Eigenvectors

Given some matrix $A$ with entries in some field $F$, we say a nonzero vector $\underset{―}{v}$ is an eigenvector of $A$ if:

$$\begin{array}{}\text{(1)}& A\underset{―}{v}=\lambda \underset{―}{v}\end{array}$$

For some $\lambda \in F$, in which case we also say $\lambda$ is an eigenvalue of $A$ (associated with $\underset{―}{v}$).

Theorem $\lambda$ is an eigenvector of $A$ iff it satisfies $\text{det}(A-\lambda I)=0$

Proof Suppose there exists some $\underset{―}{v}$ satisfying (1), then:

$$\begin{array}{rl}A\underset{―}{v}=\lambda \underset{―}{v}& ⟺(A-\lambda I)\underset{―}{v}=\underset{―}{0}\\ & ⟺(A-\lambda I)\text{ is singular (since }\underset{―}{v}\text{ is nonzero)}\\ & ⟺\text{det}(A-\lambda I)=\underset{―}{0}\end{array}$$

This theorem shows that the eigenvalues of $A$ correspond to the roots of the polynomial $\text{det}(A-\lambda I)$.

The algebraic multiplicity of an eigenvalue $\lambda$ is simply its multiplicity as a root of the polynomial $\text{det}(A-\lambda I)$.

Defining the geometric multiplicity of $\lambda$ on the other hand will first involve some set up. We define the eigenspace of $A$ corresponding to as:

$$E_{\lambda }=\{v:A\underset{―}{v}=\underset{―}{0}\}$$

Note this is a subspace of the vector space $\underset{―}{v}$ resides in. From this we define the algebraic multiplicity of $\lambda$ as the dimension of $E_{\lambda }$.

It is important to note that the algebraic and geometric multiplicities of $\lambda$ may not be the same. For example, the matrix:

$$\left(\begin{array}{cc}1& 0\\ -1& 1\end{array}\right)$$

It has characteristic polynomial $(1-\lambda )(1-\lambda )$ and hence one eigenvalue, 1, of algebraic multiplicity 2. However, we find that the eigenspace of 1 to be the set:

$$S=\{\alpha \left[\begin{array}{c}0\\ 1\end{array}\right]:\alpha \in K\}$$

Indicating that the geometric multiplicity of the eigenvalue is 1.

Theorem The algebraic multiplicity of $\lambda$ is at least as large as the geometric multiplicity of $\lambda$.

Proof First suppose $\lambda$ has geometric multiplicity $r$. Let $S=\{{\underset{―}{v}}_1...{\underset{―}{v}}_r\}$ be any set of linearly independent eigenvectors associated with $\lambda$.

We can extend $S$ to form a basis to $K^n$ with $n-r$ linearly independent vectors ${\underset{―}{v}}_{r+1}...{\underset{―}{v}}_n$. Now let $G$ be the matrix with with column ${\underset{―}{v}}_i$ (note $G$ is invertible since its columns are linearly independent vectors), and consider the product $AG$.

The first $r$ columns of $AG$ will be given by $\lambda {\underset{―}{v}}_i$, therefore the first $r$ columns of $G^{-1}AG$ will form a diagonal matrix with each diagonal entry given by $\lambda$. We now claim $G^{-1}AG$ has the same characteristic polynomial as $A$:

$$\begin{array}{rlr}\text{det}(G^{-1}AG-xI)& =\text{det}(G^{-1}AG-G^{-1}xIG)& \text{ since any multiple of the identity matrix commutes with all matrices}\\ & =\text{det}(G^{-1})\ast \text{det}(A-Ix)\ast \text{det}(G)\\ & =\text{det}(A-Ix)\end{array}$$

However, the first $r$ columns of $A-Ix$ form a diagonal matrix with entries $\lambda -x$, hence the characteristic polynomial of $A$ has a factor of at least $(\lambda -x{)}^r$, this completes the proof.