MHB Diagonalizability of Linear Transformation

[Sudharaka]

Hi everyone, :)

Here's a question I was stuck on. Hope you people can help me out. :)

Prove that \(f:\,V\rightarrow V\) over \(\mathbb{C}\) is diagonalizable iff all root vectors are eigenvectors.

The definition of root vectors is given >>here<<.

Now a \(n\times n\) matrix can be diagonalized if it has \(n\) distinct eigenvalues. So I don't see how the given condition (all root vectors are eigenvectors) can ensure that there are \(n\) distinct eigenvalues.

 

[I like Serena]

Hi everyone, :)

Here's a question I was stuck on. Hope you people can help me out. :)
The definition of root vectors is given >>here<<.

Now a \(n\times n\) matrix can be diagonalized if it has \(n\) distinct eigenvalues. So I don't see how the given condition (all root vectors are eigenvectors) can ensure that there are \(n\) distinct eigenvalues.

Hi Sudharaka!

The eigenvalues do not have to be distinct for a matrix to be diagonalizable.
(It is a sufficient condition though.)
Consider for instance the identity matrix. For instance in 2 dimensions:
\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}
All its eigenvalues are 1 and it is a diagonal matrix.
It has a unique eigenspace of dimension 2 (edited).

The difference with a root space is for instance visible in:
\begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}
This matrix also has 1 as its eigenvalue, but the corresponding eigenspace is 1-dimensional, as opposed to the root space which is 2-dimensional.
They say that the eigenvalue 1 has geometric multiplicity 1 and algebraic multiplicity 2.

 

[caffeinemachine]

Hi everyone, :)

Here's a question I was stuck on. Hope you people can help me out. :)
The definition of root vectors is given >>here<<.

Now a \(n\times n\) matrix can be diagonalized if it has \(n\) distinct eigenvalues. So I don't see how the given condition (all root vectors are eigenvectors) can ensure that there are \(n\) distinct eigenvalues.

Hey Sudharaka!
I cannot preview the definition of 'root vectors' in the link you have provided. Can you write the definition out here?

 

[Sudharaka]

Hi Sudharaka!

The eigenvalues do not have to be distinct for a matrix to be diagonalizable.
(It is a sufficient condition though.)
Consider for instance the identity matrix. For instance in 2 dimensions:
\begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}
All its eigenvalues are 1 and it is a diagonal matrix.
It has a unique eigenspace of dimension 2 (edited).

The difference with a root space is for instance visible in:
\begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}
This matrix also has 1 as its eigenvalue, but the corresponding eigenspace is 1-dimensional, as opposed to the root space which is 2-dimensional.
They say that the eigenvalue 1 has geometric multiplicity 1 and algebraic multiplicity 2.

Thanks very much for the reply. :) So we cannot use the condition "a n×n matrix can be diagonalized if it has n distinct eigenvalues" since it's only a sufficient condition. Okay, I'll try think about a different approach. Back to square one. :p

Hey Sudharaka!
I cannot preview the definition of 'root vectors' in the link you have provided. Can you write the definition out here?

Hi caffeinemachine, :)

Thanks very much for the interest in answering my question. :)

It defines the root space, the set of all root vectors. Here's the definition,

Definition: The set \(S_i\) of all vectors \(x\in V\) defined by,

\[S_i=\{x\in V\,:\,(\lambda_i I_v-L)^{m_i}x=0\}\]

where \(i=1,2,\cdots ,k\).

\(m_i\) is the multiplicity of the eigenvalue \(\lambda_i\) in the minimum polynomial \(\psi(\lambda)\) of \(L\), is called the root space \(L\) associated with \(\lambda_i\). In particular, when \(m_i=1\), the root space \(S_i\) becomes the eigenspace \(E_i\) of \(L\) associated with the eigenvalue \(\lambda_i\).

 

[caffeinemachine]

Thanks very much for the reply. :) So we cannot use the condition "a n×n matrix can be diagonalized if it has n distinct eigenvalues" since it's only a sufficient condition. Okay, I'll try think about a different approach. Back to square one. :p
Hi caffeinemachine, :)

Thanks very much for the interest in answering my question. :)

It defines the root space, the set of all root vectors. Here's the definition,

Definition: The set \(S_i\) of all vectors \(x\in V\) defined by,

\[S_i=\{x\in V\,:\,(\lambda_i I_v-L)^{m_i}x=0\}\]

where \(i=1,2,\cdots ,k\).

\(m_i\) is the multiplicity of the eigenvalue \(\lambda_i\) in the minimum polynomial \(\psi(\lambda)\) of \(L\), is called the root space \(L\) associated with \(\lambda_i\). In particular, when \(m_i=1\), the root space \(S_i\) becomes the eigenspace \(E_i\) of \(L\) associated with the eigenvalue \(\lambda_i\).

Let $\lambda_1,\ldots,\lambda_k$ be all the distinct eigenvalues of an operator $L$ on a finite dimensional complex vector space $V$, with $\lambda_i$ having multiplicity (algebraic) equal to $m_i$.

Then, you may be already knowing that:

$V=\displaystyle \bigoplus_{i=1}^k\text{ null}(L-\lambda_iI)^{m_i}$.

Now if all the root vectors are eigenvectors, then, from the above equation it is easy to see that there is a basis of $V$ consisting only of eigenvectors of $L$. Thus $L$ is diagonalizable.

Try proving the other direction. :)

 

[Sudharaka]

Let $\lambda_1,\ldots,\lambda_k$ be all the distinct eigenvalues of an operator $L$ on a finite dimensional complex vector space $V$, with $\lambda_i$ having multiplicity (algebraic) equal to $m_i$.

Then, you may be already knowing that:

$V=\displaystyle \bigoplus_{i=1}^k\text{ null}(L-\lambda_iI)^{m_i}$.

Now if all the root vectors are eigenvectors, then, from the above equation it is easy to see that there is a basis of $V$ consisting only of eigenvectors of $L$. Thus $L$ is diagonalizable.

Try proving the other direction. :)

Thanks very much for the useful idea about solving this problem. May I also know about the book that you refer for Linear Algebra. :)

 

[caffeinemachine]

Thanks very much for the useful idea about solving this problem. May I also know about the book that you refer for Linear Algebra. :)

Thanks Sudharaka. I use Axler's Linear Algebra Done Right ​for linear algebra.

 

[I like Serena]

Prove that \(f:\,V\rightarrow V\) over \(\mathbb{C}\) is diagonalizable iff all root vectors are eigenvectors.

Not sure what you can use.
Here's another method.

Put f in Jordan Normal form.
The size of each Jordan block corresponds with the multiplicity of the corresponding root space.
If they are all 1, that means each Jordan block has size 1.
In other words, we have a diagonal matrix.
If they are not, we can't diagonalize it.