94 lines
3.5 KiB
Markdown
94 lines
3.5 KiB
Markdown
# Matrix algebra
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> *Theorem*: let $A, B$ and $C$ be matrices and $\alpha$ and $\beta$ be scalars. Each of the following statements is valid
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>
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> 1. $A + B = B + A$,
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> 2. $(A + B) + C = A + (B + C)$,
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> 3. $(AB)C = A(BC)$,
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> 4. $A(B + C) = AB + AC$,
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> 5. $(A + B)C = AC + BC$,
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> 6. $(\alpha \beta) A = \alpha(\beta A)$,
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> 7. $\alpha (AB) = (\alpha A)B = A (\alpha) B$,
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> 8. $(\alpha + \beta)A = \alpha A + \beta A$,
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> 9. $\alpha (A + B) = \alpha A + \alpha B$.
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??? note "*Proof*:"
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Will be added later.
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In the case where an $n \times n$ matrix $A$ is multiplied by itself $k$ times it is convenient to use exponential notation: $AA \cdots A = A^k$.
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> *Definition*: the $n \times n$ **identity matrix** is the matrix $I = (\delta_{ij})$, where
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>
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> $$
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> \delta_{ij} = \begin{cases} 1 &\text{ if } i = j, \\ 0 &\text{ if } i \neq j.\end{cases}
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> $$
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Obtaining for the multiplication of a $n \times n$ matrix $A$ with the identitiy matrix; $A I = A$.
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> *Definition*: an $n \times n$ matrix $A$ is said to be **nonsingular** or **invertible** if there exists a matrix $A^{-1}$ such that $AA^{-1} = A^{-1}A = I$. The matrix $A^{-1}$ is said to be a **multiplicative inverse** of $A$.
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If $B$ and $C$ are both multiplicative inverses of $A$ then
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$$
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B = BI = B(AC) = (BA)C = IC = C,
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$$
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thus a matrix can have at most one multiplicative inverse.
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> *Definition*: an $n \times n$ matrix is said to be **singular** if it does not have a multiplicative inverse.
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Or similarly, an $n \times n$ matrix $A$ is singular if $A \mathbf{x} = \mathbf{0}$ for some non trivial $\mathbf{x} \in \mathbb{R}^n \backslash \{\mathbf{0}\}$. For a nonsingular matrix $A$, $\mathbf{x} = \mathbf{0}$ is the only solution to $A \mathbf{x} = \mathbf{0}$.
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> *Theorem*: if $A$ and $B$ are nonsingular $n \times n$ matrices, then $AB$ is also nonsingular and
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>
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> $$
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> (AB)^{-1} = B^{-1} A^{-1}.
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> $$
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??? note "*Proof*:"
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Let $A$ and $B$ be nonsingular $n \times n$ matrices. If we suppose $AB$ is nonsingular and $(AB)^{-1} = B^{-1} A^{-1}$ we have
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$$
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(AB)^{-1}AB = (B^{-1} A^{-1})AB = B^{-1} (A^{-1} A) B = B^{-1} B = I, \\
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AB(AB)^{-1} = AB(B^{-1} A^{-1}) = A (B B^{-1}) A^{-1} = A A^{-1} = I.
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$$
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> *Theorem*: let $A$ be a nonsingular $n \times n$ matrix, the inverse of $A$ given by $A^{-1}$ is nonsingular.
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??? note "*Proof*:"
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Let $A$ be a nonsingular $n \times n$ matrix, $A^{-1}$ its inverse and $\mathbf{x} \in \mathbb{R}^n$ a vector. Suppose $A^{-1} \mathbf{x} = \mathbf{0}$ then
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$$
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\mathbf{x} = I \mathbf{x} = (A A^{-1}) \mathbf{x} = A(A^{-1} \mathbf{x}) = \mathbf{0}.
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$$
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> *Theorem*: let $A$ be a nonsingular $n \times n$ matrix then the solution of the system $A\mathbf{x} = \mathbf{b}$ is $\mathbf{x} = A^{-1} \mathbf{b}$ with $\mathbf{x}, \mathbf{b} \in \mathbb{R}^n$.
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??? note "*Proof*:"
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Let $A$ be a nonsingular $n \times n$ matrix, $A^{-1}$ its inverse and $\mathbf{x}, \mathbf{b} \in \mathbb{R}^n$ vectors. Suppose $\mathbf{x} = A^{-1} \mathbf{b}$ then we have
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$$
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A \mathbf{x} = A (A^{-1} \mathbf{b}) = (A A^{-1}) \mathbf{b} = \mathbf{b}.
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$$
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> *Corollary*: the system $A \mathbf{x} = \mathbf{b}$ of $n$ linear equations in $n$ unknowns has a unique solution if and only if $A$ is nonsingular.
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??? note "*Proof*:"
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The proof follows from the above theorem.
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> *Theorem*: let $A$ and $B$ be matrices and $\alpha$ and $\beta$ be scalars. Each of the following statements valid
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>
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> 1. $(A^T)^T = A$,
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> 2. $(\alpha A)^T = \alpha A^T$,
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> 3. $(A + B)^T = A^T + B^T$,
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> 4. $(AB)^T = B^T A^T$.
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??? note "*Proof*:"
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Will be added later.
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