Diagonalizing singular matrices

More precisely, if \(d_{ij}\) is the \(ij^{th}\) entry of a diagonal matrix \(D\), then \(d_{ij}=0\) unless \(i=j\). 1 As you can argue by Spectral Theorem, hermitian matrices are always diagonalizable

2024-03-29
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  1. Step 4: Determine linearly independent eigenvectors
  2. Let A be an n × n matrix with the characteristic polynomial
  3. Then, A k is also easy to compute
  4. Construct the Matrix \(P\) The matrix \(P\) will have the
  5. So the
  6. Let A, B A, B be n × n n
  7. The matrix B = S−1AS has the eigenvalues in the diagonal
  8. Unit 6 Two-variable inequalities
  9. Singular Matrix
  10. 3
  11. 1 The matrix of a linear transformation
  12. 4
  13. Answer
  14. 0