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Can a non-diagonal 2x2 matrix with just one eigenvalue be diagonalizable?

A matrix with at least one (real) non-zero eigenvalueDoes non-Hermitian implies at least one complex eigenvalue?How to find 2x2 matrix with non zero elements and repeated eigenvalues?Does a diagonal matrix commute with every other matrix of the same size?Matrix with non-negative eigenvaluesDiagonalisability of 2×2 matrices with repeated eigenvaluesIs it always possible to make a non-diagonalizable matrix diagonalizable by row scaling only?To make a special non-normal matrix diagonalizableDo any even-order square non-diagonalizable matrices with strictly positive entries exist?Eigenvalues of the product of one diagonal and one regular matrix

3

1

$begingroup$

I know that some (all?) diagonal 2x2 matrices can still be diagonalisable if they have just one repeated eigenvalue, but I'm wondering if it's safe to claim that a non-diagonal matrix with just one repeated eigenvalue is definitely not diagonalisable?

I'm working in SL(2,F) with F an algebraically closed field if that makes any difference.

Thanks!

linear-algebra group-theory eigenvalues-eigenvectors diagonalization

share|cite|improve this question

edited 8 hours ago

Chris Butler

asked 8 hours ago

Chris ButlerChris Butler

162

New contributor

Chris Butler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you mean "a non-diagonal matrix .... is definitely not diagonalizable?". Also do you know about Jordan Normal Form?
  $endgroup$
  – Sean Haight
  8 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  For 2x2s the only matrices with a single two dimensional eigenspace are multiples of the identity.
  $endgroup$
  – Ian
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The question in the title seems clear, and the answer is no, such a matrix cannot be diagonalizable.
  $endgroup$
  – Derek Holt
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, sorry for not being clear enough. Is there a simple way of proving this? ie proving that a 2x2 matrix with a single two-dimensional eigenspace must be a multiple of the identity?
  $endgroup$
  – Chris Butler
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Suppose $A$ is diagonalizable, and has only one eigenvalue $lambda$. Then, you can find invertible matrix $P$ and a diagonal matrix $D = lambda I$ such that $A = P(lambda I)P^{-1} = lambda I$. Hence $A$ has to be a multiple of the identity matrix, the multiple being the eigenvalue
  $endgroup$
  – peek-a-boo
  8 hours ago
  
  
  

  
  
  
  

add a comment | 

3

1

$begingroup$

I know that some (all?) diagonal 2x2 matrices can still be diagonalisable if they have just one repeated eigenvalue, but I'm wondering if it's safe to claim that a non-diagonal matrix with just one repeated eigenvalue is definitely not diagonalisable?

I'm working in SL(2,F) with F an algebraically closed field if that makes any difference.

Thanks!

linear-algebra group-theory eigenvalues-eigenvectors diagonalization

share|cite|improve this question

edited 8 hours ago

Chris Butler

asked 8 hours ago

Chris ButlerChris Butler

162

New contributor

Chris Butler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Do you mean "a non-diagonal matrix .... is definitely not diagonalizable?". Also do you know about Jordan Normal Form?
  $endgroup$
  – Sean Haight
  8 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  For 2x2s the only matrices with a single two dimensional eigenspace are multiples of the identity.
  $endgroup$
  – Ian
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The question in the title seems clear, and the answer is no, such a matrix cannot be diagonalizable.
  $endgroup$
  – Derek Holt
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks, sorry for not being clear enough. Is there a simple way of proving this? ie proving that a 2x2 matrix with a single two-dimensional eigenspace must be a multiple of the identity?
  $endgroup$
  – Chris Butler
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Suppose $A$ is diagonalizable, and has only one eigenvalue $lambda$. Then, you can find invertible matrix $P$ and a diagonal matrix $D = lambda I$ such that $A = P(lambda I)P^{-1} = lambda I$. Hence $A$ has to be a multiple of the identity matrix, the multiple being the eigenvalue
  $endgroup$
  – peek-a-boo
  8 hours ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

I know that some (all?) diagonal 2x2 matrices can still be diagonalisable if they have just one repeated eigenvalue, but I'm wondering if it's safe to claim that a non-diagonal matrix with just one repeated eigenvalue is definitely not diagonalisable?

I'm working in SL(2,F) with F an algebraically closed field if that makes any difference.

Thanks!

linear-algebra group-theory eigenvalues-eigenvectors diagonalization

share|cite|improve this question

edited 8 hours ago

Chris Butler

asked 8 hours ago

Chris ButlerChris Butler

162

New contributor

Chris Butler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited 8 hours ago

Chris Butler

asked 8 hours ago

Chris ButlerChris Butler

162

New contributor

Chris Butler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited 8 hours ago

Chris Butler

asked 8 hours ago

Chris ButlerChris Butler

162

New contributor

Chris Butler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 8 hours ago

Chris ButlerChris Butler

162

New contributor

Chris Butler is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked 8 hours ago

Chris ButlerChris Butler

162

3 Answers
3

active

oldest

votes

4

$begingroup$

If a linear transformation (at least for a finite-dimensional vector space) has only a single eigenvalue and is diagonalizable, then in some basis (and therefore all bases) it is represented by a multiple of the identity matrix.

So a non-diagonal matrix over an algebraically closed field with only a single eigenvalue cannot be diagonalizable.

share|cite|improve this answer

answered 8 hours ago

ArthurArthur

128k7124214

$endgroup$

add a comment | 

1

$begingroup$

Take for example

$$;A=begin{pmatrix}0&-1\1&;2end{pmatrix}implies p_A(t)=t^2-2t+1=(t-1)^2;$$

If this matrix were diagonalizable then its minimal polynomial would be $;t-1;$, which clearly is false.

Thus, in general, if a matrix $;B;$ has a characteristic polynomial of the form $;(t-lambda)^2;$ , then it is diagonalizable iff $;t-lambda;$ is its minimal polynomial, which would mean that in fact

$$B=begin{pmatrix}lambda&0\0&lambdaend{pmatrix}=lambda I$$

meaning $;B;$ is a scalar multiple of the identity matrix. This is in fact true, mutatis mutandis, for any square matrix with one single eigenvalue (assuming we're on an algebraically closed field)

share|cite|improve this answer

answered 8 hours ago

DonAntonioDonAntonio

182k1497234

$endgroup$

add a comment | 

1

$begingroup$

Suppose $A$ is diagonalizable, and has only one eigenvalue $lambda$. Then, you can find invertible matrix $P$ and a diagonal matrix $D = lambda I$ such that $A = P(lambda I)P^{-1} = lambda I$. Hence $A$ has to be a multiple of the identity matrix, the multiple being the eigenvalue. (This holds for any $n times n$ matrix)

So we showed that diagonalizable and only one eigenvalue $implies$ diagonal. The negation is " diagonalizable with only one eigenvalue and non-diagonal"; this cannot happen.

share|cite|improve this answer

answered 8 hours ago

peek-a-boopeek-a-boo

1,62529

$endgroup$

add a comment | 

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4

$begingroup$

If a linear transformation (at least for a finite-dimensional vector space) has only a single eigenvalue and is diagonalizable, then in some basis (and therefore all bases) it is represented by a multiple of the identity matrix.

So a non-diagonal matrix over an algebraically closed field with only a single eigenvalue cannot be diagonalizable.

share|cite|improve this answer

answered 8 hours ago

ArthurArthur

128k7124214

$endgroup$

add a comment | 

4

$begingroup$

If a linear transformation (at least for a finite-dimensional vector space) has only a single eigenvalue and is diagonalizable, then in some basis (and therefore all bases) it is represented by a multiple of the identity matrix.

So a non-diagonal matrix over an algebraically closed field with only a single eigenvalue cannot be diagonalizable.

share|cite|improve this answer

answered 8 hours ago

ArthurArthur

128k7124214

$endgroup$

add a comment | 

$begingroup$

If a linear transformation (at least for a finite-dimensional vector space) has only a single eigenvalue and is diagonalizable, then in some basis (and therefore all bases) it is represented by a multiple of the identity matrix.

So a non-diagonal matrix over an algebraically closed field with only a single eigenvalue cannot be diagonalizable.

share|cite|improve this answer

answered 8 hours ago

ArthurArthur

128k7124214

$endgroup$

share|cite|improve this answer

answered 8 hours ago

ArthurArthur

128k7124214

answered 8 hours ago

ArthurArthur

128k7124214

answered 8 hours ago

ArthurArthur

128k7124214

1

$begingroup$

Take for example

$$;A=begin{pmatrix}0&-1\1&;2end{pmatrix}implies p_A(t)=t^2-2t+1=(t-1)^2;$$

If this matrix were diagonalizable then its minimal polynomial would be $;t-1;$, which clearly is false.

Thus, in general, if a matrix $;B;$ has a characteristic polynomial of the form $;(t-lambda)^2;$ , then it is diagonalizable iff $;t-lambda;$ is its minimal polynomial, which would mean that in fact

$$B=begin{pmatrix}lambda&0\0&lambdaend{pmatrix}=lambda I$$

meaning $;B;$ is a scalar multiple of the identity matrix. This is in fact true, mutatis mutandis, for any square matrix with one single eigenvalue (assuming we're on an algebraically closed field)

share|cite|improve this answer

answered 8 hours ago

DonAntonioDonAntonio

182k1497234

$endgroup$

add a comment | 

1

$begingroup$

Take for example

$$;A=begin{pmatrix}0&-1\1&;2end{pmatrix}implies p_A(t)=t^2-2t+1=(t-1)^2;$$

If this matrix were diagonalizable then its minimal polynomial would be $;t-1;$, which clearly is false.

Thus, in general, if a matrix $;B;$ has a characteristic polynomial of the form $;(t-lambda)^2;$ , then it is diagonalizable iff $;t-lambda;$ is its minimal polynomial, which would mean that in fact

$$B=begin{pmatrix}lambda&0\0&lambdaend{pmatrix}=lambda I$$

meaning $;B;$ is a scalar multiple of the identity matrix. This is in fact true, mutatis mutandis, for any square matrix with one single eigenvalue (assuming we're on an algebraically closed field)

share|cite|improve this answer

answered 8 hours ago

DonAntonioDonAntonio

182k1497234

$endgroup$

add a comment | 

$begingroup$

Take for example

$$;A=begin{pmatrix}0&-1\1&;2end{pmatrix}implies p_A(t)=t^2-2t+1=(t-1)^2;$$

If this matrix were diagonalizable then its minimal polynomial would be $;t-1;$, which clearly is false.

Thus, in general, if a matrix $;B;$ has a characteristic polynomial of the form $;(t-lambda)^2;$ , then it is diagonalizable iff $;t-lambda;$ is its minimal polynomial, which would mean that in fact

$$B=begin{pmatrix}lambda&0\0&lambdaend{pmatrix}=lambda I$$

meaning $;B;$ is a scalar multiple of the identity matrix. This is in fact true, mutatis mutandis, for any square matrix with one single eigenvalue (assuming we're on an algebraically closed field)

share|cite|improve this answer

answered 8 hours ago

DonAntonioDonAntonio

182k1497234

$endgroup$

share|cite|improve this answer

answered 8 hours ago

DonAntonioDonAntonio

182k1497234

answered 8 hours ago

DonAntonioDonAntonio

182k1497234

answered 8 hours ago

DonAntonioDonAntonio

182k1497234

1

$begingroup$

Suppose $A$ is diagonalizable, and has only one eigenvalue $lambda$. Then, you can find invertible matrix $P$ and a diagonal matrix $D = lambda I$ such that $A = P(lambda I)P^{-1} = lambda I$. Hence $A$ has to be a multiple of the identity matrix, the multiple being the eigenvalue. (This holds for any $n times n$ matrix)

So we showed that diagonalizable and only one eigenvalue $implies$ diagonal. The negation is " diagonalizable with only one eigenvalue and non-diagonal"; this cannot happen.

share|cite|improve this answer

answered 8 hours ago

peek-a-boopeek-a-boo

1,62529

$endgroup$

add a comment | 

1

$begingroup$

Suppose $A$ is diagonalizable, and has only one eigenvalue $lambda$. Then, you can find invertible matrix $P$ and a diagonal matrix $D = lambda I$ such that $A = P(lambda I)P^{-1} = lambda I$. Hence $A$ has to be a multiple of the identity matrix, the multiple being the eigenvalue. (This holds for any $n times n$ matrix)

So we showed that diagonalizable and only one eigenvalue $implies$ diagonal. The negation is " diagonalizable with only one eigenvalue and non-diagonal"; this cannot happen.

share|cite|improve this answer

answered 8 hours ago

peek-a-boopeek-a-boo

1,62529

$endgroup$

add a comment | 

$begingroup$

Suppose $A$ is diagonalizable, and has only one eigenvalue $lambda$. Then, you can find invertible matrix $P$ and a diagonal matrix $D = lambda I$ such that $A = P(lambda I)P^{-1} = lambda I$. Hence $A$ has to be a multiple of the identity matrix, the multiple being the eigenvalue. (This holds for any $n times n$ matrix)

So we showed that diagonalizable and only one eigenvalue $implies$ diagonal. The negation is " diagonalizable with only one eigenvalue and non-diagonal"; this cannot happen.

share|cite|improve this answer

answered 8 hours ago

peek-a-boopeek-a-boo

1,62529

$endgroup$

share|cite|improve this answer

answered 8 hours ago

peek-a-boopeek-a-boo

1,62529

answered 8 hours ago

peek-a-boopeek-a-boo

1,62529

answered 8 hours ago

peek-a-boopeek-a-boo

1,62529