Eigenvectors

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

\begin{align} A\underline{v} = \lambda \underline{v} \end{align}

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

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

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

\begin{align*} A\underline{v} = \lambda \underline{v} & \iff (A - \lambda I)\underline{v} = \underline{0} \\ & \iff (A - \lambda I) \text{ is singular (since } \underline{v} \text{ is nonzero)} \\ & \iff \text{det}(A - \lambda I) = \underline{0} \end{align*}

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:

\[ E_\lambda = \{ v : A\underline{v} = \underline{0} \} \]

Note this is a subspace of the vector space \( \underline{v} \) 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:

\begin{pmatrix} 1 & 0 \\ -1 & 1 \\ \end{pmatrix}

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:

\[ S = \{ \alpha \begin{bmatrix}0\\1\\\end{bmatrix} : \alpha \in K \} \]

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 \( S = \{ \underline{v}_1 ... \underline{v}_r \} \) 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 \( \underline{v}_{r + 1} ... \underline{v}_n \). Now let \( G \) be the matrix with with column \( \underline{v}_i \) (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 \( \lambda \underline{v}_i \), 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{align*} \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})*\text{det}(A - Ix)*\text{det}(G) \\ &= \text{det}(A - Ix) \end{align*}

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.