Finished linear transformations.
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- 'Determinants': mathematics/linear-algebra/determinants.md
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- 'Vector spaces': mathematics/linear-algebra/vector-spaces.md
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- 'Linear transformations': mathematics/linear-algebra/linear-transformations.md
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- 'Orthogonality': mathematics/linear-algebra/orthogonality.md
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- 'Diagonalization': mathematics/linear-algebra/diagonalization.md
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- 'Calculus':
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- 'Limits': mathematics/calculus/limits.md
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- 'Continuity': mathematics/calculus/continuity.md
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- 'Physics':
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- physics/index.md
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# - 'Mechanics':
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# - 'Newtonian mechanics':
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# - 'Lagrangian mechanics':
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# - 'Hamiltonian mechanics':
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# - 'Relativistic mechanics':
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# - 'Quantum mechanics':
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- 'Electromagnetism':
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# - 'Electrostatics':
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# - 'Magnetostatics':
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- 'Interference': physics/electromagnetism/optics/interference.md
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- 'Diffraction': physics/electromagnetism/optics/diffraction.md
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- 'Polarisation': physics/electromagnetism/optics/polarisation.md
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# - 'Statistical physics':
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- 'Mathematical physics':
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- 'Signal analysis':
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- 'Signals': physics/mathematical-physics/signal-analysis/signals.md
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docs/en/mathematics/linear-algebra/diagonalization.md
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docs/en/mathematics/linear-algebra/diagonalization.md
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# Diagonalization
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>
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> for all $\mathbf{v}_{1,2} \in V$ and $\lambda, \mu \in \mathbb{K}$.
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In the case that the vector spaces $V$ and $W$ are the same; $V=W$, a linear transformation $L: V \to V$ will be reffered to as a **linear operator** on $V$.
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A linear transformation may also be called a **vector space homomorphism**. If the linear transformation is a bijection then it may be called a **linear isomorphism**.
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In the case that the vector spaces $V$ and $W$ are the same; $V=W$, a linear transformation $L: V \to V$ will be referred to as a **linear operator** on $V$ or **linear endomorphism** .
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## The image and kernel
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since $\lambda \mathbf{v}_1 + \mu \mathbf{v}_2 \in S$ it follows that $\lambda \mathbf{w}_1 + \mu \mathbf{w}_2 \in L(S)$ and hence $L(S)$ is a subspace of $W$.
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## Matrix representations
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> *Theorem*: let $L: \mathbb{R}^n \to \mathbb{R}^m$ be a linear transformation, then there is an $m \times n$ matrix $A$ such that
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>
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> $$
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> L(\mathbf{x}) = A \mathbf{x},
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> $$
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>
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> for all $x \in \mathbb{R}^n$. With the $i$th column vector of $A$ given by
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>
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> $$
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> \mathbf{a}_i = L(\mathbf{e}_i),
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> $$
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>
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> for a basis $\{\mathbf{e}_1, \dots, \mathbf{e}_n\} \subset \mathbb{R}^n$ and $i \in \{1, \dots, n\}$.
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??? note "*Proof*:"
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For $i \in \{1, \dots, n\}$, define
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$$
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\mathbf{a}_i = L(\mathbf{e}_i),
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$$
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and let
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$$
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A = (\mathbf{a}_1, \dots, \mathbf{a}_n).
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$$
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If $\mathbf{x} = x_1 \mathbf{e}_1 + \dots + x_n \mathbf{e}_n$ is an arbitrary element of $\mathbb{R}^n$, then
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$$
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\begin{align*}
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L(\mathbf{x}) &= x_1 L(\mathbf{e}_1) + \dots + x_n L(\mathbf{e}_n), \\
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&= x_1 \mathbf{a}_1 + \dots + x_n \mathbf{a}_n, \\
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&= A \mathbf{x}.
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\end{align*}
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$$
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It has therefore been established that each linear transformation from $\mathbb{R}^n$ to $\mathbb{R}^m$ can be represented in terms of an $m \times n$ matrix.
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> *Theorem*: let $E = \{\mathbf{e}_1, \dots, \mathbf{e}_n\}$ and $F = \{\mathbf{f}_1, \dots, \mathbf{f}_n\}$ be two ordered bases for a vector space $V$, and let $L: V \to V$ be a linear operator on $V$, $\dim V = n \in \mathbb{N}$. Let $S$ be the $n \times n$ transition matrix representing the change from $F$ to $E$,
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> $$
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> \mathbf{e}_i = S \mathbf{f}_i,
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> $$
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>
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> for $i \in \mathbb{N}; i\leq n$.
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>
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> If $A$ is the matrix representing $L$ with respect to $E$, and $B$ is the matrix representing $L$ with respect to $F$, then
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>
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> $$
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> B = S^{-1} A S.
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> $$
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??? note "*Proof*:"
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Will be added later.
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> *Definition*: let $A$ and $B$ be $n \times n$ matrices. $B$ is said to be **similar** to $A$ if there exists a nonsingular matrix $S$ such that $B = S^{-1} A S$.
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It follows from the above theorem that if $A$ and $B$ are $n \times n$ matrices representing the same operator $L$, then $A$ and $B$ are similar.
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docs/en/mathematics/linear-algebra/orthogonality.md
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docs/en/mathematics/linear-algebra/orthogonality.md
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# Orthogonality
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Let $A$ and $B$ be two matrices, if $B$ is row equivalent to $A$ then $B$ can be formed from $A$ by a finite sequence of row operations. Thus the row vectors of $B$ must be linear combinations of the row vectors of $A$. Consequently, the row space of $B$ must be a subspace of the row space of $A$. Since $A$ is row equivalent to $B$, by the same reasoning, the row space of $A$ is a subspace of the row space of $B$.
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With the definition of a column space a theorem posed in [systems of linear equations](systems-of-linear-equations.md) may be restatated as.
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With the definition of a column space a theorem posed in [systems of linear equations](systems-of-linear-equations.md) may be restated as.
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> *Theorem*: a linear system $A \mathbf{x} = \mathbf{b}$ is consistent if and only if $\mathbf{b}$ is in the column space of $A$.
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Let $A$ be an $m \times n$ matrix. If the column vectors of $A$ span $\mathbb{R}^m$, then $n$ must be greater or equal to $m$, since no set of fewer than $m$ vectors could span $\mathbb{R}^m$. If the columns of $A$ are linearly independent, then $n$ must be less than or equal to $m$, since every set of more than $m$ vectors in $\mathbb{R}^m$ is linearly dependent. Thus, if the column vectors of $A$ form a basis for $\mathbb{R}^m$, then $n = m$.
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<br>
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> *Theorem*: if $A$ is an $m \times n$ matrix, the dimension of the row space of $A$ equals the dimension of the column space of $A$.
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??? note "*Proof*:"
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Will be added later.
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## Rank and nullity
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> *Definition*: the **rank** of a matrix $A$, denoted as $\text{rank}(A)$, is the dimension of the row space of $A$.
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??? note "*Proof*:"
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Let $U$ be the reduced echelon form of $A$. The system $A \mathbf{x} = \mathbf{0}$ is equivalent to the system $U \mathbf{x} = \mathbf{0}$. If $A$ has rank $r$, then $U$ will have $r$ nonzero rows and consequently the system $U \mathbf{x} = \mathbf{0}$ will involve $r$ pivots and $n - r$ free variables. The dimension of the null space will equal the number of free variables.
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The section of vector spaces may be finished off, with this reasonably important theorem.
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> *Theorem*: if $A$ is an $m \times n$ matrix, the dimension of the row space of $A$ equals the dimension of the column space of $A$.
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??? note "*Proof*:"
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Will be added later.
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Let $U$ be the reduced echelon form of $A$. The system $A \mathbf{x} = \mathbf{0}$ is equivalent to the system $U \mathbf{x} = \mathbf{0}$. If $A$ has rank $r$, then $U$ will have $r$ nonzero rows and consequently the system $U \mathbf{x} = \mathbf{0}$ will involve $r$ pivots and $n - r$ free variables. The dimension of the null space will equal the number of free variables.
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