Given square real matrix $A$ with $det(A) = 108$ and $(A-2I)(A^2-9I)=0$, is $A$ normal?

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Given real square matrix $A$ with $det (A) = 108$ and $(A-2I)(A^2-9I) = 0$, find:



a. The characteristic and minimal polynomial options (all options).



b. Is $A$ normal?




I think I found the minimal and characteristic polynomials but I can't tell if $A$ is normal.




linear-algebra matrices minimal-polynomials




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  • 1




    What are the polynomials you have found?
    – uniquesolution
    Aug 13 at 9:50










  • @uniquesolution $(t-2)^4(t-3)^2(t+3)$ and $(t-2)(t-3)(t+3)$
    – Jason
    Aug 13 at 10:10










  • Are you assuming that $A$ is a $7times 7$ matrix?
    – uniquesolution
    Aug 13 at 10:13










  • Thats the only way the constant matches the determinant I think
    – Jason
    Aug 13 at 10:14






  • 2




    Determinant is product of eigenvalues and $108 = 2^2cdot 3^3$. Since polynomial $(x-2)(x-3)(x+3)$ vanishes at $A$, we know that $sigma(A)subseteq2,3,-3$, thus possible characteristic polynomials are $(x-2)^2(x-3)^3$ and $(x-2)^2(x-3)(x+3)^2$ with corresponding minimal polynomials $(x-2)(x-3)$ and $(x-2)(x-3)(x+3)$.
    – Ennar
    Aug 13 at 10:27














up vote
1
down vote

favorite













Given real square matrix $A$ with $det (A) = 108$ and $(A-2I)(A^2-9I) = 0$, find:



a. The characteristic and minimal polynomial options (all options).



b. Is $A$ normal?




I think I found the minimal and characteristic polynomials but I can't tell if $A$ is normal.




linear-algebra matrices minimal-polynomials




share|cite|improve this question


















  • 1




    What are the polynomials you have found?
    – uniquesolution
    Aug 13 at 9:50










  • @uniquesolution $(t-2)^4(t-3)^2(t+3)$ and $(t-2)(t-3)(t+3)$
    – Jason
    Aug 13 at 10:10










  • Are you assuming that $A$ is a $7times 7$ matrix?
    – uniquesolution
    Aug 13 at 10:13










  • Thats the only way the constant matches the determinant I think
    – Jason
    Aug 13 at 10:14






  • 2




    Determinant is product of eigenvalues and $108 = 2^2cdot 3^3$. Since polynomial $(x-2)(x-3)(x+3)$ vanishes at $A$, we know that $sigma(A)subseteq2,3,-3$, thus possible characteristic polynomials are $(x-2)^2(x-3)^3$ and $(x-2)^2(x-3)(x+3)^2$ with corresponding minimal polynomials $(x-2)(x-3)$ and $(x-2)(x-3)(x+3)$.
    – Ennar
    Aug 13 at 10:27












up vote
1
down vote

favorite









up vote
1
down vote

favorite












Given real square matrix $A$ with $det (A) = 108$ and $(A-2I)(A^2-9I) = 0$, find:



a. The characteristic and minimal polynomial options (all options).



b. Is $A$ normal?




I think I found the minimal and characteristic polynomials but I can't tell if $A$ is normal.




linear-algebra matrices minimal-polynomials




share|cite|improve this question















Given real square matrix $A$ with $det (A) = 108$ and $(A-2I)(A^2-9I) = 0$, find:



a. The characteristic and minimal polynomial options (all options).



b. Is $A$ normal?




I think I found the minimal and characteristic polynomials but I can't tell if $A$ is normal.





linear-algebra matrices minimal-polynomials





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 13 at 12:15









Rodrigo de Azevedo

12.6k41751




12.6k41751










asked Aug 13 at 8:56









Jason

1,12711327




1,12711327







  • 1




    What are the polynomials you have found?
    – uniquesolution
    Aug 13 at 9:50










  • @uniquesolution $(t-2)^4(t-3)^2(t+3)$ and $(t-2)(t-3)(t+3)$
    – Jason
    Aug 13 at 10:10










  • Are you assuming that $A$ is a $7times 7$ matrix?
    – uniquesolution
    Aug 13 at 10:13










  • Thats the only way the constant matches the determinant I think
    – Jason
    Aug 13 at 10:14






  • 2




    Determinant is product of eigenvalues and $108 = 2^2cdot 3^3$. Since polynomial $(x-2)(x-3)(x+3)$ vanishes at $A$, we know that $sigma(A)subseteq2,3,-3$, thus possible characteristic polynomials are $(x-2)^2(x-3)^3$ and $(x-2)^2(x-3)(x+3)^2$ with corresponding minimal polynomials $(x-2)(x-3)$ and $(x-2)(x-3)(x+3)$.
    – Ennar
    Aug 13 at 10:27












  • 1




    What are the polynomials you have found?
    – uniquesolution
    Aug 13 at 9:50










  • @uniquesolution $(t-2)^4(t-3)^2(t+3)$ and $(t-2)(t-3)(t+3)$
    – Jason
    Aug 13 at 10:10










  • Are you assuming that $A$ is a $7times 7$ matrix?
    – uniquesolution
    Aug 13 at 10:13










  • Thats the only way the constant matches the determinant I think
    – Jason
    Aug 13 at 10:14






  • 2




    Determinant is product of eigenvalues and $108 = 2^2cdot 3^3$. Since polynomial $(x-2)(x-3)(x+3)$ vanishes at $A$, we know that $sigma(A)subseteq2,3,-3$, thus possible characteristic polynomials are $(x-2)^2(x-3)^3$ and $(x-2)^2(x-3)(x+3)^2$ with corresponding minimal polynomials $(x-2)(x-3)$ and $(x-2)(x-3)(x+3)$.
    – Ennar
    Aug 13 at 10:27







1




1




What are the polynomials you have found?
– uniquesolution
Aug 13 at 9:50




What are the polynomials you have found?
– uniquesolution
Aug 13 at 9:50












@uniquesolution $(t-2)^4(t-3)^2(t+3)$ and $(t-2)(t-3)(t+3)$
– Jason
Aug 13 at 10:10




@uniquesolution $(t-2)^4(t-3)^2(t+3)$ and $(t-2)(t-3)(t+3)$
– Jason
Aug 13 at 10:10












Are you assuming that $A$ is a $7times 7$ matrix?
– uniquesolution
Aug 13 at 10:13




Are you assuming that $A$ is a $7times 7$ matrix?
– uniquesolution
Aug 13 at 10:13












Thats the only way the constant matches the determinant I think
– Jason
Aug 13 at 10:14




Thats the only way the constant matches the determinant I think
– Jason
Aug 13 at 10:14




2




2




Determinant is product of eigenvalues and $108 = 2^2cdot 3^3$. Since polynomial $(x-2)(x-3)(x+3)$ vanishes at $A$, we know that $sigma(A)subseteq2,3,-3$, thus possible characteristic polynomials are $(x-2)^2(x-3)^3$ and $(x-2)^2(x-3)(x+3)^2$ with corresponding minimal polynomials $(x-2)(x-3)$ and $(x-2)(x-3)(x+3)$.
– Ennar
Aug 13 at 10:27




Determinant is product of eigenvalues and $108 = 2^2cdot 3^3$. Since polynomial $(x-2)(x-3)(x+3)$ vanishes at $A$, we know that $sigma(A)subseteq2,3,-3$, thus possible characteristic polynomials are $(x-2)^2(x-3)^3$ and $(x-2)^2(x-3)(x+3)^2$ with corresponding minimal polynomials $(x-2)(x-3)$ and $(x-2)(x-3)(x+3)$.
– Ennar
Aug 13 at 10:27










1 Answer
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Hint: Normal matrix with real eigenvalues is Hermitian which in real case just means symmetric. From the minimal polynomial you know how Jordan normal form looks like. Can you find non-symmetric matrix with such a Jordan normal form?






share|cite|improve this answer
















  • 2




    Yea I can by multiplying with a non orthogonal invertable matrix. Thanks
    – Jason
    Aug 13 at 12:22










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active

oldest

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes








up vote
3
down vote



accepted










Hint: Normal matrix with real eigenvalues is Hermitian which in real case just means symmetric. From the minimal polynomial you know how Jordan normal form looks like. Can you find non-symmetric matrix with such a Jordan normal form?






share|cite|improve this answer
















  • 2




    Yea I can by multiplying with a non orthogonal invertable matrix. Thanks
    – Jason
    Aug 13 at 12:22














up vote
3
down vote



accepted










Hint: Normal matrix with real eigenvalues is Hermitian which in real case just means symmetric. From the minimal polynomial you know how Jordan normal form looks like. Can you find non-symmetric matrix with such a Jordan normal form?






share|cite|improve this answer
















  • 2




    Yea I can by multiplying with a non orthogonal invertable matrix. Thanks
    – Jason
    Aug 13 at 12:22












up vote
3
down vote



accepted







up vote
3
down vote



accepted






Hint: Normal matrix with real eigenvalues is Hermitian which in real case just means symmetric. From the minimal polynomial you know how Jordan normal form looks like. Can you find non-symmetric matrix with such a Jordan normal form?






share|cite|improve this answer












Hint: Normal matrix with real eigenvalues is Hermitian which in real case just means symmetric. From the minimal polynomial you know how Jordan normal form looks like. Can you find non-symmetric matrix with such a Jordan normal form?







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 13 at 11:51









Ennar

13.1k32343




13.1k32343







  • 2




    Yea I can by multiplying with a non orthogonal invertable matrix. Thanks
    – Jason
    Aug 13 at 12:22












  • 2




    Yea I can by multiplying with a non orthogonal invertable matrix. Thanks
    – Jason
    Aug 13 at 12:22







2




2




Yea I can by multiplying with a non orthogonal invertable matrix. Thanks
– Jason
Aug 13 at 12:22




Yea I can by multiplying with a non orthogonal invertable matrix. Thanks
– Jason
Aug 13 at 12:22












 

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