Is the following statement is True false regarding inner product












0












$begingroup$


Is the following statement is True false



Let $V = mathbb{R}^5$ be equipped with the usual euclidean inner-product.
If $W$ and $Z$ are subspaces of $V$ such that both of them are of dimension
$3,$ then there exists $z in Z$ such that $z neq 0$ and $z ⊥ W.$



My attempt : i thinks this statement is True take $W= (1,-1,1)$ and $Z=(-1,1,-1)$ the $Z ⊥ W$ that is $langle Z. Wrangle=0$



Is my thinking is correct or not ?



Any hints/solution will be appreciated










share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    What if W=Z i.e. W and Z are the same subspace ? The conditions of the question don't seem to rule this out.
    $endgroup$
    – gandalf61
    Jan 9 at 10:46






  • 2




    $begingroup$
    The question is not clear. Your $W$ and $Z$ seem vectors of $mathbb{R}^3$, so their are not subspaces of $mathbb{R}^5$ nor elements of this vector space.
    $endgroup$
    – Emilio Novati
    Jan 9 at 10:48








  • 1




    $begingroup$
    exactly! by the way, I tried to incorporate your comments in my answer, i hope that is ok
    $endgroup$
    – Enkidu
    Jan 9 at 10:49
















0












$begingroup$


Is the following statement is True false



Let $V = mathbb{R}^5$ be equipped with the usual euclidean inner-product.
If $W$ and $Z$ are subspaces of $V$ such that both of them are of dimension
$3,$ then there exists $z in Z$ such that $z neq 0$ and $z ⊥ W.$



My attempt : i thinks this statement is True take $W= (1,-1,1)$ and $Z=(-1,1,-1)$ the $Z ⊥ W$ that is $langle Z. Wrangle=0$



Is my thinking is correct or not ?



Any hints/solution will be appreciated










share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    What if W=Z i.e. W and Z are the same subspace ? The conditions of the question don't seem to rule this out.
    $endgroup$
    – gandalf61
    Jan 9 at 10:46






  • 2




    $begingroup$
    The question is not clear. Your $W$ and $Z$ seem vectors of $mathbb{R}^3$, so their are not subspaces of $mathbb{R}^5$ nor elements of this vector space.
    $endgroup$
    – Emilio Novati
    Jan 9 at 10:48








  • 1




    $begingroup$
    exactly! by the way, I tried to incorporate your comments in my answer, i hope that is ok
    $endgroup$
    – Enkidu
    Jan 9 at 10:49














0












0








0





$begingroup$


Is the following statement is True false



Let $V = mathbb{R}^5$ be equipped with the usual euclidean inner-product.
If $W$ and $Z$ are subspaces of $V$ such that both of them are of dimension
$3,$ then there exists $z in Z$ such that $z neq 0$ and $z ⊥ W.$



My attempt : i thinks this statement is True take $W= (1,-1,1)$ and $Z=(-1,1,-1)$ the $Z ⊥ W$ that is $langle Z. Wrangle=0$



Is my thinking is correct or not ?



Any hints/solution will be appreciated










share|cite|improve this question









$endgroup$




Is the following statement is True false



Let $V = mathbb{R}^5$ be equipped with the usual euclidean inner-product.
If $W$ and $Z$ are subspaces of $V$ such that both of them are of dimension
$3,$ then there exists $z in Z$ such that $z neq 0$ and $z ⊥ W.$



My attempt : i thinks this statement is True take $W= (1,-1,1)$ and $Z=(-1,1,-1)$ the $Z ⊥ W$ that is $langle Z. Wrangle=0$



Is my thinking is correct or not ?



Any hints/solution will be appreciated







linear-algebra






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 9 at 10:38









jasminejasmine

1,969420




1,969420








  • 3




    $begingroup$
    What if W=Z i.e. W and Z are the same subspace ? The conditions of the question don't seem to rule this out.
    $endgroup$
    – gandalf61
    Jan 9 at 10:46






  • 2




    $begingroup$
    The question is not clear. Your $W$ and $Z$ seem vectors of $mathbb{R}^3$, so their are not subspaces of $mathbb{R}^5$ nor elements of this vector space.
    $endgroup$
    – Emilio Novati
    Jan 9 at 10:48








  • 1




    $begingroup$
    exactly! by the way, I tried to incorporate your comments in my answer, i hope that is ok
    $endgroup$
    – Enkidu
    Jan 9 at 10:49














  • 3




    $begingroup$
    What if W=Z i.e. W and Z are the same subspace ? The conditions of the question don't seem to rule this out.
    $endgroup$
    – gandalf61
    Jan 9 at 10:46






  • 2




    $begingroup$
    The question is not clear. Your $W$ and $Z$ seem vectors of $mathbb{R}^3$, so their are not subspaces of $mathbb{R}^5$ nor elements of this vector space.
    $endgroup$
    – Emilio Novati
    Jan 9 at 10:48








  • 1




    $begingroup$
    exactly! by the way, I tried to incorporate your comments in my answer, i hope that is ok
    $endgroup$
    – Enkidu
    Jan 9 at 10:49








3




3




$begingroup$
What if W=Z i.e. W and Z are the same subspace ? The conditions of the question don't seem to rule this out.
$endgroup$
– gandalf61
Jan 9 at 10:46




$begingroup$
What if W=Z i.e. W and Z are the same subspace ? The conditions of the question don't seem to rule this out.
$endgroup$
– gandalf61
Jan 9 at 10:46




2




2




$begingroup$
The question is not clear. Your $W$ and $Z$ seem vectors of $mathbb{R}^3$, so their are not subspaces of $mathbb{R}^5$ nor elements of this vector space.
$endgroup$
– Emilio Novati
Jan 9 at 10:48






$begingroup$
The question is not clear. Your $W$ and $Z$ seem vectors of $mathbb{R}^3$, so their are not subspaces of $mathbb{R}^5$ nor elements of this vector space.
$endgroup$
– Emilio Novati
Jan 9 at 10:48






1




1




$begingroup$
exactly! by the way, I tried to incorporate your comments in my answer, i hope that is ok
$endgroup$
– Enkidu
Jan 9 at 10:49




$begingroup$
exactly! by the way, I tried to incorporate your comments in my answer, i hope that is ok
$endgroup$
– Enkidu
Jan 9 at 10:49










1 Answer
1






active

oldest

votes


















1












$begingroup$

No, it is false, it needs further assumptions, also as stated in the comments, be aware that we are talking subspaces here, not vectors! for example (as gandalf61 said) take $Z=Wsubset V$ then we have, since $langle_,_ rangle$ is non degenerate, for all $0 neq z in Z$ a $w in W=Z$ such that $langle z , w rangle neq 0$ in particular (to give a very concrete example) you could pick $w=z$ and then we already know that $$langle z, wrangle = langle z, zrangle = 0 iff z=0$$



Also, be aware that only the concrete counterexample with $z=w$ needs an inner product, for the first one it suffices to have bilinear form $langle _ ,_ rangle$ inducing a non degenerate bilinearform on $Z=W$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    thanks u @Enkidu
    $endgroup$
    – jasmine
    Jan 9 at 10:53












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


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3067309%2fis-the-following-statement-is-true-false-regarding-inner-product%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












$begingroup$

No, it is false, it needs further assumptions, also as stated in the comments, be aware that we are talking subspaces here, not vectors! for example (as gandalf61 said) take $Z=Wsubset V$ then we have, since $langle_,_ rangle$ is non degenerate, for all $0 neq z in Z$ a $w in W=Z$ such that $langle z , w rangle neq 0$ in particular (to give a very concrete example) you could pick $w=z$ and then we already know that $$langle z, wrangle = langle z, zrangle = 0 iff z=0$$



Also, be aware that only the concrete counterexample with $z=w$ needs an inner product, for the first one it suffices to have bilinear form $langle _ ,_ rangle$ inducing a non degenerate bilinearform on $Z=W$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    thanks u @Enkidu
    $endgroup$
    – jasmine
    Jan 9 at 10:53
















1












$begingroup$

No, it is false, it needs further assumptions, also as stated in the comments, be aware that we are talking subspaces here, not vectors! for example (as gandalf61 said) take $Z=Wsubset V$ then we have, since $langle_,_ rangle$ is non degenerate, for all $0 neq z in Z$ a $w in W=Z$ such that $langle z , w rangle neq 0$ in particular (to give a very concrete example) you could pick $w=z$ and then we already know that $$langle z, wrangle = langle z, zrangle = 0 iff z=0$$



Also, be aware that only the concrete counterexample with $z=w$ needs an inner product, for the first one it suffices to have bilinear form $langle _ ,_ rangle$ inducing a non degenerate bilinearform on $Z=W$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    thanks u @Enkidu
    $endgroup$
    – jasmine
    Jan 9 at 10:53














1












1








1





$begingroup$

No, it is false, it needs further assumptions, also as stated in the comments, be aware that we are talking subspaces here, not vectors! for example (as gandalf61 said) take $Z=Wsubset V$ then we have, since $langle_,_ rangle$ is non degenerate, for all $0 neq z in Z$ a $w in W=Z$ such that $langle z , w rangle neq 0$ in particular (to give a very concrete example) you could pick $w=z$ and then we already know that $$langle z, wrangle = langle z, zrangle = 0 iff z=0$$



Also, be aware that only the concrete counterexample with $z=w$ needs an inner product, for the first one it suffices to have bilinear form $langle _ ,_ rangle$ inducing a non degenerate bilinearform on $Z=W$.






share|cite|improve this answer











$endgroup$



No, it is false, it needs further assumptions, also as stated in the comments, be aware that we are talking subspaces here, not vectors! for example (as gandalf61 said) take $Z=Wsubset V$ then we have, since $langle_,_ rangle$ is non degenerate, for all $0 neq z in Z$ a $w in W=Z$ such that $langle z , w rangle neq 0$ in particular (to give a very concrete example) you could pick $w=z$ and then we already know that $$langle z, wrangle = langle z, zrangle = 0 iff z=0$$



Also, be aware that only the concrete counterexample with $z=w$ needs an inner product, for the first one it suffices to have bilinear form $langle _ ,_ rangle$ inducing a non degenerate bilinearform on $Z=W$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 9 at 10:53

























answered Jan 9 at 10:48









EnkiduEnkidu

1,44429




1,44429












  • $begingroup$
    thanks u @Enkidu
    $endgroup$
    – jasmine
    Jan 9 at 10:53


















  • $begingroup$
    thanks u @Enkidu
    $endgroup$
    – jasmine
    Jan 9 at 10:53
















$begingroup$
thanks u @Enkidu
$endgroup$
– jasmine
Jan 9 at 10:53




$begingroup$
thanks u @Enkidu
$endgroup$
– jasmine
Jan 9 at 10:53


















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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3067309%2fis-the-following-statement-is-true-false-regarding-inner-product%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

Probability when a professor distributes a quiz and homework assignment to a class of n students.

Aardman Animations

Are they similar matrix