Diagonalizability

A matrix is diagonalizable if:

1. it admits n linearly independent eigenvectors (an eigenbasis)
2. the matrix's eigenvalues have equal algebraic (# of times in the characteristic polynomial) and geometric multiplicities (number of linearly independent eigenvectors for a given eigenvalue)

Def: Eigenvector

If T: V$\to$V is a linear operator and eigenvector of T is a non-zero v$\in$V s.t T(v)$=\lambda$v and $\lambda$ is its associated eigenvalue.

Def: determinant, trace, characteristic polynomial of a linear transformation

V is a fdvs with basis B and T is a linear operator.

1. det($T$) = det($[T{]}_B$)
2. tr($T$) = tr($[T{]}_B$)
3. $p_T(\lambda )=$ det($[T-\lambda \cdot id{]}_B$)

Thm: These do not depend on Basis

pf uses change of basis
example det$(PA{P}^{-1})=$ det($A$)

Thm: Isomorphism and Determinant

$⟹$
Isomorphism then $T^1\circ T=I$
det(I) = 1 so det(T)$\times \frac{1}{det(T)}$=1
which is not possible if det(T) = 0
$⟸$
det(T)=0 $⟹$ $[T]$ is invertible
$[T^{-1}{]}_B[T{]}_B=I_n$
$C_B^{-1}[T^{-1}oT{]}_B=Iv$

Thm: Eigenvectors and Basis

If V is fdvs and T: V$\to$V is linear:

v$\in$V is an eigenvector $⟺$for any basis B $[\text{v}{]}_B$ is an eigenvector of $[T{]}_B$

Thm: Eigenvectors and Diagonizability

T: V$\to$V is diagonalizable $⟺$ there is a basis B of eigenvectors of T

$⟸$
$[\text{T}{]}_B=[\text{T}(b_i){]}_{i=1}^n=\left[\begin{array}{ccc}{\lambda }_i& & \\ & \ddots \\ & & {\lambda }_n\end{array}\right]$