9.5 KiB
Eigenspaces
Eigenvalues and eigenvectors
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.
Definition 1: let
Abe an \times nmatrix, a scalar\lambda \in \mathbb{K}is defined as an eigenvalue ofAif and only if there exists a vector\mathbf{x} \in V \backslash \{\mathbf{0}\}such that
A \mathbf{x} = \lambda \mathbf{x},with
\mathbf{x}defined as an eigenvector belonging to\lambda.
This notion can be further generalized to a linear operator L: V \to V such that
L(\mathbf{x}) = \lambda \mathbf{x},
note that L(\mathbf{x}) = A \mathbf{x}, which implies the similarity.
Furthermore it follows from the definition that any linear combination of eigenvectors is also a eigenvector of A.
Theorem 1: let
Abe an \times nmatrix, a scalar\lambda \in \mathbb{K}is an eigenvalue ofAif and only if
\det (A - \lambda I) = 0.
??? note "Proof:"
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
$$
A \mathbf{x} = \lambda \mathbf{x},
$$
obtains
$$
A \mathbf{x} - \lambda \mathbf{x} = (A - \lambda I) \mathbf{x} = \mathbf{0},
$$
which implies that $(A - \lambda I)$ is singular and $\det(A - \lambda I) = 0$ by [definition](../determinants/#properties-of-determinants).
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.
Theorem 2: let
Abe an \times nmatrix and let\lambda \in \mathbb{K}be an eigenvalue ofA. A vector\mathbf{x} \in Vis an eigenvector ofAcorresponding to\lambdaif and only if
\mathbf{x} \in N(A - \lambda I) \backslash {\mathbf{0}}.
??? note "Proof:"
Let $A$ be a $n \times n$ matrix, $\mathbf{x} \in V$ is an eigenvector of $A$ if and only if
$$
A \mathbf{x} = \lambda \mathbf{x},
$$
for an eigenvalue $\lambda \in \mathbb{K}$. Therefore
$$
A \mathbf{x} - \lambda \mathbf{x} = (A - \lambda I) \mathbf{x} = \mathbf{0},
$$
which implies that $\mathbf{x} \in N(A - \lambda I)$.
Which implies that the eigenvectors can be obtained by determining the corresponding null space of A - \lambda I.
Definition 2: let
L: V \to Vbe a linear operator and let\lambda \in \mathbb{K}be an eigenvalue ofL. Let the eigenspaceE_\lambdaof the corresponding eigenvalue\lambdabe defined as
E_\lambda = {\mathbf{x} \in V ;|; L(\mathbf{x}) = \lambda \mathbf{x}} = N(A - \lambda I),with
L(\mathbf{x}) = A \mathbf{x}.
It may be observed that E_\lambda is a subspace of V consisting of the zero vector and the eigenvectors of L or A.
Properties
Theorem 3: if
\lambda_1, \dots, \lambda_n \in \mathbb{K}are distinct eigenvalues of ann \times nmatrixAwith corresponding eigenvectors\mathbf{x}_1, \dots \mathbf{x}_k \in V\backslash \{\mathbf{0}\}, then\mathbf{x}_1, \dots \mathbf{x}_kare linearly independent.
??? note "Proof:"
Will be added later.
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
A \mathbf{\bar x} = \overline{A \mathbf{x}} = \overline{\lambda \mathbf{x}} = \bar \lambda \mathbf{\bar x}.
The complex conjugate of an eigenvector of A is also an eigenvector of A with an eigenvalue \bar \lambda.
Theorem 4: let
Abe an \times nmatrix and let\lambda_1, \dots, \lambda_n \in \mathbb{K}be the eigenvalues ofA. It follows that
\det (A - \lambda I) = (\lambda_1 - \lambda)(\lambda_2 - \lambda) \cdots (\lambda_n - \lambda),and
\det (A) = \lambda_1 \lambda_2 \cdots \lambda_n.
??? note "Proof:"
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
$$
\det (A - \lambda I) = (\lambda_1 - \lambda)(\lambda_2 - \lambda) \cdots (\lambda_n - \lambda),
$$
by taking $\lambda = 0$ it follows that
$$
\det (A) = \lambda_1 \lambda_2 \cdots \lambda_n.
$$
From \det (A) = \lambda_1 \lambda_2 \cdots \lambda_n it must follow that
\mathrm{trace}(A) = \sum_{i=1}^n \lambda_i.
Theorem 5: let
AandBben \times nmatrices. IfBis similar toA, thenAandBhave the same eigenvalues.
??? note "Proof:"
Let $A$ and $B$ be similar $n \times n$ matrices, then there exists a nonsingular matrix $S$ such that
$$
B = S^{-1} A S.
$$
Let $\lambda \in \mathbb{K}$ be an eigenvalue of $B$ then
$$
\begin{align*}
0 &= \det(B - \lambda I), \\
&= \det(S^{-1} A S - \lambda I), \\
&= \det(S^{-1}(A - \lambda I) S), \\
&= \det(S^{-1}) \det(A - \lambda I) \det(S), \\
&= \det(A - \lambda I).
\end{align*}
$$
Diagonalization
Definition 3: an
n \times nmatrixAis diagonalizable if there exists a nonsingular diagonalizing matrixXand a diagonal matrixDsuch that
A X = X D.
We may now pose the following theorem.
Theorem 6: an
n \times nmatrixAis diagonalizable if and only ifAhasn \in \mathbb{N}linearly independent eigenvectors.
??? note "Proof:"
Will be added later.
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
A = X D X^{-1},
it follows then that
A^k = X D^k X^{-1},
for k \in \mathbb{K}.
Hermitian case
The following section is for the special case that a matrix is Hermitian.
Theorem 7: the eigenvalues of a Hermitian matrix are real.
??? note "Proof:"
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
$$
\begin{align*}
\lambda \mathbf{x}^H \mathbf{x} &= \mathbf{x}^H (\lambda \mathbf{x}), \\
&= \mathbf{x}^H (A \mathbf{x}), \\
&= (\mathbf{x}^H A) \mathbf{x}, \\
&= (A^H \mathbf{x})^H \mathbf{x} , \\
&= (A \mathbf{x})^H \mathbf{x}, \\
&= (\lambda \mathbf{x})^H \mathbf{x}, \\
&= \bar \lambda \mathbf{x}^H \mathbf{x},
\end{align*}
$$
since $\bar \lambda = \lambda$ we must have that $\lambda \in \mathbb{R}$.
Theorem 8: the eigenvectors of a Hermitian matrix corresponding to distinct eigenvalues are orthogonal.
??? note "Proof:"
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
$$
\begin{align*}
\lambda_1 \mathbf{x}_1^H \mathbf{x}_2 &= (\lambda_1 \mathbf{x}_1)^H \mathbf{x}_2, \\
&= (A \mathbf{x}_1)^H \mathbf{x}_2, \\
&= \mathbf{x}_1^H A^H \mathbf{x}_2, \\
&= \mathbf{x}_1^H A \mathbf{x}_2, \\
&= \mathbf{x}_1^H (\lambda_2 \mathbf{x}_2), \\
&= \lambda_2 \mathbf{x}_1^H \mathbf{x}_2,
\end{align*}
$$
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.
Theorem 7 and 8 impose that the following definition can be used.
Definition 4: an
n \times nmatrixUis unitary if the column vectors ofUform an orthonormal set inV.
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.
One may observe that theorem 8 implies that the diagonalizing matrix of a Hermitian matrix A is unitary when A has distinct eigenvalues.
Lemma 1: if the eigenvalues of a Hermitian matrix
Aare distinct, then there exists a unitary matrixUand a diagonal matrixDsuch that
A U = U D.
??? note "Proof:"
Will be added later.
With the column vectors of U the eigenvectors of A and the diagonal elements of D the corresponding eigenvalues of A.
Theorem 9: let
Abe ann \times nmatrix, there exists a unitary matrixUand a upper triangular matrixTsuch that
A U = U T.
??? note "Proof:"
Will be added later.
The factorization A = U T U^H is often referred to as the Schur decomposition of A.
Theorem 10: if
Ais Hermitian, then there exists a unitary matrixUand a diagonal matrixDsuch that
A U = U D.
??? note "Proof:"
Will be added later.