📂Linear Algebra

Eigen Spaces of Linear Transformations and Geometric Multiplicity

Definition¹

Let’s define $V$ and $n$ as dimension vector spaces, $T : V \to V$ as linear transformation. Let’s also define $\lambda$ as the eigenvalue of $T$. The set defined as follows, $E_{\lambda}$, is called the eigenspace of $T$ corresponding to the eigenvalue $\lambda$.

$$ E_{\lambda} = V_{\lambda} := \left\{ x \in V : Tx = \lambda x \right\} = N(T - \lambda I) $$

In this case, $N$ is the null space.

Similarly, the eigenspace of a square matrix $A$ is defined as the eigenspace of $L_{A}$.

Explanation

Although there is a condition that to be an eigenvector, one must not be the zero vector, there is no specific condition in the definition of $E_{\lambda}$ that $x$ must be an eigenvector. Therefore, $E_{\lambda}$ is the set of eigenvectors corresponding to $\lambda$ and the zero vector. It’s important to note that for $E_{\lambda}$ to be a subspace, it must include the zero vector. Indeed, $E_{\lambda}$ is a subspace of $V$ since $T$, being a linear transformation, is obviously closed under addition and scalar multiplication (subspace criterion). If we define $x, y \in E_{\lambda}$,

$$ T(ax + y) = aT(x) + T(y) = a\lambda x + \lambda y = \lambda (ax + y) $$

then $ax + y \in E_{\lambda}$ holds.

Geometric Multiplicity

The dimension of the eigenspace $E_{\lambda}$ corresponding to eigenvalue $\lambda$ is known as the geometric multiplicity of $\lambda$.

In other words, it is the number of linearly independent eigenvectors corresponding to $\lambda$, and thus, it is at least $1$. Its maximum value is related to the algebraic multiplicity.

Theorem

Let’s define the algebraic multiplicity of eigenvalue $\lambda$ of $T$ as $m$. The algebraic multiplicity is greater than or equal to the geometric multiplicity.

$$ 1 \le \dim E_{\lambda} \le m $$

Proof

Let’s denote the ordered basis of $E_{\lambda}$ by $\gamma = \left\{ v_{1}, \dots, v_{p} \right\}$. The expanded ordered basis of $V$ is denoted as $\beta = \left\{ v_{1}, \dots, v_{p}, v_{p+1}, \dots, v_{n} \right\}$. The matrix representation of $T$ is referred to as $A = \begin{bmatrix} T \end{bmatrix}_{\beta}$. Then $A$ is as follows in block matrix form:

$$ A = \begin{bmatrix} \begin{bmatrix} T|_{E_{\lambda}} \end{bmatrix}_{\gamma} & B \\ O_{n-p} & C \end{bmatrix} $$

$O_{n-p}$ is a $n-p \times n-p$ zero matrix. In this case, for $1 \le i \le p$, since $Tv_{i} = \lambda v_{i}$, the coordinate vector of $Tv_{i}$ is as follows:

$$ \begin{bmatrix} Tv_{i} \end{bmatrix}_{\gamma} = \begin{bmatrix} 0 \\ \vdots \\ \lambda \\ \vdots \\ 0 \end{bmatrix}i\text{-th row} $$

Therefore, since $\begin{bmatrix} T|_{E_{\lambda}} \end{bmatrix}_{\gamma} = \begin{bmatrix} \begin{bmatrix} Tv_{1} \end{bmatrix}_{\gamma} & \cdots & \begin{bmatrix} Tv_{p} \end{bmatrix}_{\gamma} \end{bmatrix}$,

$$ A = \begin{bmatrix} \lambda I_{p} & B \\ O & C \end{bmatrix} $$

$I_{p}$ is a $p \times p$ identity matrix.

Determinant of Block Matrices

Let’s consider $A = \begin{bmatrix} A_{1} & A_{2} \\ O & A_{3} \end{bmatrix}$ to be a block matrix. Then, the following holds:

$$ \det A = \det A_{1} \det A_{3} $$

Hence, the characteristic polynomial of $T$ is as follows:

$$ \begin{align*} f(t) = \det (A - t I_{n}) &= \det \begin{bmatrix} \lambda I_{p} - \lambda I_{p} & B \\ O & C - tI_{n-p} \end{bmatrix} \\ & = \det \left( (\lambda - t)I_{p} \right) \det \left( C - tI_{n-p} \right) \\ & = (\lambda - t)^{p} \det \left( C - tI_{n-p} \right) \\ & = (-1)^{p}(t - \lambda)^{p} \det \left( C - tI_{n-p} \right) \\ \end{align*} $$

This shows that $(t - \lambda)^{p}$ is a factor of the characteristic polynomial $f(t)$. Therefore, $f(t)$ has at least $\lambda$ as a root of multiplicity $p$, and by the definition of algebraic multiplicity, the algebraic multiplicity is greater than or equal to the geometric multiplicity.

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