Diagonalizable matricies and eigenvalues












2












$begingroup$



Let $A$,$B$,$C$ be three different real $3 times 3$ matricies with the following properties:




  • $A$ has the complex eigenvalue $lambda=3-5i$


  • $B$ has eigenvalues $lambda=0$, $lambda=5$, $lambda=-5$


  • $C=M M^T$ for some real $3 times 2$ matrix $M$.



Which of the matrices are necessarily diagonalizable? In the case of complex eigenvectors, diagonalization is over $mathbb{C}$.



$(A)$ Only $B$



$(B)$ Only $A$ and $B$



$(C)$ Only $B$ and $C$



$(D)$ All three of them



$(E)$ None of them




Okay so first of all, I am almost certain that $(B)$ is diagonalizable. There are $3$ distinct eigenvalues for a $3 times 3$ matrix so it can definitely be diagonalized.



I know for $(A)$ that because we have a real matrix, that the complex numbers must come in conjugate pairs so there must be a 2nd complex eigen value and one more real value so that $(A)$ can diagonalized too.



I'm not sure about $(C)$ though. I know that $(C)$ is a $3times 3$ matrix but I am not sure if that justifies in there being 3 distinct eigenvalues so thus I could conclude that all three are diagonalizable (Thus $(D)$ being the answer).



Is my reasoning relatively on the right track?






linear-algebra matrices eigenvalues-eigenvectors diagonalization







share|cite|improve this question









$endgroup$












  • $begingroup$
    Well, $;A;$ can be diagonalized...but not over the real numbers, and you defined the matrices to be real...
    $endgroup$
    – DonAntonio
    Dec 22 '18 at 1:30










  • $begingroup$
    There was a statement saying for the complex eigenvalues, diagonalization is over the complex numbers.
    $endgroup$
    – Future Math person
    Dec 22 '18 at 1:42
















2












$begingroup$



Let $A$,$B$,$C$ be three different real $3 times 3$ matricies with the following properties:




  • $A$ has the complex eigenvalue $lambda=3-5i$


  • $B$ has eigenvalues $lambda=0$, $lambda=5$, $lambda=-5$


  • $C=M M^T$ for some real $3 times 2$ matrix $M$.



Which of the matrices are necessarily diagonalizable? In the case of complex eigenvectors, diagonalization is over $mathbb{C}$.



$(A)$ Only $B$



$(B)$ Only $A$ and $B$



$(C)$ Only $B$ and $C$



$(D)$ All three of them



$(E)$ None of them




Okay so first of all, I am almost certain that $(B)$ is diagonalizable. There are $3$ distinct eigenvalues for a $3 times 3$ matrix so it can definitely be diagonalized.



I know for $(A)$ that because we have a real matrix, that the complex numbers must come in conjugate pairs so there must be a 2nd complex eigen value and one more real value so that $(A)$ can diagonalized too.



I'm not sure about $(C)$ though. I know that $(C)$ is a $3times 3$ matrix but I am not sure if that justifies in there being 3 distinct eigenvalues so thus I could conclude that all three are diagonalizable (Thus $(D)$ being the answer).



Is my reasoning relatively on the right track?






linear-algebra matrices eigenvalues-eigenvectors diagonalization







share|cite|improve this question









$endgroup$












  • $begingroup$
    Well, $;A;$ can be diagonalized...but not over the real numbers, and you defined the matrices to be real...
    $endgroup$
    – DonAntonio
    Dec 22 '18 at 1:30










  • $begingroup$
    There was a statement saying for the complex eigenvalues, diagonalization is over the complex numbers.
    $endgroup$
    – Future Math person
    Dec 22 '18 at 1:42














2












2








2


1



$begingroup$



Let $A$,$B$,$C$ be three different real $3 times 3$ matricies with the following properties:




  • $A$ has the complex eigenvalue $lambda=3-5i$


  • $B$ has eigenvalues $lambda=0$, $lambda=5$, $lambda=-5$


  • $C=M M^T$ for some real $3 times 2$ matrix $M$.



Which of the matrices are necessarily diagonalizable? In the case of complex eigenvectors, diagonalization is over $mathbb{C}$.



$(A)$ Only $B$



$(B)$ Only $A$ and $B$



$(C)$ Only $B$ and $C$



$(D)$ All three of them



$(E)$ None of them




Okay so first of all, I am almost certain that $(B)$ is diagonalizable. There are $3$ distinct eigenvalues for a $3 times 3$ matrix so it can definitely be diagonalized.



I know for $(A)$ that because we have a real matrix, that the complex numbers must come in conjugate pairs so there must be a 2nd complex eigen value and one more real value so that $(A)$ can diagonalized too.



I'm not sure about $(C)$ though. I know that $(C)$ is a $3times 3$ matrix but I am not sure if that justifies in there being 3 distinct eigenvalues so thus I could conclude that all three are diagonalizable (Thus $(D)$ being the answer).



Is my reasoning relatively on the right track?






linear-algebra matrices eigenvalues-eigenvectors diagonalization







share|cite|improve this question









$endgroup$





Let $A$,$B$,$C$ be three different real $3 times 3$ matricies with the following properties:




  • $A$ has the complex eigenvalue $lambda=3-5i$


  • $B$ has eigenvalues $lambda=0$, $lambda=5$, $lambda=-5$


  • $C=M M^T$ for some real $3 times 2$ matrix $M$.



Which of the matrices are necessarily diagonalizable? In the case of complex eigenvectors, diagonalization is over $mathbb{C}$.



$(A)$ Only $B$



$(B)$ Only $A$ and $B$



$(C)$ Only $B$ and $C$



$(D)$ All three of them



$(E)$ None of them




Okay so first of all, I am almost certain that $(B)$ is diagonalizable. There are $3$ distinct eigenvalues for a $3 times 3$ matrix so it can definitely be diagonalized.



I know for $(A)$ that because we have a real matrix, that the complex numbers must come in conjugate pairs so there must be a 2nd complex eigen value and one more real value so that $(A)$ can diagonalized too.



I'm not sure about $(C)$ though. I know that $(C)$ is a $3times 3$ matrix but I am not sure if that justifies in there being 3 distinct eigenvalues so thus I could conclude that all three are diagonalizable (Thus $(D)$ being the answer).



Is my reasoning relatively on the right track?






linear-algebra matrices eigenvalues-eigenvectors diagonalization




linear-algebra matrices eigenvalues-eigenvectors diagonalization






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 22 '18 at 1:03









Future Math personFuture Math person

977817




977817












  • $begingroup$
    Well, $;A;$ can be diagonalized...but not over the real numbers, and you defined the matrices to be real...
    $endgroup$
    – DonAntonio
    Dec 22 '18 at 1:30










  • $begingroup$
    There was a statement saying for the complex eigenvalues, diagonalization is over the complex numbers.
    $endgroup$
    – Future Math person
    Dec 22 '18 at 1:42


















  • $begingroup$
    Well, $;A;$ can be diagonalized...but not over the real numbers, and you defined the matrices to be real...
    $endgroup$
    – DonAntonio
    Dec 22 '18 at 1:30










  • $begingroup$
    There was a statement saying for the complex eigenvalues, diagonalization is over the complex numbers.
    $endgroup$
    – Future Math person
    Dec 22 '18 at 1:42
















$begingroup$
Well, $;A;$ can be diagonalized...but not over the real numbers, and you defined the matrices to be real...
$endgroup$
– DonAntonio
Dec 22 '18 at 1:30




$begingroup$
Well, $;A;$ can be diagonalized...but not over the real numbers, and you defined the matrices to be real...
$endgroup$
– DonAntonio
Dec 22 '18 at 1:30












$begingroup$
There was a statement saying for the complex eigenvalues, diagonalization is over the complex numbers.
$endgroup$
– Future Math person
Dec 22 '18 at 1:42




$begingroup$
There was a statement saying for the complex eigenvalues, diagonalization is over the complex numbers.
$endgroup$
– Future Math person
Dec 22 '18 at 1:42










1 Answer
1






active

oldest

votes


















1












$begingroup$

Our OP Future Math person has correctly argued that $A$ and $B$ are diagonalizable, $A$ over $Bbb C$ and $B$ over $Bbb R$; in each case for the reason that the matrix has $3$ distinct eigenvalues, hence $3$ linearly independent eigenvectors.



So what about the case



$C = MM^T? tag 1$



here we don't know too much about the eigenvalues, but we may observe that



$C^T = (MM^T)^T = (M^T)^TM^T = MM^T; tag 2$



that is, $C$ is a real symmetric matrix; as such, it too may be diagonalized, whether or not the eigenvalues are distinct; that real symmetric matrices are diagonalizable is a well-known result.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Ohh wow. Yes I didn't even realize it was a symmetric matrix. Thanks!
    $endgroup$
    – Future Math person
    Dec 22 '18 at 1:43










  • $begingroup$
    @FutureMathperson: you are most welcome! And thanks for the "acceptance"!
    $endgroup$
    – Robert Lewis
    Dec 22 '18 at 1:50






  • 1




    $begingroup$
    It may be worth noting that we do know that $0$ is an eigenvalue for $C$. This is because $M^T:mathbb{R}^3 to mathbb{R}^2$ and thus has non-trivial kernel forcing $MM^T$ to have nontrivial kernel.
    $endgroup$
    – Dionel Jaime
    Dec 22 '18 at 4:54










  • $begingroup$
    @DionelJaime: right you are! Cheers!
    $endgroup$
    – Robert Lewis
    Dec 22 '18 at 4:56











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%2f3049047%2fdiagonalizable-matricies-and-eigenvalues%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












$begingroup$

Our OP Future Math person has correctly argued that $A$ and $B$ are diagonalizable, $A$ over $Bbb C$ and $B$ over $Bbb R$; in each case for the reason that the matrix has $3$ distinct eigenvalues, hence $3$ linearly independent eigenvectors.



So what about the case



$C = MM^T? tag 1$



here we don't know too much about the eigenvalues, but we may observe that



$C^T = (MM^T)^T = (M^T)^TM^T = MM^T; tag 2$



that is, $C$ is a real symmetric matrix; as such, it too may be diagonalized, whether or not the eigenvalues are distinct; that real symmetric matrices are diagonalizable is a well-known result.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Ohh wow. Yes I didn't even realize it was a symmetric matrix. Thanks!
    $endgroup$
    – Future Math person
    Dec 22 '18 at 1:43










  • $begingroup$
    @FutureMathperson: you are most welcome! And thanks for the "acceptance"!
    $endgroup$
    – Robert Lewis
    Dec 22 '18 at 1:50






  • 1




    $begingroup$
    It may be worth noting that we do know that $0$ is an eigenvalue for $C$. This is because $M^T:mathbb{R}^3 to mathbb{R}^2$ and thus has non-trivial kernel forcing $MM^T$ to have nontrivial kernel.
    $endgroup$
    – Dionel Jaime
    Dec 22 '18 at 4:54










  • $begingroup$
    @DionelJaime: right you are! Cheers!
    $endgroup$
    – Robert Lewis
    Dec 22 '18 at 4:56
















1












$begingroup$

Our OP Future Math person has correctly argued that $A$ and $B$ are diagonalizable, $A$ over $Bbb C$ and $B$ over $Bbb R$; in each case for the reason that the matrix has $3$ distinct eigenvalues, hence $3$ linearly independent eigenvectors.



So what about the case



$C = MM^T? tag 1$



here we don't know too much about the eigenvalues, but we may observe that



$C^T = (MM^T)^T = (M^T)^TM^T = MM^T; tag 2$



that is, $C$ is a real symmetric matrix; as such, it too may be diagonalized, whether or not the eigenvalues are distinct; that real symmetric matrices are diagonalizable is a well-known result.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Ohh wow. Yes I didn't even realize it was a symmetric matrix. Thanks!
    $endgroup$
    – Future Math person
    Dec 22 '18 at 1:43










  • $begingroup$
    @FutureMathperson: you are most welcome! And thanks for the "acceptance"!
    $endgroup$
    – Robert Lewis
    Dec 22 '18 at 1:50






  • 1




    $begingroup$
    It may be worth noting that we do know that $0$ is an eigenvalue for $C$. This is because $M^T:mathbb{R}^3 to mathbb{R}^2$ and thus has non-trivial kernel forcing $MM^T$ to have nontrivial kernel.
    $endgroup$
    – Dionel Jaime
    Dec 22 '18 at 4:54










  • $begingroup$
    @DionelJaime: right you are! Cheers!
    $endgroup$
    – Robert Lewis
    Dec 22 '18 at 4:56














1












1








1





$begingroup$

Our OP Future Math person has correctly argued that $A$ and $B$ are diagonalizable, $A$ over $Bbb C$ and $B$ over $Bbb R$; in each case for the reason that the matrix has $3$ distinct eigenvalues, hence $3$ linearly independent eigenvectors.



So what about the case



$C = MM^T? tag 1$



here we don't know too much about the eigenvalues, but we may observe that



$C^T = (MM^T)^T = (M^T)^TM^T = MM^T; tag 2$



that is, $C$ is a real symmetric matrix; as such, it too may be diagonalized, whether or not the eigenvalues are distinct; that real symmetric matrices are diagonalizable is a well-known result.






share|cite|improve this answer











$endgroup$



Our OP Future Math person has correctly argued that $A$ and $B$ are diagonalizable, $A$ over $Bbb C$ and $B$ over $Bbb R$; in each case for the reason that the matrix has $3$ distinct eigenvalues, hence $3$ linearly independent eigenvectors.



So what about the case



$C = MM^T? tag 1$



here we don't know too much about the eigenvalues, but we may observe that



$C^T = (MM^T)^T = (M^T)^TM^T = MM^T; tag 2$



that is, $C$ is a real symmetric matrix; as such, it too may be diagonalized, whether or not the eigenvalues are distinct; that real symmetric matrices are diagonalizable is a well-known result.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 22 '18 at 1:52

























answered Dec 22 '18 at 1:41









Robert LewisRobert Lewis

47.5k23067




47.5k23067








  • 1




    $begingroup$
    Ohh wow. Yes I didn't even realize it was a symmetric matrix. Thanks!
    $endgroup$
    – Future Math person
    Dec 22 '18 at 1:43










  • $begingroup$
    @FutureMathperson: you are most welcome! And thanks for the "acceptance"!
    $endgroup$
    – Robert Lewis
    Dec 22 '18 at 1:50






  • 1




    $begingroup$
    It may be worth noting that we do know that $0$ is an eigenvalue for $C$. This is because $M^T:mathbb{R}^3 to mathbb{R}^2$ and thus has non-trivial kernel forcing $MM^T$ to have nontrivial kernel.
    $endgroup$
    – Dionel Jaime
    Dec 22 '18 at 4:54










  • $begingroup$
    @DionelJaime: right you are! Cheers!
    $endgroup$
    – Robert Lewis
    Dec 22 '18 at 4:56














  • 1




    $begingroup$
    Ohh wow. Yes I didn't even realize it was a symmetric matrix. Thanks!
    $endgroup$
    – Future Math person
    Dec 22 '18 at 1:43










  • $begingroup$
    @FutureMathperson: you are most welcome! And thanks for the "acceptance"!
    $endgroup$
    – Robert Lewis
    Dec 22 '18 at 1:50






  • 1




    $begingroup$
    It may be worth noting that we do know that $0$ is an eigenvalue for $C$. This is because $M^T:mathbb{R}^3 to mathbb{R}^2$ and thus has non-trivial kernel forcing $MM^T$ to have nontrivial kernel.
    $endgroup$
    – Dionel Jaime
    Dec 22 '18 at 4:54










  • $begingroup$
    @DionelJaime: right you are! Cheers!
    $endgroup$
    – Robert Lewis
    Dec 22 '18 at 4:56








1




1




$begingroup$
Ohh wow. Yes I didn't even realize it was a symmetric matrix. Thanks!
$endgroup$
– Future Math person
Dec 22 '18 at 1:43




$begingroup$
Ohh wow. Yes I didn't even realize it was a symmetric matrix. Thanks!
$endgroup$
– Future Math person
Dec 22 '18 at 1:43












$begingroup$
@FutureMathperson: you are most welcome! And thanks for the "acceptance"!
$endgroup$
– Robert Lewis
Dec 22 '18 at 1:50




$begingroup$
@FutureMathperson: you are most welcome! And thanks for the "acceptance"!
$endgroup$
– Robert Lewis
Dec 22 '18 at 1:50




1




1




$begingroup$
It may be worth noting that we do know that $0$ is an eigenvalue for $C$. This is because $M^T:mathbb{R}^3 to mathbb{R}^2$ and thus has non-trivial kernel forcing $MM^T$ to have nontrivial kernel.
$endgroup$
– Dionel Jaime
Dec 22 '18 at 4:54




$begingroup$
It may be worth noting that we do know that $0$ is an eigenvalue for $C$. This is because $M^T:mathbb{R}^3 to mathbb{R}^2$ and thus has non-trivial kernel forcing $MM^T$ to have nontrivial kernel.
$endgroup$
– Dionel Jaime
Dec 22 '18 at 4:54












$begingroup$
@DionelJaime: right you are! Cheers!
$endgroup$
– Robert Lewis
Dec 22 '18 at 4:56




$begingroup$
@DionelJaime: right you are! Cheers!
$endgroup$
– Robert Lewis
Dec 22 '18 at 4:56


















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%2f3049047%2fdiagonalizable-matricies-and-eigenvalues%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

Wiesbaden

Image
Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
Read more

Marschland

Image
Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
Read more

Dieringhausen

Image
Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
Read more