TR3.5

Diagonal Matrix and Diagonalizability

Taproot 2021-09-27 Mon 11:51

1 diagonal matrix   def

A diagonal matrix is a square matrix that is zero everywhere except possibly along the diagonal.

1.1 results

1.1.1 every diagonal matrix is upper triangular

2 diagonalizable   def

An operator \(T \in \mathcal{L} (V)\) is called diagonalizable if the operator has a diagonal matrix with respect to some basis of \(V\).

2.1 results

2.1.1 Axler5.41 conditions equivalent to diagonalizability

Suppose \(V\) is finite-dimensional and \(T \in \mathcal{L} (V)\). Let \(\lambda_1, \ldots, \lambda_m\) denote the distinct eigenvalues of \(T\). Then the following are equivalent:

  1. \(T\) is diagonalizable
  2. \(V\) has a basis consisting of eigenvalues of \(T\)
  3. there exist 1-dimensional subspaces \(U_1, \ldots, U_n\) of \(V\), each invariant under \(T\), s.t.

\[\begin{aligned} V = U_1 \oplus \cdots \oplus U_n \end{aligned}\]

  1. \(V = E(\lambda_1, T) \oplus \cdots \oplus E(\lambda_m, T)\) (\(V\) is the (direct) sum of eigenspaces)
  2. \(\odim V = \odim E(\lambda_1, T) + \cdots + \odim E(\lambda_m, T)\)

2.1.2 Axler5.44 Enough eigenvalues implies diagonalizability

If \(T\in \mathcal{L} (V)\) has \(\odim V\) distinct eigenvalues, then \(T\) is diagonalizable.

  1. intuition

    Because distinct eigenvalues correspond to linearly independent eigenvectors, so there will be enough linearly independent eigenvecs to form a basis and thus a diagonal matrix.

    In fact, we just need the geometric multiplicities to add up (a result Axler promises in later chapters)

2.1.3 Relationship to non-diagonal matrix (in class 31 March 2021)

Suppose \(A\) is the original map (not diagonal), and that \(P\) is the matrix where each column is an eigenvector written in terms of the original basis (standard basis, usually). Then \[\begin{aligned} AP = PD \end{aligned}\] where \(D\) is the diagonal matrix.

  1. this (or a conjugation??) forms a similarity transform, which is a type of equivalence relation