guess this has nothing to do with algebraic geometry. Construct a matrix as follows. The first column is any nonzero vector, $v_1 in k^m$. There exists $x_1in X$ such that $(f_1(x_1),...,f(x_m))cdot v_1 neq 0 $ (by linear independence). Now choose $v_2$ orthogonal to $(f_1(x_1),...,f_m(x_1))$. There exists $x_2$ such that $(f_1(x_2),...,f_m(x_2))cdot v_2 neq 0$. Now choose $v_3$ orthongoal to $(f_1(x_1),...,f_m(x_1))$ and $(f_1(x_2),...,f_m(x_2))$. There exists $x_3$ such that $(f_1(x_2),...,f_m(x_2))cdot v_3 neq 0$... Continuing in that way gives a matrix $(v_1,...,v_m)$ whose product with the matriix $(f_i(x_j))$ is lower triangular with nonzero diagonal elements.
answered Nov 30 '17 at 17:31
Pliny the illPliny the ill
2991316
$endgroup$
add a comment |
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
});
}
});
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%2f2544724%2fwhy-cant-a-matrix-of-linearly-independent-functions-have-determinant-vanishing%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here is an answer to your question:
By hypothesis, span$lbrace left(f_1(x),dots,f_m(x)right)in k^m:xin X rbrace$ cannot have dimension $<m$. Hence this set contains a basis.
answered Dec 5 '18 at 18:14
UnpudUnpud
452419
$endgroup$
add a comment |
$begingroup$
Here is an answer to your question:
By hypothesis, span$lbrace left(f_1(x),dots,f_m(x)right)in k^m:xin X rbrace$ cannot have dimension $<m$. Hence this set contains a basis.
answered Dec 5 '18 at 18:14
UnpudUnpud
452419
$endgroup$
add a comment |
$begingroup$
Here is an answer to your question:
By hypothesis, span$lbrace left(f_1(x),dots,f_m(x)right)in k^m:xin X rbrace$ cannot have dimension $<m$. Hence this set contains a basis.
answered Dec 5 '18 at 18:14
UnpudUnpud
452419
$endgroup$
Here is an answer to your question:
By hypothesis, span$lbrace left(f_1(x),dots,f_m(x)right)in k^m:xin X rbrace$ cannot have dimension $<m$. Hence this set contains a basis.
answered Dec 5 '18 at 18:14
UnpudUnpud
452419
edited Dec 5 '18 at 20:09
answered Dec 5 '18 at 18:14
UnpudUnpud
452419
answered Dec 5 '18 at 18:14
UnpudUnpud
452419
answered Dec 5 '18 at 18:14
UnpudUnpud
452419
452419
add a comment |
add a comment |
$begingroup$
Ok I guess this has nothing to do with algebraic geometry. Construct a matrix as follows. The first column is any nonzero vector, $v_1 in k^m$. There exists $x_1in X$ such that $(f_1(x_1),...,f(x_m))cdot v_1 neq 0 $ (by linear independence). Now choose $v_2$ orthogonal to $(f_1(x_1),...,f_m(x_1))$. There exists $x_2$ such that $(f_1(x_2),...,f_m(x_2))cdot v_2 neq 0$. Now choose $v_3$ orthongoal to $(f_1(x_1),...,f_m(x_1))$ and $(f_1(x_2),...,f_m(x_2))$. There exists $x_3$ such that $(f_1(x_2),...,f_m(x_2))cdot v_3 neq 0$... Continuing in that way gives a matrix $(v_1,...,v_m)$ whose product with the matriix $(f_i(x_j))$ is lower triangular with nonzero diagonal elements.
answered Nov 30 '17 at 17:31
Pliny the illPliny the ill
2991316
$endgroup$
add a comment |
$begingroup$
Ok I guess this has nothing to do with algebraic geometry. Construct a matrix as follows. The first column is any nonzero vector, $v_1 in k^m$. There exists $x_1in X$ such that $(f_1(x_1),...,f(x_m))cdot v_1 neq 0 $ (by linear independence). Now choose $v_2$ orthogonal to $(f_1(x_1),...,f_m(x_1))$. There exists $x_2$ such that $(f_1(x_2),...,f_m(x_2))cdot v_2 neq 0$. Now choose $v_3$ orthongoal to $(f_1(x_1),...,f_m(x_1))$ and $(f_1(x_2),...,f_m(x_2))$. There exists $x_3$ such that $(f_1(x_2),...,f_m(x_2))cdot v_3 neq 0$... Continuing in that way gives a matrix $(v_1,...,v_m)$ whose product with the matriix $(f_i(x_j))$ is lower triangular with nonzero diagonal elements.
answered Nov 30 '17 at 17:31
Pliny the illPliny the ill
2991316
$endgroup$
add a comment |
$begingroup$
Ok I guess this has nothing to do with algebraic geometry. Construct a matrix as follows. The first column is any nonzero vector, $v_1 in k^m$. There exists $x_1in X$ such that $(f_1(x_1),...,f(x_m))cdot v_1 neq 0 $ (by linear independence). Now choose $v_2$ orthogonal to $(f_1(x_1),...,f_m(x_1))$. There exists $x_2$ such that $(f_1(x_2),...,f_m(x_2))cdot v_2 neq 0$. Now choose $v_3$ orthongoal to $(f_1(x_1),...,f_m(x_1))$ and $(f_1(x_2),...,f_m(x_2))$. There exists $x_3$ such that $(f_1(x_2),...,f_m(x_2))cdot v_3 neq 0$... Continuing in that way gives a matrix $(v_1,...,v_m)$ whose product with the matriix $(f_i(x_j))$ is lower triangular with nonzero diagonal elements.
answered Nov 30 '17 at 17:31
Pliny the illPliny the ill
2991316
$endgroup$
Ok I guess this has nothing to do with algebraic geometry. Construct a matrix as follows. The first column is any nonzero vector, $v_1 in k^m$. There exists $x_1in X$ such that $(f_1(x_1),...,f(x_m))cdot v_1 neq 0 $ (by linear independence). Now choose $v_2$ orthogonal to $(f_1(x_1),...,f_m(x_1))$. There exists $x_2$ such that $(f_1(x_2),...,f_m(x_2))cdot v_2 neq 0$. Now choose $v_3$ orthongoal to $(f_1(x_1),...,f_m(x_1))$ and $(f_1(x_2),...,f_m(x_2))$. There exists $x_3$ such that $(f_1(x_2),...,f_m(x_2))cdot v_3 neq 0$... Continuing in that way gives a matrix $(v_1,...,v_m)$ whose product with the matriix $(f_i(x_j))$ is lower triangular with nonzero diagonal elements.
answered Nov 30 '17 at 17:31
Pliny the illPliny the ill
2991316
edited Nov 30 '17 at 17:42
answered Nov 30 '17 at 17:31
Pliny the illPliny the ill
2991316
answered Nov 30 '17 at 17:31
Pliny the illPliny the ill
2991316
answered Nov 30 '17 at 17:31
Pliny the illPliny the ill
2991316
2991316
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%2f2544724%2fwhy-cant-a-matrix-of-linearly-independent-functions-have-determinant-vanishing%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
