Diagonalizing

Definition

Suppose that $A$ has eigenvectors $v_1,\dots v_n\in {\mathbb{R}}^n$ that form a basis of ${\mathbb{R}}^n$. ($n$ linearly independent eigenvectors)

$$A=V\Lambda V^{-1}$$

where $V=[v_1,v_2,\dots ,v_n]$ are the eigenvectors $\Lambda =\left[\begin{array}{ccc}{\lambda }_1& 0& 0\\ 0& \ddots & 0\\ 0& 0& {\lambda }_n\end{array}\right]$ are the eigenvalues

Proof: $AV=\left[\begin{array}{ccccc}{\lambda }_1{v}_1& {\lambda }_2{v}_2& \dots & {\lambda }_n{v}_n& \end{array}\right]=\left[\begin{array}{cccc}{v}_1& v_2& \dots & v_n\end{array}\right]\left[\begin{array}{ccc}{\lambda }_1& 0& 0\\ 0& \ddots & 0\\ 0& 0& {\lambda }_n\end{array}\right]=V\Lambda$

Requirement

• $A$ is real symmetric $⟹$ $A$ is diagonalizable over $\mathbb{R}$ by an orthogonal matrix
• $A$ is normal ($A^{\ast }A=A{A}^{\ast }$) $⟹$ $A$ is unitary diagonalizable over $\mathbb{C}$
• All eigenvalues are distinct $⟹$ diagonalizable over the field containing them

Definition: normal

• over $\mathbb{C}$
  • $A^{\ast }A=A{A}^{\ast }$
• over $\mathbb{R}$
  • $A^{\top }A=A{A}^{\top }$

Definition: multiplicity

• geometric multiplicity: $dim(\mathbf{N}(A-\lambda I))$
• algebraic multiplicity: the number of times $\lambda$ appears as a root of the characteristic polynomial $P_A(\lambda )$

We always have: $1\le$ geometric multiplicity $\le$ algebraic multiplicity

NOTE

A matrix is diagonalizable if for all $\lambda$ we have geometric multiplicity $=$ algebraic multiplicity