Mathematics for Applied Sciences (Osnabrück 2023-2024)/Part I/Lecture 24/refcontrol

Base change

We know, due to Theorem 23.15 , that in a finite-dimensional vector space, any two bases have the same length, the same number of vectors. Every vector has, with respect to every basis, unique coordinates (the coefficient tuple). How do these coordinates behave when we change the bases? This is answered by the following statement.

LemmaCreate referencenumber

Let ${}K$ be a field,MDLD/field and let ${}V$ be a ${}K$-vector spaceMDLD/vector space of dimensionMDLD/dimension (fgvs) ${}n$. Let ${}{\mathfrak {v}}=v_{1},\ldots ,v_{n}$ and ${}{\mathfrak {w}}=w_{1},\ldots ,w_{n}$ denote basesMDLD/bases (vs) of ${}V$. Suppose that

${}v_{j}=\sum _{i=1}^{n}c_{ij}w_{i}\,$

with coefficients ${}c_{ij}\in K$, which we collect into the ${}n\times n$-matrixMDLD/matrix

${}M_{\mathfrak {w}}^{\mathfrak {v}}={\left(c_{ij}\right)}_{ij}\,.$

Then a vector ${}u$, which has the coordinates ${}{\begin{pmatrix}s_{1}\\\vdots \\s_{n}\end{pmatrix}}$ with respect to the basis ${}{\mathfrak {v}}$, has the coordinates

${}{\begin{pmatrix}t_{1}\\\vdots \\t_{n}\end{pmatrix}}=M_{\mathfrak {w}}^{\mathfrak {v}}{\begin{pmatrix}s_{1}\\\vdots \\s_{n}\end{pmatrix}}={\begin{pmatrix}c_{11}&c_{12}&\ldots &c_{1n}\\c_{21}&c_{22}&\ldots &c_{2n}\\\vdots &\vdots &\ddots &\vdots \\c_{n1}&c_{n2}&\ldots &c_{nn}\end{pmatrix}}{\begin{pmatrix}s_{1}\\\vdots \\s_{n}\end{pmatrix}}\,$

with respect to the basis ${}{\mathfrak {w}}$.

Proof

This follows directly from

${}u=\sum _{j=1}^{n}s_{j}v_{j}=\sum _{j=1}^{n}s_{j}{\left(\sum _{i=1}^{n}c_{ij}w_{i}\right)}=\sum _{i=1}^{n}{\left(\sum _{j=1}^{n}s_{j}c_{ij}\right)}w_{i}\,,$

and the definition of matrix multiplication.MDLD/matrix multiplication

$\Box$

The matrix ${}M_{\mathfrak {w}}^{\mathfrak {v}}$, which describes the base change from ${}{\mathfrak {v}}$ to ${}{\mathfrak {w}}$, is called the transformation matrix. In the ${}j$-th column of the transformation matrix, there are the coordinates of ${}v_{j}$ with respect to the basis ${}{\mathfrak {w}}$. When we denote, for a vector ${}u\in V$ and a basis ${}{\mathfrak {v}}$, the corresponding coordinate tuple by ${}\Psi _{\mathfrak {v}}(u)$, then the transformation can be quickly written as

${}\Psi _{\mathfrak {w}}(u)=M_{\mathfrak {w}}^{\mathfrak {v}}(\Psi _{\mathfrak {v}}(u))\,.$

Example Create referencenumber

We consider in ${}\mathbb {R} ^{2}$ the standard basis,MDLD/standard basis

${}{\mathfrak {u}}={\begin{pmatrix}1\\0\end{pmatrix}},\,{\begin{pmatrix}0\\1\end{pmatrix}}\,,$

and the basis

${}{\mathfrak {v}}={\begin{pmatrix}1\\2\end{pmatrix}},\,{\begin{pmatrix}-2\\3\end{pmatrix}}\,.$

The basis vectors of ${}{\mathfrak {v}}$ can be expressed directly with the standard basis, namely

$v_{1}={\begin{pmatrix}1\\2\end{pmatrix}}=1{\begin{pmatrix}1\\0\end{pmatrix}}+2{\begin{pmatrix}0\\1\end{pmatrix}}\,\,{\text{ and }}\,\,v_{2}={\begin{pmatrix}-2\\3\end{pmatrix}}=-2{\begin{pmatrix}1\\0\end{pmatrix}}+3{\begin{pmatrix}0\\1\end{pmatrix}}.$

Therefore, we get immediately

${}M_{\mathfrak {u}}^{\mathfrak {v}}={\begin{pmatrix}1&-2\\2&3\end{pmatrix}}\,.$

For example, the vector which has with respect to ${}{\mathfrak {v}}$ the coordinatesMDLD/coordinates ${}(4,-3)$, has the coordinates

${}M_{\mathfrak {u}}^{\mathfrak {v}}{\begin{pmatrix}4\\-3\end{pmatrix}}={\begin{pmatrix}1&-2\\2&3\end{pmatrix}}{\begin{pmatrix}4\\-3\end{pmatrix}}={\begin{pmatrix}10\\-1\end{pmatrix}}\,$

with respect to the standard basis ${}{\mathfrak {u}}$. The transformation matrix ${}M_{\mathfrak {v}}^{\mathfrak {u}}$ is more difficult to compute: We have to write the standard vectors as linear combinationsMDLD/linear combinations of ${}v_{1}$ and ${}v_{2}$. A direct computation (solving two linear systems) yields

${}{\begin{pmatrix}1\\0\end{pmatrix}}={\frac {3}{7}}{\begin{pmatrix}1\\2\end{pmatrix}}-{\frac {2}{7}}{\begin{pmatrix}-2\\3\end{pmatrix}}\,$

and

${}{\begin{pmatrix}0\\1\end{pmatrix}}={\frac {2}{7}}{\begin{pmatrix}1\\2\end{pmatrix}}+{\frac {1}{7}}{\begin{pmatrix}-2\\3\end{pmatrix}}\,.$

Hence,

${}M_{\mathfrak {v}}^{\mathfrak {u}}={\begin{pmatrix}{\frac {3}{7}}&{\frac {2}{7}}\\-{\frac {2}{7}}&{\frac {1}{7}}\end{pmatrix}}\,.$

Linear mappings

Definition

Let ${}K$ be a field,MDLD/field and let ${}V$ and ${}W$ be ${}K$-vector spaces.MDLD/vector spaces A mappingMDLD/mapping

$\varphi \colon V\longrightarrow W$

is called a linear mapping, if the following two properties are fulfilled.

1. ${}\varphi (u+v)=\varphi (u)+\varphi (v)$ for all ${}u,v\in V$.
2. ${}\varphi (sv)=s\varphi (v)$ for all ${}s\in K$ and ${}v\in V$.

Here, the first property is called additivity and the second property is called compatibility with scaling. When we want to stress the base field, then we say ${}K$-linearity. The identity ${}\operatorname {Id} _{V}\colon V\rightarrow V$, the null mapping ${}V\rightarrow 0$ and the inclusion ${}U\subseteq V$ of a linear subspace are the simplest examples for linear mappings.

Example Create referencenumber

Let ${}K$ denote a field,MDLD/field and let ${}K^{n}$ be the ${}n$-dimensionalMDLD/dimensional (fgvs) standard space.MDLD/standard space Then the ${}i$-th projection, this is the mappingMDLD/mapping

$K^{n}\longrightarrow K,\left(x_{1},\,\ldots ,\,x_{i-1},\,x_{i},\,x_{i+1},\,\ldots ,\,x_{n}\right)\longmapsto x_{i},$

is a ${}K$-linear mapping.MDLD/linear mapping This follows immediately from componentwise addition and scalar multiplication on the standard space. The ${}i$-th projection is also called the ${}i$-th coordinate function.

LemmaCreate referencenumber

Let ${}K$ denote a field,MDLD/field and let ${}U,V,W$ denote vector spacesMDLD/vector spaces over ${}K$. Suppose that

$\varphi :U\longrightarrow V\,\,{\text{ and }}\,\,\psi :V\longrightarrow W$

are linear mappings.MDLD/linear mappings Then also the compositionMDLD/composition (map)

$\psi \circ \varphi \colon U\longrightarrow W$

is a linear mapping.

Proof

See Exercise 24.17 .

$\Box$

LemmaCreate referencenumber

Let ${}K$ be a field,MDLD/field and let ${}V$ and ${}W$ be ${}K$-vector spaces.MDLD/vector spaces Let

$\varphi \colon V\longrightarrow W$

be a bijectiveMDLD/bijective linear map.MDLD/linear map Then also the inverse mappingMDLD/inverse mapping

$\varphi ^{-1}\colon W\longrightarrow V$

is linear.

Proof

See Exercise 24.18 .

$\Box$

Determination on a basis

Behind the following statement (the determination theorem), there is the important principle that in linear algebra (of finite-dimensional vector spaces), the objects are determined by finitely many data.

TheoremTheorem 24.7 change

Let ${}K$ be a field,MDLD/field and let ${}V$ and ${}W$ be ${}K$-vector spaces.MDLD/vector spaces Let ${}v_{i}$, ${}i\in I$, denote a basisMDLD/basis (vs) of ${}V$, and let ${}w_{i}$, ${}i\in I$, denote elements in ${}W$. Then there exists a unique linear mappingMDLD/linear mapping

$f\colon V\longrightarrow W$

with

$f(v_{i})=w_{i}{\text{ for all }}i\in I.$

Proof

This proof was not presented in the lecture.

$\Box$

The graph of a linear mapping from ${}\mathbb {R}$ to ${}\mathbb {R}$, the mapping is determined by the proportionality factor ${}k$ alone.

Example Create referencenumber

The easiest linear mappingsMDLD/linear mappings are (beside the null mapping) the linear maps from ${}K$ to ${}K$. Such a linear mapping

$\varphi \colon K\longrightarrow K,x\longmapsto \varphi (x),$

is determined (by Theorem 24.7 , but this is also directly clear) by ${}\varphi (1)$, or by the value ${}\varphi (t)$ for an single element ${}t\in K$, ${}t\neq 0$. In particular, ${}\varphi (x)=ax$, with a uniquely determined ${}a\in K$. In the context of physics, for ${}K=\mathbb {R}$, and if there is a linear relation between two measurable quantities, we talk about proportionality, and ${}a$ is called the proportionality factor. In school, such a linear relation occurs as "rule of three“.

Linear mappings and matrices

The effect of several linear mappings from ${}\mathbb {R} ^{2}$ to itself, represented on a brain cell.

Due to Theorem 24.7 , a linear mapping

$\varphi \colon K^{n}\longrightarrow K^{m}$

is determined by the images ${}\varphi (e_{j})$, ${}j=1,\ldots ,n$, of the standard vectors. Every ${}\varphi (e_{j})$ is a linear combination

${}\varphi (e_{j})=\sum _{i=1}^{m}a_{ij}e_{i}\,,$

and therefore the linear mapping is determined by the elements ${}a_{ij}$. So, such a linear map is determined by the ${}mn$ elements ${}a_{ij}$, ${}1\leq i\leq m$, ${}1\leq j\leq n$, from the field. We can write such a data set as a matrix. Because of the determination theorem, this holds for linear maps in general, as soon as in both vector spaces bases are fixed.

Definition

Let ${}K$ denote a field,MDLD/field and let ${}V$ be an ${}n$-dimensionalMDLD/dimensional (vs) vector spaceMDLD/vector space with a basisMDLD/basis (vs) ${}{\mathfrak {v}}=v_{1},\ldots ,v_{n}$, and let ${}W$ be an ${}m$-dimensional vector space with a basis ${}{\mathfrak {w}}=w_{1},\ldots ,w_{m}$.

For a linear mappingMDLD/linear mapping

$\varphi \colon V\longrightarrow W,$

the matrixMDLD/matrix

${}M=M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )=(a_{ij})_{ij}\,,$

where ${}a_{ij}$ is the ${}i$-th coordinateMDLD/coordinate (vs) of ${}\varphi (v_{j})$ with respect to the basis ${}{\mathfrak {w}}$, is called the describing matrix for ${}\varphi$ with respect to the bases.

For a matrix ${}M=(a_{ij})_{ij}\in \operatorname {Mat} _{m\times n}(K)$, the linear mapping ${}\varphi _{\mathfrak {w}}^{\mathfrak {v}}(M)$ determined by

$v_{j}\longmapsto \sum _{i=1}^{m}a_{ij}w_{i}$

in the sense of Theorem 24.7 ,

is called the linear mapping determined by the matrix ${}M$.

For a linear mapping ${}\varphi \colon K^{n}\rightarrow K^{m}$, we always assume that everything is with respect to the standard bases, unless otherwise stated. For a linear mapping ${}\varphi \colon V\rightarrow V$ from a vector space in itself (what is called an endomorphism), one usually takes the same bases on both sides. The identity on a vector space of dimension ${}n$ is described by the identity matrix, with respect to every basis.

TheoremCreate referencenumber

Let ${}K$ be a field, and let ${}V$ be an ${}n$-dimensionalMDLD/dimensional (vs) vector spaceMDLD/vector space with a basisMDLD/basis (vs) ${}{\mathfrak {v}}=v_{1},\ldots ,v_{n}$, and let ${}W$ be an ${}m$-dimensional vector space with a basis ${}{\mathfrak {w}}=w_{1},\ldots ,w_{m}$. Then the mappings

$\varphi \longmapsto M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi ){\text{ and }}M\longmapsto \varphi _{\mathfrak {w}}^{\mathfrak {v}}(M),$

defined in definition,MDLD/definition are inverseMDLD/inverse to each other.

Proof

This proof was not presented in the lecture.

$\Box$

Example Create referencenumber

A linear mappingMDLD/linear mapping

$\varphi \colon K^{n}\longrightarrow K^{m}$

is usually described by the matrix ${}M$ with respect to the standard basesMDLD/standard bases (vs) on the left and on the right. The result of the matrix multiplication

${}{\begin{pmatrix}y_{1}\\\vdots \\y_{m}\end{pmatrix}}=M{\begin{pmatrix}x_{1}\\\vdots \\x_{n}\end{pmatrix}}\,$

can be interpreted directly as a point in ${}K^{m}$. The ${}j$-th column of ${}M$ is the image of the ${}j$-th standard vector ${}e_{j}$.

Rotations

A rotation of the real plane ${}\mathbb {R} ^{2}$ around the origin, given the angle ${}\alpha$ counterclockwise, maps ${}{\begin{pmatrix}1\\0\end{pmatrix}}$ to ${}{\begin{pmatrix}\cos \alpha \\\sin \alpha \end{pmatrix}}$ and ${}{\begin{pmatrix}0\\1\end{pmatrix}}$ to ${}{\begin{pmatrix}-\sin \alpha \\\cos \alpha \end{pmatrix}}$. Therefore, plane rotations are described in the following way.

Definition

A linear mappingMDLD/linear mapping

$D(\alpha )\colon \mathbb {R} ^{2}\longrightarrow \mathbb {R} ^{2},$

which is given by a rotation matrix ${}{\begin{pmatrix}\operatorname {cos} \,\alpha &-\operatorname {sin} \,\alpha \\\operatorname {sin} \,\alpha &\operatorname {cos} \,\alpha \end{pmatrix}}$ (with some ${}\alpha \in \mathbb {R}$)with respect to the standard basis is called

rotation.

A space rotation is a linear mapping of the space ${}\mathbb {R} ^{3}$ in itself around a rotation axis (a line through the origin) with an certain angle ${}\alpha$. If the vector ${}v_{1}\neq 0$ defines the axis, and ${}u_{2}$ and ${}u_{3}$ are orthogonal to ${}v_{1}$ and to each other, and all have length ${}1$, then the rotation is described by the matrix

${\begin{pmatrix}1&0&0\\0&\operatorname {cos} \,\alpha &-\operatorname {sin} \,\alpha \\0&\operatorname {sin} \,\alpha &\operatorname {cos} \,\alpha \end{pmatrix}}$

with respect to the basis ${}v_{1},u_{2},u_{3}$.

The kernel of a linear mapping

Definition

Let ${}K$ denote a field,MDLD/field let ${}V$ and ${}W$ denote ${}K$-vector spaces,MDLD/vector spaces and let

$\varphi \colon V\longrightarrow W$

denote a ${}K$-linear mapping.MDLD/linear mapping Then

${}\operatorname {kern} \varphi :=\varphi ^{-1}(0)={\left\{v\in V\mid \varphi (v)=0\right\}}\,$

is called the kernel of ${}\varphi$.

The kernel is a linear subspace of ${}V$.

The following criterion for injectivity is important.

LemmaLemma 24.14 change

Let ${}K$ denote a field,MDLD/field let ${}V$ and ${}W$ denote ${}K$-vector spaces,MDLD/vector spaces and let

$\varphi \colon V\longrightarrow W$

denote a ${}K$-linear mapping.MDLD/linear mapping Then ${}\varphi$ is injectiveMDLD/injective if and only if ${}\operatorname {kern} \varphi =0$ holds.

Proof

If the mapping is injective, then there can exist, apart from ${}0\in V$, no other vector ${}v\in V$ with ${}\varphi (v)=0$. Hence, ${}\varphi ^{-1}(0)=\{0\}$.
So suppose that ${}\operatorname {kern} \varphi =0$, and let ${}v_{1},v_{2}\in V$ be given with ${}\varphi (v_{1})=\varphi (v_{2})$. Then, due to linearity,

${}\varphi (v_{1}-v_{2})=\varphi (v_{1})-\varphi (v_{2})=0\,.$

Therefore, ${}v_{1}-v_{2}\in \operatorname {kern} \varphi$, and so ${}v_{1}=v_{2}$.

$\Box$

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