OpenLLM-France/wikimedia huggingface dataset:
# Derivation as operator/Linear/Eigenvectors/Exercise

Let $V$ denote the real vector space that consists of all functions from $\mathbb {R}$ to $\mathbb {R}$ that are arbitrarily often differentiable.
a) Show that the derivation $f\mapsto f'$ is a linear mapping from $V$ to $V$.
b) Determine the eigenvalues of the derivation and determine, for each eigenvalue, at least one eigenvector.
c) Determine for every real number the eigenspace and its dimension.
# Vector/Value/As eigenvector/Linear system/Exercise

Let $K$ be a field, $c\in K$ and let $a={\begin{pmatrix}a_{1}\\\vdots \\a_{n}\end{pmatrix}}\in K^{n}$ denote a vector different from $0$. Establish an inhomogeneous system of linear equations such that its solution set is the set of all $n\times n$-matrices, for that $a$ is an eigenvector with eigenvalue $c$. What is special about this system, and what is the dimension of its solution set?
# Eigenvectors/2 1 -2+i 0 i 1+i 0 0 -1+2i/Characteristic polynomial/Eigenvalues and eigenspaces/Diagonal matrix/Exercise

We consider the linear mapping
    $\varphi \colon \mathbb {C} ^{3}\longrightarrow \mathbb {C} ^{3}$
that is given by the matrix
    $A={\begin{pmatrix}2&1&-2+{\mathrm {i} }\\0&{\mathrm {i} }&1+{\mathrm {i} }\\0&0&-1+2{\mathrm {i} }\end{pmatrix}}\,,$
with respect to the standard basis.
a) Determine the characteristic polynomial and the eigenvalues of $A$.
b) Compute, for every eigenvalue, an eigenvector.
c) Establish a matrix for $\varphi$ with respect to a basis of eigenvectors.
# Four tuple/Modulus of differences/R/Infinite dilation/Exercise

We consider the mapping
    $\Psi \colon \mathbb {R} _{\geq 0}^{4}\longrightarrow \mathbb {R} _{\geq 0}^{4}$
that assigns to a four tuple $(a,b,c,d)$ the four tuple
    $(\vert {b-a}\vert ,\vert {c-b}\vert ,\vert {d-c}\vert ,\vert {a-d}\vert ).$
Show that there exists a tuple $(a,b,c,d)$, for that arbitrary iterations of the mapping do never reach the zero tuple.
# Vector space/Endomorphism/Smallest invariant subspace/Exercise

Let $\varphi \colon V\rightarrow V$ a linear mapping on a $K$-vector space $V$ over a field $K$, and let $v\in V$. Show that the smallest $\varphi$-invariant linear subspace of $V$ that contains $v$, equals
    $\langle \varphi ^{n}(v),\,n\in \mathbb {N} \rangle .$
# Permutation matrix/Invariant standard subspaces/Exercise

Let $\pi \in S_{n}$ be a permutation, and let $M_{\pi }$ denote the corresponding permutation matrix over a field $K$. For $J\subseteq \{1,\ldots ,n\}$, let
    $V_{J}=\langle e_{j},\,j\in J\rangle \subseteq K^{n}\,.$
a) Show that $V_{J}$ is $M_{\pi }$-invariant if and only if $\pi (J)\subseteq J$.
b) Show that there might exist $M_{\pi }$-invariant subspaces that are not of the form $V_{J}$.
# 2x2-matrix/At least one 0/Not two 0/Exercise

Let
    $M={\begin{pmatrix}a&b\\c&d\end{pmatrix}}\,$
be a matrix over a field $K$.
a) Show that there exists a matrix that is similar to $M$, and where at least one entry equals $0$.
b) Show that, in general, there does not exist a matrix that is similar to $M$, and where at least two entries equal $0$.
# Square matrix/Real not trigonalizable/Complex diagonalizable/Exercise

Let $M$ be a real matrix that is over $\mathbb {R}$ not trigonalizable. Show that $M$ is over $\mathbb {C}$ diagonalizable.
# 2x2-matrix/Trace is 0/Anti diagonal/Exercise

Let
    $M={\begin{pmatrix}a&b\\c&d\end{pmatrix}}\,$
be a matrix over $\mathbb {Q}$, whose trace is $0$. Show that there exists a matrix $N$ of the form
    $N={\begin{pmatrix}0&r\\s&0\end{pmatrix}}\,$
that is similar to $M$.
# Matrix/Trigonalizable/Not diagonalizable/Not bijective/Nilpotent/Exercise

a) Let $M$ be a $2\times 2$-matrix that is trigonalizable, but neither diagonalizable nor invertible. Show that $M$ is nilpotent.
b) Give an example of a $3\times 3$-matrix $M$ that is trigonalizable, but neither diagonalizable, nor invertible, nor nilpotent.
# Endomorphism/3x3/Not nilpotent/Trigonalizable/Determinant and Trace/Exercise

Give an example of a linear mapping $\varphi \colon K^{3}\rightarrow K^{3}$ that is trigonalizable, where the trace and the determinant are $0$, and that is not nilpotent.
# Matrix/ 2 3 4 0 2 5 0 0 2/Jordan/Exercise

We consider the linear mapping
    $\varphi \colon \mathbb {R} ^{3}\longrightarrow \mathbb {R} ^{3}$
that is described, with respect to the standard basis, by the matrix
    ${\begin{pmatrix}2&3&4\\0&2&5\\0&0&2\end{pmatrix}}.$
Find a basis such that $\varphi$ is described, with respect to this basis, by the matrix
    ${\begin{pmatrix}2&1&0\\0&2&1\\0&0&2\end{pmatrix}}.$
# Jordan matrix/No direct sum decomposition/Exercise

Let
    $\varphi \colon V\longrightarrow V$
be an endomorphism that is described, with respect to a suitable basis, by a $n\times n$-Jordan matrix. Show that there does not exist a direct sum decomposition
    $V=U\oplus W\,$
into $\varphi$-invariant linear subspaces $U,W\subset V$.
# R^2/Points/Barycentric combination/Sketch/1/Exercise

Determine by a drawing the point that is given by the barycentric combination
    $0,2P_{1}+0,4P_{2}-0,3P_{3}+0,7P_{4}$
in the image on the right. Start with different starting points.
# Polynomials/1/Affine-linear/Exercise

Determine the polynomials $P\in \mathbb {R} [X]$ that define an affine-linear mapping
    $P\colon \mathbb {R} \longrightarrow \mathbb {R} .$
# Affine mapping/Affine plane/Points/Image/1/Exercise

Determine, by a drawing, the image point of $P$ under the affine mapping $\varphi$ that is determined by $\varphi (P_{i})=Q_{i}$.
# Affine mapping/Linear part/Homomorphism/Exercise

Let $E$ be an affine space over the $K$-vector space $V$. Show that the mapping that assigns to an affine mapping
    $\psi \colon E\longrightarrow E$
its linear part $\psi _{0}$, satisfies the following properties.
    ${\left(\operatorname {Id} _{E}\right)}_{0}=\operatorname {Id} _{V}\,,$
    $(\psi \circ \varphi )_{0}=\psi _{0}\circ \varphi _{0}\,.$
# Vector space with inner product/Orthogonal/Length symmetry/Remark

The following reasoning shows that the orthogonality, defined via the inner product, corresponds to the intuitively given orthogonality . For orthogonal vectors $u,v\in V$, we get that $v$ has the same distance to the points $u$ and $-u$. This holds because of
    ${\begin{aligned}\Vert {v-u}\Vert ^{2}&=\left\langle v-u,v-u\right\rangle \\&=\left\langle v,v\right\rangle +\left\langle u,u\right\rangle \\&=\left\langle v+u,v+u\right\rangle \\&=\Vert {v+u}\Vert ^{2}.\end{aligned}}$
The reverse statement holds as well; see exercise.
# Vector space with inner product/Orthogonal complement/Definition

Let $V$ be a ${\mathbb {K} }$-vector space, endowed with an inner product, and let $U\subseteq V$ denote a linear subspace. Then
    $[[:Template:Op:Orthogonalcomplement]]={\left\{v\in V\mid \left\langle v,u\right\rangle =0{\text{ for all }}u\in U\right\}}\,$
is called the orthogonal complement of $U$.
# Russell's paradox

Russell's paradox resembles Pinocchio paradox in that it involves self-referencing. Pinocchio's words are in conflict with his nose: If his words are true, his nose would not grow. But since his nose is growing, we know his words are false. In short:
* His words are false if his words are true.
* His words are true if his words are false.
With Russell's paradox, the self-referencing conflict is between a set's definition and the items that are in the set. The paradox is important when sets are allowed be contain sets as members, as is the case with the pair of sets used to define surreal numbers.
In naive set theory, the universe is a collection that contains all the entities under consideration for membership in a set. The figure to the right corresponds to a universe that consists of three elements: $a$, $b$, and $c.$ The powerset consists of all sets that one can construct from these three elements (including the null set, $\Phi \equiv \{\;\},$ and the universe, $U\equiv \{a,b,c\}.$

### When set theory gets complicated

We begin with a simple (naive) version of set theory based on the integers $1$ and $2$. For a universe $U$ with only two elements, the powerset contains four elements: $P(U)=\{\Phi ,A,B,U\}.$
    $\Phi =\{\,\},\;$$\,\;A=\{1\},\;$ $\;B=\{2\},\;$ and $\;U=\{1,2\}=\{2,1\}.$
Great complexity results from what will turn out to be a failed attempt to define an alternate "universe" that we shall call $U^{*},$ that includes not only $1$ and $2,$ but all sets associated with this pair of integers. It is obvious that $U^{*}$ contains an infinite number of elements. For example, consider $A_{1}=\{A\}=\{\{1\}\},$ $A_{2}=\{A_{1}\}=\{\{\{1\}\}\},$ and $A_{3}=\{\{\{\{1\}\}\}\}.$This brings us to a collection of entities that can be defined recursively
$A_{n+1}=\{A_{n}\}.$
Letting $n$ go to $\infty$ yields:
$A_{\infty }=\lim _{n\to \infty }A_{n}=\{\{\{\ldots \{\{\{\,1\,\}\}\}\ldots \}\}\}$
This set contains itself, in that $A_{\infty }$ is an element of $A_{\infty }.$ Another set that contains itself is,
$b=\{2,b\}=\{2,\{2,b\}\}=\{2,\{2,\{2,b\}\}\}=\ldots .$
Curiously, Russell's paradox involves not sets that contain themselves as members (e.g., $a$ and $b,$) but sets where no member of the set in question is that set. This collection of sets that do not contain themselves includes not only the aforementioned power set $(\{1\},\{2\},\{1,2\},\{\,\}),\,$ but an infinite number of other sets that do not contain themselves. For example,
$c=\{2,\{2\},b\}$
does not contain itself, despite the fact that the third member of $c$ does contain itself.

### Defining R as the set of all sets that do not contain themselves

Given the original elements 1 and 2, permitting sets to be elements greatly expands the "universe". Russell's paradox involves the set we shall call $R,$ which is the set of all sets that do not contain themselves as members.
The paradoxical question is whether $R$ is a member of $R.$ In other words: Is$R\in R?$ Or is $R\not \in R?$ Both statements violate the definition of $R.$
In symbols:
    ${\text{Let }}R=\{x\mid x\not \in x\}{\text{, then }}R\in R\iff R\not \in R$
# Base change/Transformation matrix/Definition

Let $K$ denote a field, and let $V$ denote a $K$-vector space of dimension $n$. Let ${\mathfrak {v}}=v_{1},\ldots ,v_{n}$ and ${\mathfrak {w}}=w_{1},\ldots ,w_{n}$ denote two bases of $V$. Let
    $v_{j}=\sum _{i=1}^{n}c_{ij}w_{i}\,$
with coefficients $c_{ij}\in K$. Then the $n\times n$-matrix
    $M_{\mathfrak {w}}^{\mathfrak {v}}=(c_{ij})_{ij}\,$
is called the transformation matrix of the base change from ${\mathfrak {v}}$ to ${\mathfrak {w}}$.
# MDLD/change of basis

Start=change of basis Target=Base change/Transformation matrix/Definition
# MDLD/transformation matrix

Start=transformation matrix Target=Base change/Transformation matrix/Definition
# Base change/Three bases/Composition/Fact

Let $K$ be a field, and let $V$ be a $K$-vector space of dimension $n$. Let ${\mathfrak {u}}=u_{1},\ldots ,u_{n},\,{\mathfrak {v}}=v_{1},\ldots ,v_{n},\,$ and ${\mathfrak {w}}=w_{1},\ldots ,w_{n}$ denote bases of $V$.
transformation matrices fulfill the relation
    $M_{\mathfrak {w}}^{\mathfrak {u}}=M_{\mathfrak {w}}^{\mathfrak {v}}\circ M_{\mathfrak {v}}^{\mathfrak {u}}\,.$
In particular, we have
    $M_{\mathfrak {v}}^{\mathfrak {u}}\circ M_{\mathfrak {u}}^{\mathfrak {v}}=E_{n}\,.$
# Vector space/Linear subspaces/Finite/Direct sum/Definition

Let $K$ denote a field, and let $V$ denote a $K$-vector space. Let $U_{1},\ldots ,U_{m}$ be a family of linear subspaces of $V$. We say that $V$ is the direct sum of the $U_{i}$ if the following conditions are fulfilled.
* Every vector $v\in V$ has a representation where $u_{i}\in U_{i}$.
    $v=u_{1}+u_{2}+\cdots +u_{m}\,,$
* $U_{i}\cap {\left(\sum _{j\neq i}U_{j}\right)}=0$ for all $i$.
# Vector space/Finite dimensional/Linear subspace/Direct complement/Fact

Let $V$ be a finite-dimensional $K$-vector space, and let $U\subseteq V$ be a linear subspace.
$W\subseteq V$ such that we have the direct sum decomposition
    $V=U\oplus W\,.$
# Product set/Arbitrary/Vector spaces/Direct sum/Definition

Let $I$ denote a set, and let $K$ denote a field. Suppose that, for every $i\in I$, a $K$-vector space $V_{i}$ is given. Then the set
    $\bigoplus _{i\in I}V_{i}={\left\{(v_{i})_{i\in I}\mid v_{i}\in V_{i},\,v_{i}\neq 0{\text{ for only finitely many }}i\right\}}\,$
is called the direct sum of the $V_{i}$.
# Base change/Transformation matrix/Varia/Remark

The $j$-th column of a transformation matrix $M_{\mathfrak {w}}^{\mathfrak {v}}$ consists of the coordinates of $v_{j}$ with respect to the basis ${\mathfrak {w}}$. The vector $v_{j}$ has the coordinate tuple $e_{j}$ with respect to the basis ${\mathfrak {v}}$, and when we apply the matrix to $e_{j}$, we get the $j$-th column of the matrix, and this is just the coordinate tuple of $v_{j}$ with respect to the basis ${\mathfrak {w}}$.
For a one-dimensional space and
    $v=cw\,,$
we have $M_{\mathfrak {w}}^{\mathfrak {v}}=c={\frac {v}{w}}$, where the fraction is well-defined. This might help in memorizing the order of the bases in this notation.
Another important relation is
    ${\mathfrak {v}}={{\left(M_{\mathfrak {w}}^{\mathfrak {v}}\right)}^{\text{tr}}}{\mathfrak {w}}\,.$
Note that here, the matrix is not applied to an $n$-tuple of $K$ but to an $n$-tuple of $V$, yielding a new $n$-tuple of $V$. This equation might be an argument to define the transformation matrix the other way around; however, we consider the behavior in fact as decisive.
In case
    $V=K^{n}\,,$
if ${\mathfrak {e}}$ is the standard basis, and ${\mathfrak {v}}$ some further basis, we obtain the transformation matrix $M_{\mathfrak {v}}^{\mathfrak {e}}$ of the base change from ${\mathfrak {e}}$ to ${\mathfrak {v}}$ by expressing each $e_{j}$ as a linear combination of the basis vectors $v_{1},\ldots ,v_{n}$, and writing down the corresponding tuples as columns. The inverse transformation matrix, $M_{\mathfrak {e}}^{\mathfrak {v}}$, consists simply in $v_{1},\ldots ,v_{n}$, written as columns.
# Change of basis/R^2/Standard and 12,-23/Example

We consider in $\mathbb {R} ^{2}$ the 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 that has the coordinates $(4,-3)$ with respect to ${\mathfrak {v}}$, 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 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 subspaces/Sum/Definition

For a $K$-vector space $V$ and a family of linear subspaces $U_{1},\ldots ,U_{n}\subseteq V$, we define the sum of these linear subspaces by
    $U_{1}+\cdots +U_{n}={\left\{u_{1}+\cdots +u_{n}\mid u_{i}\in U_{i}\right\}}\,.$
# Linear subspace/Intersection/Dimension estimate/Fact

Let $K$ be a field, and let $V$ be a $K$-vector space of dimension $n$. Let $U_{1},U_{2}\subseteq V$ denote linear subspaces of dimensions $\dim _{K}{\left(U_{1}\right)}=n-k_{1}$ and $\dim _{K}{\left(U_{2}\right)}=n-k_{2}$.
    $\dim _{K}{\left(U_{1}\cap U_{2}\right)}\geq n-k_{1}-k_{2}\,.$
# Vector space/Direct sum/Introduction/Section
## Definition

Let $K$ denote a field, and let $V$ denote a $K$-vector space. Let $U_{1},\ldots ,U_{m}$ be a family of linear subspaces of $V$. We say that $V$ is the direct sum of the $U_{i}$ if the following conditions are fulfilled.
* Every vector $v\in V$ has a representation where $u_{i}\in U_{i}$.
    $v=u_{1}+u_{2}+\cdots +u_{m}\,,$
* $U_{i}\cap {\left(\sum _{j\neq i}U_{j}\right)}=0$ for all $i$.
If the sum of the $U_{i}$ is direct, then we also write $U_{1}\oplus \cdots \oplus U_{m}$ instead of $U_{1}+\cdots +U_{m}$. For two linear subspaces
    $U_{1},U_{2}\subseteq V\,,$
the second condition just means $U_{1}\cap U_{2}=0$.

## Example

Let $V$ denote a finite-dimensional $K$-vector space together with a basis $v_{1},\ldots ,v_{n}$. Let
    $\{1,\ldots ,n\}=I_{1}\uplus \ldots \uplus I_{k}\,$
be a partition of the index set. Let
    $U_{j}=\langle v_{i},\,i\in I_{j}\rangle \,$
be the linear subspaces generated by the subfamilies. Then
    $V=U_{1}\oplus \cdots \oplus U_{k}\,.$
The extreme case $I_{j}=\{j\}$ yields the direct sum
    $V=Kv_{1}\oplus \cdots \oplus Kv_{n}\,$
with one-dimensional linear subspaces.

## Lemma

Let $V$ be a finite-dimensional $K$-vector space, and let $U\subseteq V$ be a linear subspace. Then there exists a linear subspace $W\subseteq V$ such that we have the direct sum decomposition
    $V=U\oplus W\,.$

### Proof

Let $v_{1},\ldots ,v_{k}$ denote a basis of $U$. We can extend this basis, according to fact, to a basis $v_{1},\ldots ,v_{k},v_{k+1},\ldots ,v_{n}$ of $V$. Then
    $W=\langle v_{k+1},\ldots ,v_{n}\rangle \,$
fulfills all the properties of a direct sum.
$\Box$
In the preceding statement, the linear subspace $W$ is called a direct complement for $U$ (in $V$). In general, there are many different direct complements.
# Real numbers/Series/Definition

Let ${\left(a_{k}\right)}_{k\in \mathbb {N} }$ be a sequence of real numbers. The series $\sum _{k=0}^{\infty }a_{k}$ is the sequence ${\left(s_{n}\right)}_{n\in \mathbb {N} }$ of the partial sums
    $s_{n}:=\sum _{k=0}^{n}a_{k}\,.$
If the sequence ${\left(s_{n}\right)}_{n\in \mathbb {N} }$ converges, then we say that the series converges. In this case, we also write
    $\sum _{k=0}^{\infty }a_{k}$
for its limit, and this limit is called the sum of the series.
# Direct product/Direct sum/Introduction/Section

Recall that, for a family $M_{i}$, $i\in I$, of sets $M_{i}$, the product set $\prod _{i\in I}M_{i}$ is defined. If all $M_{i}=V_{i}$ are $K$-vector spaces over a field $K$, then this is, using componentwise addition and scalar multiplication, again a $K$-vector space. This is called the direct product of vector spaces. If it is always the same space, say $M_{i}=V$, then we also write $V^{I}$. This is just the mapping space $\operatorname {Map} \,{\left(I,V\right)}$.
Each vector space $V_{j}$ is a linear subspace inside the direct product, namely as the set of all tuples
    $(x_{i})_{i\in I}{\text{ with }}x_{i}=0{\text{ for all }}i\neq j.$
The set of all these tuples that are only (at most) at one place different from $0$ generates a linear subspace of the direct product. For $I$ infinite, it is not the direct product.

## Definition

Let $I$ denote a set, and let $K$ denote a field. Suppose that, for every $i\in I$, a $K$-vector space $V_{i}$ is given. Then the set
    $\bigoplus _{i\in I}V_{i}={\left\{(v_{i})_{i\in I}\mid v_{i}\in V_{i},\,v_{i}\neq 0{\text{ for only finitely many }}i\right\}}\,$
$V_{i}$
We have the linear subspace relation
    $\bigoplus _{i\in I}V_{i}\subseteq \prod _{i\in I}V_{i}\,.$
If we always have the same vector space, then we write $V^{(I)}$ for this direct sum. In particular,
    $V^{(I)}\subseteq V^{I}\,$
is a linear subspace. For $I$ finite, there is no difference, but for an infinite index set, this inclusion is strict. For example, $\mathbb {R} ^{\mathbb {N} }$ is the space of all real sequences, but $\mathbb {R} ^{(\mathbb {N} )}$ consists only of those sequences satisfying the property that only finitely many members are different from $0$. The polynomial ring $K[X]$ is the direct sum of the vector spaces $KX^{n},\,n\in \mathbb {N}$. Every $K$-vector space with a basis $v_{i}$, $i\in I$, is "isomorphic“ to the direct sum $\bigoplus _{i\in I}Kv_{i}$.
# Matrix/Addition and scalar multiplication/Remark

Two matrices $A,B\in \operatorname {Mat} _{m\times n}(K)$ are added by adding corresponding entries. The multiplication of a matrix $A$ with an element $r\in K$ (a scalar) is also defined entrywise, so
    ${\begin{pmatrix}a_{11}&a_{12}&\ldots &a_{1n}\\a_{21}&a_{22}&\ldots &a_{2n}\\\vdots &\vdots &\ddots &\vdots \\a_{m1}&a_{m2}&\ldots &a_{mn}\end{pmatrix}}+{\begin{pmatrix}b_{11}&b_{12}&\ldots &b_{1n}\\b_{21}&b_{22}&\ldots &b_{2n}\\\vdots &\vdots &\ddots &\vdots \\b_{m1}&b_{m2}&\ldots &b_{mn}\end{pmatrix}}={\begin{pmatrix}a_{11}+b_{11}&a_{12}+b_{12}&\ldots &a_{1n}+b_{1n}\\a_{21}+b_{21}&a_{22}+b_{22}&\ldots &a_{2n}+b_{2n}\\\vdots &\vdots &\ddots &\vdots \\a_{m1}+b_{m1}&a_{m2}+b_{m2}&\ldots &a_{mn}+b_{mn}\end{pmatrix}}\,$
and
    $r{\begin{pmatrix}a_{11}&a_{12}&\ldots &a_{1n}\\a_{21}&a_{22}&\ldots &a_{2n}\\\vdots &\vdots &\ddots &\vdots \\a_{m1}&a_{m2}&\ldots &a_{mn}\end{pmatrix}}={\begin{pmatrix}ra_{11}&ra_{12}&\ldots &ra_{1n}\\ra_{21}&ra_{22}&\ldots &ra_{2n}\\\vdots &\vdots &\ddots &\vdots \\ra_{m1}&ra_{m2}&\ldots &ra_{mn}\end{pmatrix}}\,.$
# Illustrated Companion to the Latin Dictionary/Parastas
## Illustrated Companion to the Latin Dictionary, and Greek Lexicon (Rich, 1849)

PARAS'TAS, PARAS'TATA, PARASTAT'ICA (παραστάς, παραστάτης, παραστατικὴ). A flat column or pilaster, used to decorate the angular terminations of a square building, where it has two faces, as in the annexed example (Parastas/1.1) of the temple of Pandrosus at Athens, in which the parastas is seen behind the last figure on the extreme left; or placed against the walls of the cell (cella), with one flat face to correspond with the opposite column which supports the entablature of the colonnade. Vitruv. v. 1.
* Parastas/1.1
# Open data portal

Open data portal is a way to make open data available for fair use. Governments or non-government entities can provide data through the open data portals, and citizens, researchers, or any other entity to whom this data is of interest can make use of this data.

## Examples

Several open data portals are developed in the open government data field:
* The official portal for European data
* The Home of the U.S. Government's Open Data
* Greek Open Data portal
* Open platform for French public data
* German Open Data portal
* Swedish Open Data Portal
You can find full list of open data portals comprehensive list from https://dataportals.org/

## Further study

For further study of open data portals, please visit
* https://dataportals.org/
* https://data.worldbank.org/
* https://data.europa.eu/en/academy/introducing-open-data
* https://zenodo.org/communities/odeco-research/records?q=&l=list&p=1&s=10&sort=newest
# Science

Science is a systematic enterprise that builds and organizes knowledge in the form of testable explanations and predictions about the universe.
According to Albert Einstein, arguably the most famous scientist, "the whole of science is nothing more than a refinement of everyday thinking" (p. 349).

## Resources

* WikiJournal of Science