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

993818




993818












  • $begingroup$
    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.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

49k23168




49k23168








  • 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

Plaza Victoria

Image
Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
Read more

Puebla de Zaragoza

Image
Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
Read more

Musa

Image
Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
Read more