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












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1 Answer
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1 Answer
1






active

oldest

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active

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active

oldest

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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


















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