General Case of Diagonalization

Theorem

1. Suppose $T:V\to V$ is a linear transform of a finite dimensional vector space
2. Suppose $T$ has distinct eigenvalues ${\lambda }_1,\dots ,{\lambda }_k$
3. Let $m_i$ be the Algebraic Multiplicity of ${\lambda }_i$
4. Then, $T$ is diagonizable $⟺$
   1. $dim(V)=m_1+\cdots +m_k$
   2. $dim(E_{{\lambda }_i})=m_i$

Examples

• General Diagonalization Example
• General Diagonalization Example 2

Proof

1. Suppose $T:V\to V$ is a linear transform of a finite dimensional vector space
2. Suppose $T$ has distinct eigenvalues ${\lambda }_1,\dots ,{\lambda }_k$
3. Let $m_i$ be the Algebraic Multiplicity of ${\lambda }_i$

Proving $⟹$

1. Suppose $T:V\to V$ is diagonazible
2. In the eigenbasis $\alpha$, we have:

$$[T{]}_{\alpha }^{\alpha }=\left[\begin{array}{cccc}{\lambda }_1& 0& \dots & 0\\ 0& {\lambda }_1& \dots & 0\\ 0& \dots & 0& {\lambda }_n\end{array}\right]$$

3. $Geo({\lambda }_1)=Alg({\lambda }_1)$
4. If we compare $X(\lambda )$ in this basis, we get $X(\lambda )=({\lambda }_1-\lambda )Alg({\lambda }_1)$
5. In this matrix, we have $dim(V)=n=Alg({\lambda }_1)+\cdots +Alg({\lambda }_n)$
6. $=m_1=\cdots +m_k$

Proving $⟸$

1. Suppose $T$ satisfies $dim(V)=m_1+\cdots +m_n=Alg({\lambda }_1)+\cdots +Alg({\lambda }_n)$
2. Then, $Geo({\lambda }_i)=Alg({\lambda }_i)$ for each $i\in 1,\dots ,n$
3. We build an eigenbasis of V with the right size
4. Consider eigenspace $E_{{\lambda }_i}$ for all $i\in 1,\dots ,n$
5. Each of these is a subspace of $V$, so they are all finite dimensional and have basis
6. We have $dim(E_{{\lambda }_i})=Geo({\lambda }_i)$. We pick a basis of each $E_{{\lambda }_i}$ s.t $E_{{\lambda }_i}=span({\alpha }_i)$
7. Now, consider $\alpha =U_{i=1}^k{\alpha }_i$ (Refer to Union Notation)
8. Note that the size of this basis $|\alpha |={\sum }_{i=1}^{k}Geo({\lambda }_i)=dim(V)$
9. Moreover, $\alpha$ is indep, and contained in a distinct eigenspace