Eigenvalues and Eigenvectors of a Non-Diagonalizable Matrix

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SUMMARY

The matrix provided, represented as [[0, 1, 0], [0, 0, 0], [0, 0, 1]], has eigenvalues 0 and 1. The eigenvalue 0 has a geometric multiplicity of 1, indicating only one linearly independent eigenvector, specifically (1, 0, 0). Consequently, the matrix cannot be diagonalized, as it lacks the requisite number of independent eigenvectors to form the matrices P and D in the equation D = P^-1 * A * P.

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  • Understanding of eigenvalues and eigenvectors
  • Familiarity with matrix diagonalization concepts
  • Knowledge of Gaussian elimination techniques
  • Basic linear algebra principles
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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.
 

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