267 lines
No EOL
9.5 KiB
Markdown
267 lines
No EOL
9.5 KiB
Markdown
# Eigenspaces
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## Eigenvalues and eigenvectors
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If a linear transformation is represented by an $n \times n$ matrix $A$ and there exists a nonzero vector $\mathbf{x} \in V$ such that $A \mathbf{x} = \lambda \mathbf{x}$ for some $\lambda \in \mathbb{K}$, then for this transformation $\mathbf{x}$ is a natural choice to use as a basis vector for $V$.
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> *Definition 1*: let $A$ be a $n \times n$ matrix, a scalar $\lambda \in \mathbb{K}$ is defined as an **eigenvalue** of $A$ if and only if there exists a vector $\mathbf{x} \in V \backslash \{\mathbf{0}\}$ such that
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>
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> $$
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> A \mathbf{x} = \lambda \mathbf{x},
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> $$
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>
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> with $\mathbf{x}$ defined as an **eigenvector** belonging to $\lambda$.
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This notion can be further generalized to a linear operator $L: V \to V$ such that
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$$
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L(\mathbf{x}) = \lambda \mathbf{x},
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$$
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note that $L(\mathbf{x}) = A \mathbf{x}$, which implies the similarity.
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Furthermore it follows from the definition that any linear combination of eigenvectors is also a eigenvector of $A$.
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> *Theorem 1*: let $A$ be a $n \times n$ matrix, a scalar $\lambda \in \mathbb{K}$ is an eigenvalue of $A$ if and only if
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>
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> $$
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> \det (A - \lambda I) = 0.
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> $$
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??? note "*Proof*:"
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A scalar $\lambda \in \mathbb{K}$ is an eigenvalue of $A$ if and only if there exists a vector $\mathbf{x} \in V \backslash \{\mathbf{0}\}$ such that
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$$
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A \mathbf{x} = \lambda \mathbf{x},
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$$
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obtains
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$$
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A \mathbf{x} - \lambda \mathbf{x} = (A - \lambda I) \mathbf{x} = \mathbf{0},
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$$
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which implies that $(A - \lambda I)$ is singular and $\det(A - \lambda I) = 0$ by [definition](../determinants/#properties-of-determinants).
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The eigenvalues $\lambda$ may thus be determined from the **characteristic polynomial** of degree $n$ that is obtained from $\det (A - \lambda I) = 0$. In particular, the eigenvalues are the roots of this polynomial.
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> *Theorem 2*: let $A$ be a $n \times n$ matrix and let $\lambda \in \mathbb{K}$ be an eigenvalue of $A$. A vector $\mathbf{x} \in V$ is an eigenvector of $A$ corresponding to $\lambda$ if and only if
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>
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> $$
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> \mathbf{x} \in N(A - \lambda I) \backslash \{\mathbf{0}\}.
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> $$
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??? note "*Proof*:"
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Let $A$ be a $n \times n$ matrix, $\mathbf{x} \in V$ is an eigenvector of $A$ if and only if
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$$
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A \mathbf{x} = \lambda \mathbf{x},
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$$
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for an eigenvalue $\lambda \in \mathbb{K}$. Therefore
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$$
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A \mathbf{x} - \lambda \mathbf{x} = (A - \lambda I) \mathbf{x} = \mathbf{0},
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$$
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which implies that $\mathbf{x} \in N(A - \lambda I)$.
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Which implies that the eigenvectors can be obtained by determining the corresponding null space of $A - \lambda I$.
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> *Definition 2*: let $L: V \to V$ be a linear operator and let $\lambda \in \mathbb{K}$ be an eigenvalue of $L$. Let the **eigenspace** $E_\lambda$ of the corresponding eigenvalue $\lambda$ be defined as
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>
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> $$
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> E_\lambda = \{\mathbf{x} \in V \;|\; L(\mathbf{x}) = \lambda \mathbf{x}\} = N(A - \lambda I),
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> $$
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>
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> with $L(\mathbf{x}) = A \mathbf{x}$.
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It may be observed that $E_\lambda$ is a subspace of $V$ consisting of the zero vector and the eigenvectors of $L$ or $A.$
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### Properties
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> *Theorem 3*: if $\lambda_1, \dots, \lambda_n \in \mathbb{K}$ are distinct eigenvalues of an $n \times n$ matrix $A$ with corresponding eigenvectors $\mathbf{x}_1, \dots \mathbf{x}_k \in V\backslash \{\mathbf{0}\}$, then $\mathbf{x}_1, \dots \mathbf{x}_k$ are linearly independent.
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??? note "*Proof*:"
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Will be added later.
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If $A \in \mathbb{R}^{n \times n}$ and $A \mathbf{x} = \lambda \mathbf{x}$ for some $\mathbf{x} \in V$ and $\lambda \in \mathbb{K}$. Then
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$$
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A \mathbf{\bar x} = \overline{A \mathbf{x}} = \overline{\lambda \mathbf{x}} = \bar \lambda \mathbf{\bar x}.
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$$
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The complex conjugate of an eigenvector of $A$ is also an eigenvector of $A$ with an eigenvalue $\bar \lambda$.
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> *Theorem 4*: let $A$ be a $n \times n$ matrix and let $\lambda_1, \dots, \lambda_n \in \mathbb{K}$ be the eigenvalues of $A$. It follows that
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>
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> $$
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> \det (A - \lambda I) = (\lambda_1 - \lambda)(\lambda_2 - \lambda) \cdots (\lambda_n - \lambda),
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> $$
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>
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> and
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>
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> $$
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> \det (A) = \lambda_1 \lambda_2 \cdots \lambda_n.
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> $$
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??? note "*Proof*:"
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Let $A$ be a $n \times n$ matrix and let $\lambda_1, \dots, \lambda_n \in \mathbb{K}$ be the eigenvalues of $A$. It follows from the [fundamental theorem of algebra](../../number-theory/complex-numbers/#roots-of-polynomials) that
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$$
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\det (A - \lambda I) = (\lambda_1 - \lambda)(\lambda_2 - \lambda) \cdots (\lambda_n - \lambda),
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$$
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by taking $\lambda = 0$ it follows that
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$$
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\det (A) = \lambda_1 \lambda_2 \cdots \lambda_n.
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$$
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From $\det (A) = \lambda_1 \lambda_2 \cdots \lambda_n$ it must follow that
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$$
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\mathrm{trace}(A) = \sum_{i=1}^n \lambda_i.
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$$
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> *Theorem 5*: let $A$ and $B$ be $n \times n$ matrices. If $B$ is similar to $A$, then $A$ and $B$ have the same eigenvalues.
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??? note "*Proof*:"
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Let $A$ and $B$ be similar $n \times n$ matrices, then there exists a nonsingular matrix $S$ such that
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$$
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B = S^{-1} A S.
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$$
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Let $\lambda \in \mathbb{K}$ be an eigenvalue of $B$ then
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$$
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\begin{align*}
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0 &= \det(B - \lambda I), \\
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&= \det(S^{-1} A S - \lambda I), \\
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&= \det(S^{-1}(A - \lambda I) S), \\
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&= \det(S^{-1}) \det(A - \lambda I) \det(S), \\
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&= \det(A - \lambda I).
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\end{align*}
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$$
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## Diagonalization
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> *Definition 3*: an $n \times n$ matrix $A$ is **diagonalizable** if there exists a nonsingular diagonalizing matrix $X$ and a diagonal matrix $D$ such that
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> $$
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> A X = X D.
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> $$
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We may now pose the following theorem.
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> *Theorem 6*: an $n \times n$ matrix $A$ is diagonalizable if and only if $A$ has $n \in \mathbb{N}$ linearly independent eigenvectors.
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??? note "*Proof*:"
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Will be added later.
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It follows from the proof that the column vectors of the diagonalizing matrix $X$ are eigenvectors of $A$ and the diagonal elements of $D$ are the corresponding eigenvalues of $A$. If $A$ is diagonalizable, then
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$$
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A = X D X^{-1},
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$$
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it follows then that
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$$
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A^k = X D^k X^{-1},
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$$
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for $k \in \mathbb{K}$.
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### Hermitian case
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The following section is for the special case that a matrix is [Hermitian](../matrices/matrix-arithmatic/#hermitian-matrix).
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> *Theorem 7*: the eigenvalues of a Hermitian matrix are real.
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??? note "*Proof*:"
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Let $A$ be a Hermitian matrix and let $\mathbf{x} \in V \backslash \{\mathbf{0}\}$ be an eigenvector of $A$ with corresponding eigenvalue $\lambda \in \mathbb{C}$. We have
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$$
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\begin{align*}
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\lambda \mathbf{x}^H \mathbf{x} &= \mathbf{x}^H (\lambda \mathbf{x}), \\
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&= \mathbf{x}^H (A \mathbf{x}), \\
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&= (\mathbf{x}^H A) \mathbf{x}, \\
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&= (A^H \mathbf{x})^H \mathbf{x} , \\
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&= (A \mathbf{x})^H \mathbf{x}, \\
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&= (\lambda \mathbf{x})^H \mathbf{x}, \\
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&= \bar \lambda \mathbf{x}^H \mathbf{x},
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\end{align*}
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$$
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since $\bar \lambda = \lambda$ we must have that $\lambda \in \mathbb{R}$.
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> *Theorem 8*: the eigenvectors of a Hermitian matrix corresponding to distinct eigenvalues are orthogonal.
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??? note "*Proof*:"
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Let $A$ be a Hermitian matrix and let $\mathbf{x}_1, \mathbf{x}_2 \in V \backslash \{\mathbf{0}\}$ be two eigenvectors of $A$ with corresponding eigenvalues $\lambda_1, \lambda_2 \in \mathbb{C}[\lambda_1 \neq \lambda_2]$. We have
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$$
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\begin{align*}
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\lambda_1 \mathbf{x}_1^H \mathbf{x}_2 &= (\lambda_1 \mathbf{x}_1)^H \mathbf{x}_2, \\
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&= (A \mathbf{x}_1)^H \mathbf{x}_2, \\
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&= \mathbf{x}_1^H A^H \mathbf{x}_2, \\
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&= \mathbf{x}_1^H A \mathbf{x}_2, \\
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&= \mathbf{x}_1^H (\lambda_2 \mathbf{x}_2), \\
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&= \lambda_2 \mathbf{x}_1^H \mathbf{x}_2,
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\end{align*}
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$$
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since $\lambda_1 \neq \lambda_2$ this must imply that $\mathbf{x}_1^H \mathbf{x}_2 = 0$, implying orthogonality in terms of the Hermite scalar product.
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Theorem 7 and 8 impose that the following definition can be used.
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> *Definition 4*: an $n \times n$ matrix $U$ is **unitary** if the column vectors of $U$ form an orthonormal set in $V$.
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Thus, $U$ is unitary if and only if $U^H U = I$. Then it also follows that $U^{-1} = U^H$. A real unitary matrix is an orthogonal matrix.
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One may observe that theorem 8 implies that the diagonalizing matrix of a Hermitian matrix $A$ is unitary when $A$ has distinct eigenvalues.
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> *Lemma 1*: if the eigenvalues of a Hermitian matrix $A$ are distinct, then there exists a unitary matrix $U$ and a diagonal matrix $D$ such that
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>
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> $$
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> A U = U D.
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> $$
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??? note "*Proof*:"
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Will be added later.
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With the column vectors of $U$ the eigenvectors of $A$ and the diagonal elements of $D$ the corresponding eigenvalues of $A$.
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> *Theorem 9*: let $A$ be an $n \times n$ matrix, there exists a unitary matrix $U$ and a upper triangular matrix $T$ such that
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> $$
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> A U = U T.
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> $$
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??? note "*Proof*:"
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Will be added later.
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The factorization $A = U T U^H$ is often referred to as the *Schur decomposition* of $A$.
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> *Theorem 10*: if $A$ is Hermitian, then there exists a unitary matrix $U$ and a diagonal matrix $D$ such that
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>
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> $$
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> A U = U D.
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> $$
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??? note "*Proof*:"
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Will be added later. |