Dtjytyk

Is there a documented rationale why the House Ways and Means chairman can demand tax info?

Why are there no cargo aircraft with "flying wing" design?

Were Kohanim forbidden from serving in King David's army?

Dominant seventh chord in the major scale contains diminished triad of the seventh?

Can inflation occur in a positive-sum game currency system such as the Stack Exchange reputation system?

How do I keep my slimes from escaping their pens?

Models of set theory where not every set can be linearly ordered

do i need a schengen visa for a direct flight to amsterdam?

List *all* the tuples!

How do I mention the quality of my school without bragging

How can I make names more distinctive without making them longer?

The logistics of corpse disposal

What does the "x" in "x86" represent?

"Seemed to had" is it correct?

Why is black pepper both grey and black?

What is the longest distance a 13th-level monk can jump while attacking on the same turn?

What LEGO pieces have "real-world" functionality?

Disable hyphenation for an entire paragraph

Output the ŋarâþ crîþ alphabet song without using (m)any letters

What would be the ideal power source for a cybernetic eye?

Gastric acid as a weapon

How does a Death Domain cleric's Touch of Death feature work with Touch-range spells delivered by familiars?

When -s is used with third person singular. What's its use in this context?

Is it true that "carbohydrates are of no use for the basal metabolic need"?

If $A$ is an anti-symmetric matrix of size $N$ where $N$ is odd, then $lambda=0$ is an eigenvalue of $A$

1

1

$begingroup$

True or False: If $A$ is an anti-symmetric matrix of size $N$ where $N$ is odd then $lambda=0$ is an eigenvalue of $A$.

Through observation, it is false. However, how can I prove that this statement is false?

What part of the definition of an anti-symmetric matrix would allow me to do this?

This is for freshman linear algebra.

linear-algebra

share|cite|improve this question

edited Mar 23 at 22:49

Rócherz

3,0263823

asked Mar 23 at 22:41

MCCMCC

368

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Why do you say "Through observation, it is false"? Do you have an example to share?
  $endgroup$
  – hardmath
  Mar 23 at 22:45
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  To prove this is false, it would be sufficient for you to present a particular $Ntimes N$ matrix where $N$ is odd (e.g. a particular $3times 3$ matrix) that is anti symmetric and does not have $0$ as an eigenvalue. (You would need to show that it does not have $0$ as an eigenvalue.) (This is due to the general principle that to show that something is not always true, a specific counterexample is sufficient.) But do you have such a matrix?
  $endgroup$
  – Minus One-Twelfth
  Mar 23 at 22:46
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  To prove a universal statement is false, it suffices to produce a single instance of it failing, as Minus One-Twelfth says. But I think that you are incorrect in your observations... Because note that $0$ is an eigenvalue of $A$ if and only if $A$ is not invertible, if and only if its determinant is $0$. And in general, the determinant of a matrix $M$ and of its transpose $M^t$ are equal... So....
  $endgroup$
  – Arturo Magidin
  Mar 23 at 22:48
  

  
  
  
  

add a comment | 

1

1

$begingroup$

True or False: If $A$ is an anti-symmetric matrix of size $N$ where $N$ is odd then $lambda=0$ is an eigenvalue of $A$.

Through observation, it is false. However, how can I prove that this statement is false?

What part of the definition of an anti-symmetric matrix would allow me to do this?

This is for freshman linear algebra.

linear-algebra

share|cite|improve this question

edited Mar 23 at 22:49

Rócherz

3,0263823

asked Mar 23 at 22:41

MCCMCC

368

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Why do you say "Through observation, it is false"? Do you have an example to share?
  $endgroup$
  – hardmath
  Mar 23 at 22:45
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  To prove this is false, it would be sufficient for you to present a particular $Ntimes N$ matrix where $N$ is odd (e.g. a particular $3times 3$ matrix) that is anti symmetric and does not have $0$ as an eigenvalue. (You would need to show that it does not have $0$ as an eigenvalue.) (This is due to the general principle that to show that something is not always true, a specific counterexample is sufficient.) But do you have such a matrix?
  $endgroup$
  – Minus One-Twelfth
  Mar 23 at 22:46
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  To prove a universal statement is false, it suffices to produce a single instance of it failing, as Minus One-Twelfth says. But I think that you are incorrect in your observations... Because note that $0$ is an eigenvalue of $A$ if and only if $A$ is not invertible, if and only if its determinant is $0$. And in general, the determinant of a matrix $M$ and of its transpose $M^t$ are equal... So....
  $endgroup$
  – Arturo Magidin
  Mar 23 at 22:48
  

  
  
  
  

add a comment | 

$begingroup$

True or False: If $A$ is an anti-symmetric matrix of size $N$ where $N$ is odd then $lambda=0$ is an eigenvalue of $A$.

Through observation, it is false. However, how can I prove that this statement is false?

What part of the definition of an anti-symmetric matrix would allow me to do this?

This is for freshman linear algebra.

linear-algebra

share|cite|improve this question

edited Mar 23 at 22:49

Rócherz

3,0263823

asked Mar 23 at 22:41

MCCMCC

368

$endgroup$

share|cite|improve this question

edited Mar 23 at 22:49

Rócherz

3,0263823

asked Mar 23 at 22:41

MCCMCC

368

share|cite|improve this question

edited Mar 23 at 22:49

Rócherz

3,0263823

asked Mar 23 at 22:41

MCCMCC

368

edited Mar 23 at 22:49

Rócherz

3,0263823

edited Mar 23 at 22:49

Rócherz

3,0263823

asked Mar 23 at 22:41

MCCMCC

368

asked Mar 23 at 22:41

MCCMCC

368

2 Answers
2

active

oldest

votes

4

$begingroup$

The statement is true, you only need to prove that $det A=0$.

$det A=det (A^T)=det (-A)=(-1)^Ndet A=-det A$, then $det A$ has to be $0$.

share|cite|improve this answer

edited Mar 23 at 23:01

Arturo Magidin

266k34591922

answered Mar 23 at 22:50

GaoGao

986

$endgroup$

add a comment | 

0

$begingroup$

It is true for real skew-symmetric matrices $A = -A^T$ of odd size $N$.

Since

$text{size} A = N ; text{odd}, tag 1$

the characteristic polynomial

$det(A - lambda I) in Bbb R[lambda], tag 2$

which satisfies

$deg(det(A - lambda I)) = N tag 3$

has at least one real root; thus our proof will be completed if we show that the only real eigenvalue of a skew-symmetric matrix is $0$; now if

$A v = lambda v, tag 4$

we have

$lambda langle v, v rangle = langle v, lambda v rangle =langle v, Av rangle = langle A^T v, v rangle = langle -Av, v rangle = langle -lambda v, v rangle = -lambda langle v, v rangle; tag 5$

with $v ne 0$, $langle v, v rangle ne 0$, whence

$lambda = -lambda Longrightarrow lambda = 0. tag 6$

$OEDelta$.

share|cite|improve this answer

edited Mar 23 at 23:36

answered Mar 23 at 23:03

Robert LewisRobert Lewis

49k23168

$endgroup$

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%2f3159889%2fif-a-is-an-anti-symmetric-matrix-of-size-n-where-n-is-odd-then-lambda-0%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

4

$begingroup$

The statement is true, you only need to prove that $det A=0$.

$det A=det (A^T)=det (-A)=(-1)^Ndet A=-det A$, then $det A$ has to be $0$.

share|cite|improve this answer

edited Mar 23 at 23:01

Arturo Magidin

266k34591922

answered Mar 23 at 22:50

GaoGao

986

$endgroup$

add a comment | 

4

$begingroup$

The statement is true, you only need to prove that $det A=0$.

$det A=det (A^T)=det (-A)=(-1)^Ndet A=-det A$, then $det A$ has to be $0$.

share|cite|improve this answer

edited Mar 23 at 23:01

Arturo Magidin

266k34591922

answered Mar 23 at 22:50

GaoGao

986

$endgroup$

add a comment | 

$begingroup$

The statement is true, you only need to prove that $det A=0$.

$det A=det (A^T)=det (-A)=(-1)^Ndet A=-det A$, then $det A$ has to be $0$.

share|cite|improve this answer

edited Mar 23 at 23:01

Arturo Magidin

266k34591922

answered Mar 23 at 22:50

GaoGao

986

$endgroup$

share|cite|improve this answer

edited Mar 23 at 23:01

Arturo Magidin

266k34591922

answered Mar 23 at 22:50

GaoGao

986

edited Mar 23 at 23:01

Arturo Magidin

266k34591922

edited Mar 23 at 23:01

Arturo Magidin

266k34591922

answered Mar 23 at 22:50

GaoGao

986

answered Mar 23 at 22:50

GaoGao

986

0

$begingroup$

It is true for real skew-symmetric matrices $A = -A^T$ of odd size $N$.

Since

$text{size} A = N ; text{odd}, tag 1$

the characteristic polynomial

$det(A - lambda I) in Bbb R[lambda], tag 2$

which satisfies

$deg(det(A - lambda I)) = N tag 3$

has at least one real root; thus our proof will be completed if we show that the only real eigenvalue of a skew-symmetric matrix is $0$; now if

$A v = lambda v, tag 4$

we have

$lambda langle v, v rangle = langle v, lambda v rangle =langle v, Av rangle = langle A^T v, v rangle = langle -Av, v rangle = langle -lambda v, v rangle = -lambda langle v, v rangle; tag 5$

with $v ne 0$, $langle v, v rangle ne 0$, whence

$lambda = -lambda Longrightarrow lambda = 0. tag 6$

$OEDelta$.

share|cite|improve this answer

edited Mar 23 at 23:36

answered Mar 23 at 23:03

Robert LewisRobert Lewis

49k23168

$endgroup$

add a comment | 

0

$begingroup$

It is true for real skew-symmetric matrices $A = -A^T$ of odd size $N$.

Since

$text{size} A = N ; text{odd}, tag 1$

the characteristic polynomial

$det(A - lambda I) in Bbb R[lambda], tag 2$

which satisfies

$deg(det(A - lambda I)) = N tag 3$

has at least one real root; thus our proof will be completed if we show that the only real eigenvalue of a skew-symmetric matrix is $0$; now if

$A v = lambda v, tag 4$

we have

$lambda langle v, v rangle = langle v, lambda v rangle =langle v, Av rangle = langle A^T v, v rangle = langle -Av, v rangle = langle -lambda v, v rangle = -lambda langle v, v rangle; tag 5$

with $v ne 0$, $langle v, v rangle ne 0$, whence

$lambda = -lambda Longrightarrow lambda = 0. tag 6$

$OEDelta$.

share|cite|improve this answer

edited Mar 23 at 23:36

answered Mar 23 at 23:03

Robert LewisRobert Lewis

49k23168

$endgroup$

add a comment | 

$begingroup$

It is true for real skew-symmetric matrices $A = -A^T$ of odd size $N$.

Since

$text{size} A = N ; text{odd}, tag 1$

the characteristic polynomial

$det(A - lambda I) in Bbb R[lambda], tag 2$

which satisfies

$deg(det(A - lambda I)) = N tag 3$

has at least one real root; thus our proof will be completed if we show that the only real eigenvalue of a skew-symmetric matrix is $0$; now if

$A v = lambda v, tag 4$

we have

$lambda langle v, v rangle = langle v, lambda v rangle =langle v, Av rangle = langle A^T v, v rangle = langle -Av, v rangle = langle -lambda v, v rangle = -lambda langle v, v rangle; tag 5$

with $v ne 0$, $langle v, v rangle ne 0$, whence

$lambda = -lambda Longrightarrow lambda = 0. tag 6$

$OEDelta$.

share|cite|improve this answer

edited Mar 23 at 23:36

answered Mar 23 at 23:03

Robert LewisRobert Lewis

49k23168

$endgroup$

share|cite|improve this answer

edited Mar 23 at 23:36

answered Mar 23 at 23:03

Robert LewisRobert Lewis

49k23168

answered Mar 23 at 23:03

Robert LewisRobert Lewis

49k23168

answered Mar 23 at 23:03

Robert LewisRobert Lewis

49k23168