diff --git a/config/en/mkdocs.yaml b/config/en/mkdocs.yaml index a91a90e..1458272 100755 --- a/config/en/mkdocs.yaml +++ b/config/en/mkdocs.yaml @@ -90,7 +90,7 @@ nav: - 'Linear transformations': mathematics/linear-algebra/linear-transformations.md - 'Inner product spaces': mathematics/linear-algebra/inner-product-spaces.md - 'Orthogonality': mathematics/linear-algebra/orthogonality.md - - 'Diagonalization': mathematics/linear-algebra/diagonalization.md + - 'Eigenspaces': mathematics/linear-algebra/eigenspaces.md - 'Calculus': - 'Limits': mathematics/calculus/limits.md - 'Continuity': mathematics/calculus/continuity.md diff --git a/docs/en/mathematics/linear-algebra/diagonalization.md b/docs/en/mathematics/linear-algebra/diagonalization.md deleted file mode 100644 index b90827f..0000000 --- a/docs/en/mathematics/linear-algebra/diagonalization.md +++ /dev/null @@ -1 +0,0 @@ -# Diagonalization \ No newline at end of file diff --git a/docs/en/mathematics/linear-algebra/eigenspaces.md b/docs/en/mathematics/linear-algebra/eigenspaces.md new file mode 100644 index 0000000..9589f20 --- /dev/null +++ b/docs/en/mathematics/linear-algebra/eigenspaces.md @@ -0,0 +1,267 @@ +# 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 $A$ 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 +> +> $$ +> A \mathbf{x} = \lambda \mathbf{x}, +> $$ +> +> 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 $A$ be a $n \times n$ matrix, a scalar $\lambda \in \mathbb{K}$ is an eigenvalue of $A$ if 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 $A$ 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 +> +> $$ +> \mathbf{x} \in N(A - \lambda I) \backslash \{\mathbf{0}\}. +> $$ + +??? 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 $L: V \to V$ 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 +> +> $$ +> E_\lambda = \{\mathbf{x} \in V \;|\; L(\mathbf{x}) = \lambda \mathbf{x}\} = N(A - \lambda I), +> $$ +> +> 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 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. + +??? 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 $A$ be a $n \times n$ matrix and let $\lambda_1, \dots, \lambda_n \in \mathbb{K}$ be the eigenvalues of $A$. It follows that +> +> $$ +> \det (A - \lambda I) = (\lambda_1 - \lambda)(\lambda_2 - \lambda) \cdots (\lambda_n - \lambda), +> $$ +> +> 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 $A$ and $B$ be $n \times n$ matrices. If $B$ is similar to $A$, then $A$ and $B$ have 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 $n \times n$ matrix $A$ is **diagonalizable** if there exists a nonsingular diagonalizing matrix $X$ and a diagonal matrix $D$ such that +> +> $$ +> A X = X D. +> $$ + +We may now pose the following theorem. + +> *Theorem 6*: an $n \times n$ matrix $A$ is diagonalizable if and only if $A$ has $n \in \mathbb{N}$ 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](../matrices/matrix-arithmatic/#hermitian-matrix). + +> *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 $n \times n$ matrix $U$ is **unitary** if the column vectors of $U$ form an orthonormal set in $V$. + +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 $A$ are distinct, then there exists a unitary matrix $U$ and a diagonal matrix $D$ such 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 $A$ be an $n \times n$ matrix, there exists a unitary matrix $U$ and a upper triangular matrix $T$ such 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 $A$ is Hermitian, then there exists a unitary matrix $U$ and a diagonal matrix $D$ such that +> +> $$ +> A U = U D. +> $$ + +??? note "*Proof*:" + + Will be added later. \ No newline at end of file diff --git a/docs/en/mathematics/linear-algebra/matrices/matrix-arithmetic.md b/docs/en/mathematics/linear-algebra/matrices/matrix-arithmetic.md index 0d76ff3..5b960f8 100644 --- a/docs/en/mathematics/linear-algebra/matrices/matrix-arithmetic.md +++ b/docs/en/mathematics/linear-algebra/matrices/matrix-arithmetic.md @@ -70,10 +70,10 @@ with $A = (\mathbf{a_1}, \mathbf{a_2}, \dots, \mathbf{a_n})$. ## Transpose matrix -> *Definition*: the transpose of an $m \times n$ matrix A is the $n \times m$ matrix $B$ defined by +> *Definition*: the **transpose** of an $m \times n$ matrix A is the $n \times m$ matrix $B$ defined by > > $$ -> b_{ji} = a_{ij} +> b_{ji} = a_{ij}, > $$ > > for $j \in \{1, \dots, n\}$ and $i \in \{1, \dots m\}$. The transpose of $A$ is denoted by $A^T$. @@ -82,6 +82,20 @@ with $A = (\mathbf{a_1}, \mathbf{a_2}, \dots, \mathbf{a_n})$. > *Definition*: an $n \times n$ matrix $A$ is said to be **symmetric** if $A^T = A$. +## Hermitian matrix + +> *Definition*: the **conjugate transpose** of an $m \times n$ matrix A is the $n \times m$ matrix $B$ defined by +> +> $$ +> b_{ji} = \bar a_{ij}, +> $$ +> +> for $j \in \{1, \dots, n\}$ and $i \in \{1, \dots m\}$. The **conjugate transpose** of $A$ is denoted by $A^H$. + +
+ +> *Definition*: an $n \times n$ matrix $A$ is said to be **Hermitian** if $A^H = A$. + ## Matrix multiplication > *Definition*: if $A = (a_{ij})$ is an $m \times n$ matrix and $B = (b_{ij})$ is an $n \times r$ matrix, then the product $A B = C = (c_{ij})$ is the $m \times r$ matrix whose entries are defined by