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Diagnolization of a matrix

Let us consider a matrix \(A,\) and there exists an invertible matrix \(S,\) such that \(S^{-1}AS=B,\) where \(B\) is a diagonal matrix. Then \(A\) is said to be diagonalizable.

Relation between Diagonalization and EigenVectors

  • Claim: If eigenvalues are distinct then eigenvectors are linearly independent

Observations

  1. For a given matrix \(A,\) \(S\) is not unique. This is because the eigenvectors can be scaled by any factor.
  2. Using the power property from eigenvectors and eigenvalues, we can get \(S^{-1}A^kS=B^k\)
  3. Not all matrices are diagonalisable.

A quick tip

If you ponder upon the process, you can see that you \(S\) is a set of all the eigen vectors, i.e., \(S=\begin{bmatrix}x_1\ x_2\ x_3\ ...\ x_n\end{bmatrix}\)

And \(B\) is just a diagonal matrix with all its diagonal elements as eigen values, i.e., \(B=\lambda I,\) where \(\lambda=[\lambda_1\ \lambda_2\dots\ \lambda_n],\) where \(\lambda_1\) is the eigen value corresponding to eigen vector \(x_1\) and so on.