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prove that $A$ is diagonalizable if $A^3-3A^2-A+3I_n = 0$ [on hold]

Hermitian Matrices are DiagonalizableA question about diagonalizable.Show that matrix $A$ is NOT diagonalizable.Prove that A … is not diagonalizableProve a matrix is not diagonalizableHow to use inner products in C(n) to prove normal matrix is unitarily diagonalizable after knowing that normal matrix is diagonalizable?Prove that $A$ is diagonalizable.Prove that a general matrix is diagonalizableDetermine $a$ to make matrix $A$ diagonalizableDiagonalizable block-diagonal matrix

-1

$begingroup$

We have :

$A^3-3A^2-A+3I_n = 0 $

how can i prove that A is diagonalizable .

I don't know how to do when A is written this way

linear-algebra matrices eigenvalues-eigenvectors diagonalization

share|cite|improve this question

edited Apr 5 at 2:23

user21820

40.1k544162

asked Apr 4 at 20:34

JoshuaKJoshuaK

305

$endgroup$

put on hold as off-topic by user21820, RRL, Saad, Cesareo, mrtaurho 2 days ago

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, RRL, Saad, Cesareo, mrtaurho

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Note that if $A=operatornameId_n$, then $A^3-3A^2-A+3operatornameId_n=0$, in spite of the fact that the only root of the characteristic polynomial of $operatornameId_n$ has multiplicity $n$. So, no, you don't have to prove that all roots of the characteristic polynomial of $A$ have multiplicity $1$.
  $endgroup$
  – José Carlos Santos
  Apr 4 at 20:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes , you're right i edit my mistake
  $endgroup$
  – JoshuaK
  Apr 4 at 20:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What does "when $A$ is written this way" mean?
  $endgroup$
  – anomaly
  Apr 5 at 0:42
  

  
  
  
  

add a comment | 

-1

$begingroup$

We have :

$A^3-3A^2-A+3I_n = 0 $

how can i prove that A is diagonalizable .

I don't know how to do when A is written this way

linear-algebra matrices eigenvalues-eigenvectors diagonalization

share|cite|improve this question

edited Apr 5 at 2:23

user21820

40.1k544162

asked Apr 4 at 20:34

JoshuaKJoshuaK

305

$endgroup$

put on hold as off-topic by user21820, RRL, Saad, Cesareo, mrtaurho 2 days ago

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, RRL, Saad, Cesareo, mrtaurho

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Note that if $A=operatornameId_n$, then $A^3-3A^2-A+3operatornameId_n=0$, in spite of the fact that the only root of the characteristic polynomial of $operatornameId_n$ has multiplicity $n$. So, no, you don't have to prove that all roots of the characteristic polynomial of $A$ have multiplicity $1$.
  $endgroup$
  – José Carlos Santos
  Apr 4 at 20:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  yes , you're right i edit my mistake
  $endgroup$
  – JoshuaK
  Apr 4 at 20:39
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What does "when $A$ is written this way" mean?
  $endgroup$
  – anomaly
  Apr 5 at 0:42
  

  
  
  
  

add a comment | 

$begingroup$

We have :

$A^3-3A^2-A+3I_n = 0 $

how can i prove that A is diagonalizable .

I don't know how to do when A is written this way

linear-algebra matrices eigenvalues-eigenvectors diagonalization

share|cite|improve this question

edited Apr 5 at 2:23

user21820

40.1k544162

asked Apr 4 at 20:34

JoshuaKJoshuaK

305

$endgroup$

share|cite|improve this question

edited Apr 5 at 2:23

user21820

40.1k544162

asked Apr 4 at 20:34

JoshuaKJoshuaK

305

share|cite|improve this question

edited Apr 5 at 2:23

user21820

40.1k544162

asked Apr 4 at 20:34

JoshuaKJoshuaK

305

edited Apr 5 at 2:23

user21820

40.1k544162

edited Apr 5 at 2:23

user21820

40.1k544162

asked Apr 4 at 20:34

JoshuaKJoshuaK

305

asked Apr 4 at 20:34

JoshuaKJoshuaK

305

3 Answers
3

active

oldest

votes

3

$begingroup$

The polynomial $P(X)=X^3-3X^2-X+3 = (X-1)(X-3)(X+1)$ has three distincts real roots and you have $P(A)=0$, so $A$ is diagonalizable over $mathbbR$.

share|cite|improve this answer

answered Apr 4 at 20:39

TheSilverDoeTheSilverDoe

5,455216

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  I think you should provide more explanation for how you go from "three [distinct] real roots" to "diagonalizable".
  $endgroup$
  – Acccumulation
  Apr 4 at 23:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Acccumulation In my country, when we learn things about diagonalization, one of the first criterias we learn is that "if there exists a polynomial $P$ with distincts simple zeros such that $P(A)=0$, then $A$ is diagonalizable". That's why I thought it was a well-known criteria... If you want an explanation, then one can say that the minimal polynomial has to divide this one, so it has also distinct simple zeros, and it is enough to conclude... no ?
  $endgroup$
  – TheSilverDoe
  Apr 5 at 8:14
  

  
  
  
  

add a comment | 

2

$begingroup$

Solving a simpler example, $A-cI_n=0$, it's clear that $a_(i,i)=c$ because $a_(i,j) - cI_(i,j) = 0$ for all $i,j in 1,dots,n$. From here, a slightly more complicated example is $(A-cI)(A-dI)=0$ forces two conditions (for diagonal elements of $A$ and off-diagonal elements of $A$) that will lead you to a solution for general matrix polynomials.

share|cite|improve this answer

answered Apr 4 at 20:44

EricEric

613

$endgroup$

add a comment | 

1

$begingroup$

We know that the minimal polynomial divides any polynomial that $A$ is a root of. It's pretty easy to guess that $x=1$ is a root, and using polynomial division you can find that the other two roots are $x=-1, x=3$. Since all the roots are of multiplicity $1$, all of the roots of the minimal polynomial are of multiplicity $1$, and so $A$ is diagonalizable.

share|cite|improve this answer

answered Apr 4 at 20:41

GSoferGSofer

8831314

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice way to do it using this prorpiety , i was wondering what can i say about A being inversible
  $endgroup$
  – JoshuaK
  Apr 4 at 21:00
  

  
  
  
  

add a comment | 

3

$begingroup$

The polynomial $P(X)=X^3-3X^2-X+3 = (X-1)(X-3)(X+1)$ has three distincts real roots and you have $P(A)=0$, so $A$ is diagonalizable over $mathbbR$.

share|cite|improve this answer

answered Apr 4 at 20:39

TheSilverDoeTheSilverDoe

5,455216

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  I think you should provide more explanation for how you go from "three [distinct] real roots" to "diagonalizable".
  $endgroup$
  – Acccumulation
  Apr 4 at 23:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Acccumulation In my country, when we learn things about diagonalization, one of the first criterias we learn is that "if there exists a polynomial $P$ with distincts simple zeros such that $P(A)=0$, then $A$ is diagonalizable". That's why I thought it was a well-known criteria... If you want an explanation, then one can say that the minimal polynomial has to divide this one, so it has also distinct simple zeros, and it is enough to conclude... no ?
  $endgroup$
  – TheSilverDoe
  Apr 5 at 8:14
  

  
  
  
  

add a comment | 

3

$begingroup$

The polynomial $P(X)=X^3-3X^2-X+3 = (X-1)(X-3)(X+1)$ has three distincts real roots and you have $P(A)=0$, so $A$ is diagonalizable over $mathbbR$.

share|cite|improve this answer

answered Apr 4 at 20:39

TheSilverDoeTheSilverDoe

5,455216

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  I think you should provide more explanation for how you go from "three [distinct] real roots" to "diagonalizable".
  $endgroup$
  – Acccumulation
  Apr 4 at 23:09
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Acccumulation In my country, when we learn things about diagonalization, one of the first criterias we learn is that "if there exists a polynomial $P$ with distincts simple zeros such that $P(A)=0$, then $A$ is diagonalizable". That's why I thought it was a well-known criteria... If you want an explanation, then one can say that the minimal polynomial has to divide this one, so it has also distinct simple zeros, and it is enough to conclude... no ?
  $endgroup$
  – TheSilverDoe
  Apr 5 at 8:14
  

  
  
  
  

add a comment | 

$begingroup$

The polynomial $P(X)=X^3-3X^2-X+3 = (X-1)(X-3)(X+1)$ has three distincts real roots and you have $P(A)=0$, so $A$ is diagonalizable over $mathbbR$.

share|cite|improve this answer

answered Apr 4 at 20:39

TheSilverDoeTheSilverDoe

5,455216

$endgroup$

share|cite|improve this answer

answered Apr 4 at 20:39

TheSilverDoeTheSilverDoe

5,455216

answered Apr 4 at 20:39

TheSilverDoeTheSilverDoe

5,455216

answered Apr 4 at 20:39

TheSilverDoeTheSilverDoe

5,455216

2

$begingroup$

Solving a simpler example, $A-cI_n=0$, it's clear that $a_(i,i)=c$ because $a_(i,j) - cI_(i,j) = 0$ for all $i,j in 1,dots,n$. From here, a slightly more complicated example is $(A-cI)(A-dI)=0$ forces two conditions (for diagonal elements of $A$ and off-diagonal elements of $A$) that will lead you to a solution for general matrix polynomials.

share|cite|improve this answer

answered Apr 4 at 20:44

EricEric

613

$endgroup$

add a comment | 

2

$begingroup$

Solving a simpler example, $A-cI_n=0$, it's clear that $a_(i,i)=c$ because $a_(i,j) - cI_(i,j) = 0$ for all $i,j in 1,dots,n$. From here, a slightly more complicated example is $(A-cI)(A-dI)=0$ forces two conditions (for diagonal elements of $A$ and off-diagonal elements of $A$) that will lead you to a solution for general matrix polynomials.

share|cite|improve this answer

answered Apr 4 at 20:44

EricEric

613

$endgroup$

add a comment | 

$begingroup$

Solving a simpler example, $A-cI_n=0$, it's clear that $a_(i,i)=c$ because $a_(i,j) - cI_(i,j) = 0$ for all $i,j in 1,dots,n$. From here, a slightly more complicated example is $(A-cI)(A-dI)=0$ forces two conditions (for diagonal elements of $A$ and off-diagonal elements of $A$) that will lead you to a solution for general matrix polynomials.

share|cite|improve this answer

answered Apr 4 at 20:44

EricEric

613

$endgroup$

share|cite|improve this answer

answered Apr 4 at 20:44

EricEric

613

answered Apr 4 at 20:44

EricEric

613

answered Apr 4 at 20:44

EricEric

613

1

$begingroup$

We know that the minimal polynomial divides any polynomial that $A$ is a root of. It's pretty easy to guess that $x=1$ is a root, and using polynomial division you can find that the other two roots are $x=-1, x=3$. Since all the roots are of multiplicity $1$, all of the roots of the minimal polynomial are of multiplicity $1$, and so $A$ is diagonalizable.

share|cite|improve this answer

answered Apr 4 at 20:41

GSoferGSofer

8831314

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice way to do it using this prorpiety , i was wondering what can i say about A being inversible
  $endgroup$
  – JoshuaK
  Apr 4 at 21:00
  

  
  
  
  

add a comment | 

1

$begingroup$

We know that the minimal polynomial divides any polynomial that $A$ is a root of. It's pretty easy to guess that $x=1$ is a root, and using polynomial division you can find that the other two roots are $x=-1, x=3$. Since all the roots are of multiplicity $1$, all of the roots of the minimal polynomial are of multiplicity $1$, and so $A$ is diagonalizable.

share|cite|improve this answer

answered Apr 4 at 20:41

GSoferGSofer

8831314

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Nice way to do it using this prorpiety , i was wondering what can i say about A being inversible
  $endgroup$
  – JoshuaK
  Apr 4 at 21:00
  

  
  
  
  

add a comment | 

$begingroup$

We know that the minimal polynomial divides any polynomial that $A$ is a root of. It's pretty easy to guess that $x=1$ is a root, and using polynomial division you can find that the other two roots are $x=-1, x=3$. Since all the roots are of multiplicity $1$, all of the roots of the minimal polynomial are of multiplicity $1$, and so $A$ is diagonalizable.

share|cite|improve this answer

answered Apr 4 at 20:41

GSoferGSofer

8831314

$endgroup$

share|cite|improve this answer

answered Apr 4 at 20:41

GSoferGSofer

8831314

answered Apr 4 at 20:41

GSoferGSofer

8831314

answered Apr 4 at 20:41

GSoferGSofer

8831314