Eigenvalues, eigenvectors, and defectiveness

Eigenvalues and eigenvectors

$$ Av = \lambda v, \quad v \neq 0. $$

$\lambda$: eigenvalue. $v$: eigenvector. Found from $\det(A - \lambda I) = 0$ (characteristic equation).

Intuition: eigenvectors are the directions $A$ doesn't transform at all, it only scales them.

Shapes

$$ \underbrace{A}_{n\times n}\ \underbrace{v}_{n\times1} = \lambda\ \underbrace{v}_{n\times1}. $$

Intuition: an $n\times n$ matrix always sends an $n\times1$ vector to another $n\times1$ vector; an eigenvector is the special case where that output vector is just a rescaled copy of the input, same shape, same line.

Example

$$ A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}: \quad \det(A-\lambda I) = \lambda^2 - 4\lambda + 3 = 0 \;\Rightarrow\; \lambda = 1, 3. $$

$$ \lambda=3: (A-3I)v=0 \Rightarrow v = \begin{pmatrix}1\\1\end{pmatrix}, \qquad \lambda=1: (A-I)v=0 \Rightarrow v = \begin{pmatrix}1\\-1\end{pmatrix}. $$

Eigenvalues and rank

$$ \dim N(A) = n - \operatorname{rank}(A) \quad \text{(rank–nullity)}. $$

$N(A) = N(A - 0I)$ is the eigenspace of $\lambda = 0$, so $\dim N(A)$ is $0$'s geometric multiplicity. Hence $\operatorname{rank}(A) = n - (\text{geometric multiplicity of } 0)$, always true. If $A$ is diagonalizable, this equals the count of nonzero eigenvalues.

Intuition: $\lambda=0$ marks directions $A$ collapses; everything else contributes to rank.

Example

$$ A = \begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix}\ (\operatorname{rank} 1): \quad \det(A-\lambda I) = \lambda^2 - 5\lambda = 0 \;\Rightarrow\; \lambda = 0, 5. $$

$$ N(A) = \operatorname{span}\begin{pmatrix}2\\-1\end{pmatrix}, \ \dim = 1 = n - \operatorname{rank}(A). $$

$$ \text{1 nonzero eigenvalue} = \operatorname{rank}(A). \ \checkmark $$

Algebraic vs. geometric multiplicity

Algebraic multiplicity $\operatorname{AM}(\lambda)$: multiplicity of $\lambda$ as a root of $\det(A-\lambda I)=0$. Geometric multiplicity $\operatorname{GM}(\lambda) = \dim N(A - \lambda I)$: dimension of its eigenspace. Always $1 \le \operatorname{GM}(\lambda) \le \operatorname{AM}(\lambda)$.

Intuition: AM counts roots on paper; GM counts how many independent directions actually get that eigenvalue.

Example

$$ A = \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}: \quad \det(A-\lambda I) = (2-\lambda)^2 = 0 \;\Rightarrow\; \lambda = 2,\ \operatorname{AM}(2) = 2. $$

$$ A - 2I = \begin{pmatrix} 0&1\\0&0\end{pmatrix} \Rightarrow N(A-2I) = \operatorname{span}\begin{pmatrix}1\\0\end{pmatrix}, \ \operatorname{GM}(2) = 1 < \operatorname{AM}(2) = 2. $$

Diagonalization

Stack eigenvectors as columns of $V$, eigenvalues on the diagonal of $\Lambda$:

$$ AV = A\begin{pmatrix} v_1 & \cdots & v_n \end{pmatrix} = \begin{pmatrix} \lambda_1 v_1 & \cdots & \lambda_n v_n \end{pmatrix} = V\Lambda. $$

This always holds. But $A = V\Lambda V^{-1}$ needs $V$ invertible $\iff$ its columns (the eigenvectors) are linearly independent $\iff$ $\operatorname{rank}(V) = n \iff C(V) = \mathbb{R}^n$.

Shapes

Every factor is $n\times n$. Multiplying by a diagonal matrix on the right scales the columns of $V$, which is exactly $\lambda_iv_i$:

$$ \underbrace{A}_{n\times n}\underbrace{\begin{pmatrix}\vert&&\vert\\v_1&\cdots&v_n\\\vert&&\vert\end{pmatrix}}_{V,\ n\times n} =\underbrace{\begin{pmatrix}\vert&&\vert\\v_1&\cdots&v_n\\\vert&&\vert\end{pmatrix}}_{V,\ n\times n} \underbrace{\begin{pmatrix}\lambda_1&&\\&\ddots&\\&&\lambda_n\end{pmatrix}}_{\Lambda,\ n\times n}. $$

Intuition: independent eigenvectors give a new coordinate system in which $A$ is just a scaling; that's what diagonalizable means.

Example

Reusing $A = \begin{pmatrix}2&1\\1&2\end{pmatrix}$ with eigenvectors $(1,1), (1,-1)$ from above:

$$ V = \begin{pmatrix}1&1\\1&-1\end{pmatrix}, \quad \det(V) = -2 \neq 0 \;\Rightarrow\; \text{invertible}, \ \operatorname{rank}(V) = 2 = n. $$

So $A = V\Lambda V^{-1}$ with $\Lambda = \operatorname{diag}(3,1)$.

Defectiveness

$A$ is defective if $\operatorname{GM}(\lambda) < \operatorname{AM}(\lambda)$ for some $\lambda$: too few eigenvectors to form a basis, so $A$ is not diagonalizable. Diagonalizable $\iff$ $\operatorname{GM}(\lambda) = \operatorname{AM}(\lambda)$ for every $\lambda$.

Intuition: a repeated eigenvalue doesn't hand you a second independent eigenvector for free.

Example

$\begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix}$ above is defective (GM $<$ AM). Contrast with $\begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}$: same $\operatorname{AM}(2)=2$, but $A - 2I = 0 \Rightarrow \operatorname{GM}(2) = 2 = \operatorname{AM}(2)$: not defective (already diagonal).