MHB Eigenvalues and Eigenvectors of a Non-Diagonalizable Matrix

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The matrix in question has eigenvalues 0 and 1. For the eigenvalue 0, the reduced system reveals a dimension of 2, but only one linearly independent eigenvector exists, specifically (1, 0, 0). This lack of sufficient independent eigenvectors indicates that the matrix cannot be diagonalized. Therefore, there are no matrices P and D such that D = P^-1 * A * P for this matrix. The conclusion is that the matrix is non-diagonalizable.
Yankel
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Hello,

sorry that I am asking too many questions, I am preparing for an exam...

I have a matrix,

0 1 0
0 0 0
0 0 1

and I need to say if it has a diagonal form (I mean, if there are P and D such that D=P^-1*D*P)

I found that the eigenvalues are 0 and 1. I also know that if I use 0, I get the system

0 1 0 0
0 0 0 0
0 0 0 0

(after Gaussian process)

What can I say about the eigenvectors, do they exist ? the eigenvalue 0 had a dimension of 2. so I need 2 eigenvectors in order to say that P and D exist...
 
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There is only one linearly independent eigenvector (1, 0, 0) corresponding to the eigenvalue of 0. So, the matrix is not diagonalizable.
 
Thread 'How to define a vector field?'

Thread 'How to define a vector field?'

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