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title: 8-6

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5.2

Review the box
Algebraic multiplicity bounds geometric multiplicity. Take a basis for the eigenspace. Extend.
Let $T$ be a linear operator and $\lambda_i$ be distinct eigenvalues. For each $\lambda_i$ , let $v_i\in E_{\lambda_i}$ , then if $\sum v_i=0$ , each $v_i$ is zero.
Unions of linearly independent subsets from eigenspaces are still linearly independent.
Diagonalizable if and only if algebraic multiplicity equals geometric multiplicity.
- Proof: Let $m_i$ be algebraic multiplicity, $d_i$ be geometric multiplicity, and $n=\dim V$ .
  Suppose diagonalizable. Then let B be a basis consisting of eigenvectors. Intersect this basis with each eigenspace. Call it $B_i$ . Let $n_i=B_i$ . Then $n\leq \sum n_i \leq \sum d_i \leq \sum m_i \leq n$ .
  So $\sum (m_i-d_i)=0$ . But each term must be nonnegative.
  Conversely, suppose the multiplicties are equal. Then take union of eigenbasis. They form a basis for full space.
Testing diagonalizability. Check splitting, check multiplicities. This also computes the diagonalization along the way.
What are eigenvalues of derivative?
Finite dimensional matrices always have eigenvalues if you extend the field.
Give example of infinite dimension transform with no eigenvalues.
Work through example 6 in book.

Direct sums