A matrix is similar to a diagonal matrix if and only if there is a basis of eigenvectors for the associated linear transformation

Let F be a field, V a finite dimensional vector space over F, and \varphi : V \rightarrow V a linear transformation.

  1. Prove that if V has a basis B consisting of eigenvectors for \varphi, then the matrix M^B_B(\varphi) is diagonal and moreover the diagonal entries are eigenvalues.
  2. If E is a basis for V, then M^E_E(\varphi) is similar to a diagonal matrix D (having nonzero diagonal entries) if and only if there is a basis B for V consisting of eigenvectors of \varphi.

  1. Let B = \{v_i\}_{i=1}^n be a basis of eigenvectors for \varphi. Suppose that for each i, \lambda_i is the eigenvalue associated to v_i. That is, \varphi(v_i) = \lambda_i v_i. In particular, we have M^B_B(\varphi) = [a_{i,j}], where a_{i,j} = \lambda_i if j = i and 0 otherwise. So M^B_B(\varphi) is diagonal, with eigenvalues as diagonal entries.
  2. Suppose M^E_E(\varphi) is similar to a diagonal matrix D; say P M^E_E(\varphi) P^{-1} = D. Let \lambda_i be the diagonal entries of D. We may take the columns of P to be the coordinates (in terms of E) of a basis B = \{v_i\} for V. Then D = P M^E_E(\varphi) P^{-1} = M^B_B(\varphi) is the matrix of \varphi with respect to the basis B. In particular, \varphi(v_i) = \lambda_i v_i, so that each v_i is an eigenvector of \varphi.

    Conversely, if there is a basis B of V consisting of eigenvectors, then by part 1, M^B_B(\varphi) is a diagonal matrix. For any other basis E, we have an invertible change of basis matrix P such that P M^E_E(\varphi) P^{-1} = M^B_B(\varphi).

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