Geometry <= Algebra Theorem

Theorem

• Suppose $V$ is a $n$ dimensional $\mathbb{R}$ vector space
• If $T:V\to V$ has a real eigenvalue $\lambda$
• Then, $Geo(\lambda )\le Alg(\lambda )$

Proof

Intuition

Pick a basis of $E_{\lambda }$ then extend it to a basis $\alpha$ of $V$, and compute $X(\lambda )$ in that basis

Proof

1. We have $dim(E_{\lambda })=Geo(\lambda )=k$
2. We pick basis $E_{\lambda }=span\{v_1,\dots ,v_k\}$
3. We extend this to basis of $V$
4. $v=span(\alpha )=\{v_1,\dots ,v_k,v_{k+1},\dots ,v_n\}$
5. In this basis: $T(v_1)=\lambda v_1,\dots ,T(v_k)=\lambda v_k$ meaning everything in our $v$ from $v_1$ to $v_k$ is a eigenvector
6. We also know that $v_{k+1}$ to $v_n$ are no longer Eigenvectors. $T(v_{k+1})=a_{11}{v}_1+\cdots +a_{1n}{v}_n,\dots ,T(v_n)=a_{n1}{v}_1+\cdots +a_{nn}{v}_n$
7. In this basis, we get:

$$[T{]}_{\alpha }^{\alpha }=\left[\begin{array}{cccccc}\lambda & \dots & 0& a_{11}& \dots & a_{1n}\\ ⋮& & & & & \\ 0& \dots & \lambda & a_{n1}& \dots & a_{nn}\end{array}\right]$$

8. Expanding $X(l)=\det ([T{]}_{\alpha }^{\alpha }-lI)=(\lambda -l{)}^k\det (\text{smaller (n-k) * (n-k) matrix})$
9. This will contain more $l$ terms. We might get more factors $\lambda -l$ or we might not
10. This gives $Alg(\lambda )\ge Geo(\lambda )$