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Alternative definition of diagonalisable transformation

Linear map is diagonalizable iff its adjoint is diagonalizableDistinct eigenvalues and matrices problemHow to show $T$ is bijective in the following condition?A condition for two linear operators to commuteWhy doesn't this linear transformation exist?linear transformation of a basis where $T(v_1) = w_1, …, T(v_n) = w_n$Let $phi$ and $psi$ be two commutable diagonal endomorphisms. Then there is a basis consisting of eigenvectors of both $phi$ and $psi$.Proof on DiagonalizabilityFind the eingvalues of a Linear Transformationexistence of linear transformation doesn't hold

0

$begingroup$

Supposedly a transformation $T: V to V$ is diagonalisable iff there exists a basis of $V$ consisting only of eigenvectors of $T$. Can someone show me why this is true? I don't really know where to even start other than "Let $[v_1, ..., v_n]$ be a basis of $V$ with $v_i$ eigenvectors".

vector-spaces eigenvalues-eigenvectors linear-transformations diagonalization

share|cite|improve this question

asked Mar 11 at 8:24

cb7cb7

1426

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If you want to show that two definitions are equivalent, it's useful to include the other definition you are using - what is that?
  $endgroup$
  – StackTD
  Mar 11 at 8:26
  

  
  
  
  

add a comment | 

0

$begingroup$

Supposedly a transformation $T: V to V$ is diagonalisable iff there exists a basis of $V$ consisting only of eigenvectors of $T$. Can someone show me why this is true? I don't really know where to even start other than "Let $[v_1, ..., v_n]$ be a basis of $V$ with $v_i$ eigenvectors".

vector-spaces eigenvalues-eigenvectors linear-transformations diagonalization

share|cite|improve this question

asked Mar 11 at 8:24

cb7cb7

1426

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If you want to show that two definitions are equivalent, it's useful to include the other definition you are using - what is that?
  $endgroup$
  – StackTD
  Mar 11 at 8:26
  

  
  
  
  

add a comment | 

$begingroup$

Supposedly a transformation $T: V to V$ is diagonalisable iff there exists a basis of $V$ consisting only of eigenvectors of $T$. Can someone show me why this is true? I don't really know where to even start other than "Let $[v_1, ..., v_n]$ be a basis of $V$ with $v_i$ eigenvectors".

vector-spaces eigenvalues-eigenvectors linear-transformations diagonalization

share|cite|improve this question

asked Mar 11 at 8:24

cb7cb7

1426

$endgroup$

share|cite|improve this question

asked Mar 11 at 8:24

cb7cb7

1426

share|cite|improve this question

asked Mar 11 at 8:24

cb7cb7

1426

asked Mar 11 at 8:24

cb7cb7

1426

asked Mar 11 at 8:24

cb7cb7

1426

1 Answer
1

active

oldest

votes

2

$begingroup$

You have the right idea. If $[v_1, ..., v_n]$ is a basis of $V$ with $v_i$ eigenvectors, let us try to understand how $T$ would look in this basis (I'm assuming you already know how to find a representing matrix of a linear transofrmation, given some basis).
Since $T$ acts on the basis element $v_1$ by scaling it by a factor of $lambda_1$ (since it is an eigenvector), the first column of $T$ would be - $beginpmatrix lambda_1 \ 0 \ vdots \ 0 endpmatrix$. Similarly, the second column of $T$ would be - $beginpmatrix 0 \ lambda_2 \ vdots \ 0 endpmatrix$, and so on. Overall, you will get a diagonal matrix.

For the other way around - if $T$ is diagonal in some basis, you know that it acts on the first basis vector by scaling it by the factor on the corresponding part of the diagonal (again, I'm assuming you know the general method of finding a representing matrix, given some basis). Similarly, each basis vector is scaled by the corresponding part of the diagonal. Since $T$ acts on each basis vector by scaling it, we conclude that the basis consists only of eigenvectors of $T$.

share|cite|improve this answer

answered Mar 11 at 8:50

GSoferGSofer

664311

$endgroup$

add a comment | 

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2

$begingroup$

You have the right idea. If $[v_1, ..., v_n]$ is a basis of $V$ with $v_i$ eigenvectors, let us try to understand how $T$ would look in this basis (I'm assuming you already know how to find a representing matrix of a linear transofrmation, given some basis).
Since $T$ acts on the basis element $v_1$ by scaling it by a factor of $lambda_1$ (since it is an eigenvector), the first column of $T$ would be - $beginpmatrix lambda_1 \ 0 \ vdots \ 0 endpmatrix$. Similarly, the second column of $T$ would be - $beginpmatrix 0 \ lambda_2 \ vdots \ 0 endpmatrix$, and so on. Overall, you will get a diagonal matrix.

For the other way around - if $T$ is diagonal in some basis, you know that it acts on the first basis vector by scaling it by the factor on the corresponding part of the diagonal (again, I'm assuming you know the general method of finding a representing matrix, given some basis). Similarly, each basis vector is scaled by the corresponding part of the diagonal. Since $T$ acts on each basis vector by scaling it, we conclude that the basis consists only of eigenvectors of $T$.

share|cite|improve this answer

answered Mar 11 at 8:50

GSoferGSofer

664311

$endgroup$

add a comment | 

2

$begingroup$

You have the right idea. If $[v_1, ..., v_n]$ is a basis of $V$ with $v_i$ eigenvectors, let us try to understand how $T$ would look in this basis (I'm assuming you already know how to find a representing matrix of a linear transofrmation, given some basis).
Since $T$ acts on the basis element $v_1$ by scaling it by a factor of $lambda_1$ (since it is an eigenvector), the first column of $T$ would be - $beginpmatrix lambda_1 \ 0 \ vdots \ 0 endpmatrix$. Similarly, the second column of $T$ would be - $beginpmatrix 0 \ lambda_2 \ vdots \ 0 endpmatrix$, and so on. Overall, you will get a diagonal matrix.

For the other way around - if $T$ is diagonal in some basis, you know that it acts on the first basis vector by scaling it by the factor on the corresponding part of the diagonal (again, I'm assuming you know the general method of finding a representing matrix, given some basis). Similarly, each basis vector is scaled by the corresponding part of the diagonal. Since $T$ acts on each basis vector by scaling it, we conclude that the basis consists only of eigenvectors of $T$.

share|cite|improve this answer

answered Mar 11 at 8:50

GSoferGSofer

664311

$endgroup$

add a comment | 

$begingroup$

You have the right idea. If $[v_1, ..., v_n]$ is a basis of $V$ with $v_i$ eigenvectors, let us try to understand how $T$ would look in this basis (I'm assuming you already know how to find a representing matrix of a linear transofrmation, given some basis).
Since $T$ acts on the basis element $v_1$ by scaling it by a factor of $lambda_1$ (since it is an eigenvector), the first column of $T$ would be - $beginpmatrix lambda_1 \ 0 \ vdots \ 0 endpmatrix$. Similarly, the second column of $T$ would be - $beginpmatrix 0 \ lambda_2 \ vdots \ 0 endpmatrix$, and so on. Overall, you will get a diagonal matrix.

For the other way around - if $T$ is diagonal in some basis, you know that it acts on the first basis vector by scaling it by the factor on the corresponding part of the diagonal (again, I'm assuming you know the general method of finding a representing matrix, given some basis). Similarly, each basis vector is scaled by the corresponding part of the diagonal. Since $T$ acts on each basis vector by scaling it, we conclude that the basis consists only of eigenvectors of $T$.

share|cite|improve this answer

answered Mar 11 at 8:50

GSoferGSofer

664311

$endgroup$

share|cite|improve this answer

answered Mar 11 at 8:50

GSoferGSofer

664311

answered Mar 11 at 8:50

GSoferGSofer

664311

answered Mar 11 at 8:50

GSoferGSofer

664311