## Diagonalizable matrix

Hello,

What steps do we follow to calculate the diagonalize matrix of 3x3

(1) If we get a matrix of 3x3 do we first calculate the eigenvalues of lambda1,l2 and l3?

(2) Then form a matrix with the eigen vectors?

(3) Then apply P inv.AP?

Is that the way to calculate?

-- Shounak
 Hey shounakbhatta. The process you have described is correct but your diagonalization will be PDP^(-1) = A where A is your original matrix.

 Quote by shounakbhatta Hello, What steps do we follow to calculate the diagonalize matrix of 3x3 (1) If we get a matrix of 3x3 do we first calculate the eigenvalues of lambda1,l2 and l3? (2) Then form a matrix with the eigen vectors? (3) Then apply P inv.AP? Is that the way to calculate? -- Shounak
Find the eigenvalues and put that eigenvalues as diagonal elements (preferably in increasing order). And you are done. It worked for the almost every problem that i worked. please inform whether there is any exception to it...

## Diagonalizable matrix

If you already found the eigenvalues, then it is unnecessary to calculate (P^(-1))AP, for the result will be the diagonal matrix D with the eigenvalues on the main diagonal, which can just be written down without calculations.

With "diagonalization" it is ususally meant that A should be written in terms of D, that is: A=PD(P^(-1)). Note that this is equivalent to D=(P^(-1))AP, but it is the other form that is usually required. Still, it could be useful to calculate D=(P^(-1))AP to check the answer, so that no mistake was made in earlier calculations.
 Hello Sreerajit, Thanks for the wonderful help. Yes, it worked out. Thanks all for the help.
 Well, In that case where do we need to calculate the eigenvectors or do we don't need at all?

 Quote by shounakbhatta Well, In that case where do we need to calculate the eigenvectors or do we don't need at all?
It depends. If you only need the diagonal matrix, you don't need the eigenvectors. But in most cases, one also wants a basis of eigenvectors to obtain the connection between A and D: A=PD(P(^1)), where the columns in P are eigenvectors. One cannot use diagonalization to very much if one does not know both D and P.
 So, to summarize if I have separate eigen values I can just put them diagonally and get the diagonal matrix. If any lambda's value is a multiplicity then I need to find the corresponding eigen vectors and see whether the eigen vectors are different, if they are different, then it is diagonalizable, otherwsie not? Am I right?
 If I do have a eignvalue which has multiplicity, then how to know that the matrix is diagonalizable or not?

 Quote by shounakbhatta So, to summarize if I have separate eigen values I can just put them diagonally and get the diagonal matrix. If any lambda's value is a multiplicity then I need to find the corresponding eigen vectors and see whether the eigen vectors are different, if they are different, then it is diagonalizable, otherwsie not? Am I right?
For an eigenvalue with multiplicity > 1 (as a root of the characteristic equation), you must find the eigenvectors to this eigenvalue and check if the subspace they consitute (the eigenspace) has the same dimension as the multiplicity. If it is so for all (multiple) eigenvalues, then the matrix is diagonalizable, otherwise not. (For a simple eigevalue, with multiplicity 1, the eigenspace will always have dimension 1.)

Sorry I was a little imprecise in my previous post. I did not consider multiple eigenvalues there.
 Dimension of eigenspace means?

 Quote by shounakbhatta Dimension of eigenspace means?
Let l be an eigenvalue of an n x n-matrix A. The set of all eigenvectors belonging to l, together with the zero vector, is a subspace of R^n (or C^n, if we allow complex elements in vectors and matrices). This subspace is called the eigenspace belonging to l, and, as a subspace of R^n, it has a dimension.

What you do is that you solve the homogenous system (A-lI)x=0. The (nontrivial) solutions to this are the eigenvectors belonging to l. The eigenspace is the null space of A-lI, and if you use Gauss-Jordan elimination to find the solution, this gives a basis for this nullspace, i.e. a maximal linearly independent set of eigenvectors, and the number of vectors in the basis is the dimension of the eigenspace.

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 Quote by sreerajt Find the eigenvalues and put that eigenvalues as diagonal elements (preferably in increasing order). And you are done. It worked for the almost every problem that i worked. please inform whether there is any exception to it...
The exception is if your characteristic polynomial is inseparable that is it has repeated roots. In this case you can not diagonalise the matrix, but you can get it into a canonical block diagonal form known as the Jordan form using what are known as generalized eigenvectors.
 Thank you Jim. Thanks for the help.

 Quote by Jim Kata The exception is if your characteristic polynomial is inseparable that is it has repeated roots. In this case you can not diagonalise the matrix, but you can get it into a canonical block diagonal form known as the Jordan form using what are known as generalized eigenvectors.
The matrix can be diagonalizable even if the characteristic equation has repeated roots. This is the case if the dimension of each eigenspace is the same as the multiplicity of the corresponding root.
But it might not be so, and then the Jordan block matrix is not a diagonal matrix.
 Hello Erland, It would be very kind of you if you please post an example to illustrate. Thanks, -- Shounak

 Quote by shounakbhatta Hello Erland, It would be very kind of you if you please post an example to illustrate. Thanks, -- Shounak
The problem is that I don't know how to write matrices in TeX-notation here...

But one can always borrow from others. Look up the examples here:

http://tutorial.math.lamar.edu/Class...alization.aspx