OpenLLM-France/wikimedia huggingface dataset:
# Matrix/R/Mapping/Example

Let an $m\times n$-matrix
    ${\begin{pmatrix}a_{11}&a_{12}&\ldots &a_{1n}\\a_{21}&a_{22}&\ldots &a_{2n}\\\vdots &\vdots &\ddots &\vdots \\a_{m1}&a_{m2}&\ldots &a_{mn}\end{pmatrix}}$
be given, where the entries $a_{ij}$ are real numbers. Such a matrix defines a mapping
    $\varphi \colon \mathbb {R} ^{n}\longrightarrow \mathbb {R} ^{m},$
by sending an $n$-tuple $x={\begin{pmatrix}x_{1}\\x_{2}\\\vdots \\x_{n}\end{pmatrix}}\in \mathbb {R} ^{n}$ to the $m$-tuple Template:Relationchain/display/printalign The $i$-th component of the image vector is
    $y_{i}=\left(a_{i1},\,a_{i2},\,\ldots ,\,a_{in}\right){\begin{pmatrix}x_{1}\\x_{2}\\\vdots \\x_{n}\end{pmatrix}}=\sum _{j=1}^{n}a_{ij}x_{j}\,.$
So one has to apply the $i$-th row of the matrix to the column vector $x$ in the described way.
# Linear algebra (Osnabrück 2024-2025)/Group/Inverse element/Unique/Fact/Factreferencenumber

Lemma 3.2
# Linear algebra (Osnabrück 2024-2025)/Ring/Basic properties/Fact/Factreferencenumber

Lemma 3.5
# Linear algebra (Osnabrück 2024-2025)/Field/Basic properties/Fact/Factreferencenumber

Lemma 3.5
# Linear algebra (Osnabrück 2024-2025)/Field/N and Z/Canonical mapping/Exercise/Exercisereferencenumber

Exercise 3.23
# Fruit salad/Minerals and vitamins/Table and matrix/Linear system/Example

A healthy breakfast starts with a fruit salad. The following table shows how much vitamin C, calcium and magnesium various fruits have (in milligram with respect to 100 gram of the fruit).
My fruit salad today consists of the mentioned fruits with portions ${\begin{pmatrix}3\\2\\7\\6\end{pmatrix}}$ (meaning $300$ gram apple and so on). From that, one can calculate the total vitamin-C-amount, the calcium-amount and the magnesium-amount of the fruit salad, by simply multiplying for each fruit its portion with its specific amount, and summing up everything. The vitamin-C-amount of the complete fruit salad is thus
    $12\cdot 3+53\cdot 2+4\cdot 7+9\cdot 6=224\,.$
This operation is an example for how a matrix operates. The table yields immediately a $3\times 4$-matrix, namely ${\begin{pmatrix}12&53&4&9\\7&40&12&5\\6&10&8&27\end{pmatrix}}$, and the above calculation is realized by the matrix multiplication
    ${\begin{pmatrix}12&53&4&9\\7&40&12&5\\6&10&8&27\end{pmatrix}}{\begin{pmatrix}3\\2\\7\\6\end{pmatrix}}={\begin{pmatrix}224\\215\\256\end{pmatrix}}\,.$
One can also ask for a fruit salad which has certain amounts of vitamin C, calcium and magnesium, say ${\begin{pmatrix}180\\110\\140\end{pmatrix}}$. This leads to the linear system of linear equations in matrix form,
    ${\begin{pmatrix}12&53&4&9\\7&40&12&5\\6&10&8&27\end{pmatrix}}{\begin{pmatrix}x_{1}\\x_{2}\\x_{3}\\x_{4}\end{pmatrix}}={\begin{pmatrix}180\\110\\140\end{pmatrix}}\,.$
# Linear algebra (Osnabrück 2024-2025)/Vector space/Introductory example/Mulled wine/Example/Examplereferencenumber

Example 4.1
# Linear algebra (Osnabrück 2024-2025)/Linear inhomogeneous system/Elimination/2x+5y+2z-v is 3, 3x-4y+u+2v is 1, 4x -2z+2u is 7/Example/Examplereferencenumber

Example 5.1
# Linear algebra (Osnabrück 2024-2025)/System of linear equations/Set of variables/Equivalent system/Manipulations/Fact/Factreferencenumber

Lemma 5.3
# Linear algebra (Osnabrück 2024-2025)/Linear inhomogeneous system/Elimination/Echelon form and triangular form/Fact/Factreferencenumber

Theorem 5.5
# Linear algebra (Osnabrück 2024-2025)/Linear system/Substitution method/Remark/Remarkreferencenumber

Remark 5.8
# Linear algebra (Osnabrück 2024-2025)/Linear system/Equating method/Remark/Remarkreferencenumber

Remark 5.9
# Linear algebra (Osnabrück 2024-2025)/Linear system/Superposition principle/Fact/Factreferencenumber

Theorem 5.13
# Linear algebra (Osnabrück 2024-2025)/Invertible matrix/Find inverse matrix/Table/Method/Factreferencenumber

Method 12.11
# Linear algebra (Osnabrück 2024-2025)/Linear system/Superposition principle/Fact/Proof/Exercise/Exercisereferencenumber

Exercise 5.19
# Linear algebra (Osnabrück 2024-2025)/System of linear equations/Set of variables/Solution space is vector space/Fact/Factreferencenumber

Lemma 6.11
# Linear algebra (Osnabrück 2024-2025)/System of linear equations/Set of variables/Solution space is vector space/Fact/Proof/Exercise/Exercisereferencenumber

Exercise 6.3
# Linear algebra (Osnabrück 2024-2025)/Vector space/Simple properties/Fact/Proof/Exercise/Exercisereferencenumber

Exercise 6.27
# Linear algebra (Osnabrück 2024-2025)/Linear subspace/Closed/Vector space/Exercise/Exercisereferencenumber

Exercise 6.10
# Linear algebra (Osnabrück 2024-2025)/Vector space/Generating system and spanning subspace/Fact/Proof/Exercise/Exercisereferencenumber

Exercise 6.26
# Linear algebra (Osnabrück 2024-2025)/Vector space/Generating system and spanning subspace/Fact/Factreferencenumber

Lemma 6.16
# Linear algebra (Osnabrück 2024-2025)/Standard vectors/Linearly independent/Example/Examplereferencenumber

Example 7.2
# Linear algebra (Osnabrück 2024-2025)/Linearly independent/Simple properties/Fact/Proof/Exercise/Exercisereferencenumber

Exercise 7.14
# Linear algebra (Osnabrück 2024-2025)/Polynomial ring/Basis/Example/Examplereferencenumber

Example 7.10
# Linear algebra (Osnabrück 2024-2025)/Vector space/Characterizations of basis/Maximal/Minimal/Fact/Factreferencenumber

Theorem 7.11
# Linear algebra (Osnabrück 2024-2025)/Basis/Unique representation/Coordinates/Bijection/Remark/Remarkreferencenumber

Remark 7.12
# Linear algebra (Osnabrück 2024-2025)/vector space/Finitely generated/Basis/Fact/Factreferencenumber

Theorem 7.13
# Linear algebra (Osnabrück 2024-2025)/Matrices/Standard matrices/Basis/Exercise/Exercisereferencenumber

Exercise 7.27
# Linear algebra (Osnabrück 2024-2025)/Vector space/Generating subspace/Remove representable vector/Exercise/Exercisereferencenumber

Exercise 6.25
# Field/Polynomial ring/Not finitely generated/Example

The polynomial ring $R=K[X]$ over a field $K$ is not a finite-dimensional vector space. To see this, we have to show that there is no finite generating system for the polynomial ring. Consider $n$ polynomials $P_{1},\ldots ,P_{n}$. Let $d$ be the maximum of the degrees of these polynomials. Then every $K$-linear combination $\sum _{i=1}^{n}a_{i}P_{i}$ has at most degree $d$. In particular, polynomials of larger degree can not be presented by $P_{1},\ldots ,P_{n}$, so these do not form a generating system for all polynomials.
# Linear algebra (Osnabrück 2024-2025)/Vector space/Basis/Exchange theorem/Fact/Factreferencenumber

Theorem 8.2
# Linear algebra (Osnabrück 2024-2025)/Vector space/Finitely generated/Length of every basis/Fact/Factreferencenumber

Theorem 8.4
# Linear algebra (Osnabrück 2024-2025)/Vector space/Linear subspace/Dimensions/Compare/Fact/Factreferencenumber

Corollary 8.9
# Linear algebra (Osnabrück 2024-2025)/Vector space/Basis complement/Fact/Factreferencenumber

Theorem 8.10
# Linear algebra (Osnabrück 2024-2025)/Vector space/Dimension n and n vectors/Equivalences/Fact/Proof/Exercise/Exercisereferencenumber

Exercise 8.6
# Linear algebra (Osnabrück 2024-2025)/Infinite basis/Finite basis/Exercise/Exercisereferencenumber

Exercise 8.17
# Linear algebra (Osnabrück 2024-2025)/Vector space/Finite dimensional/Change of basis/Fact/Factreferencenumber

Lemma 9.1
# Linear algebra (Osnabrück 2024-2025)/Base change/Three bases/Composition/Fact/Proof/Exercise/Exercisereferencenumber

Exercise 9.9
# Linear algebra (Osnabrück 2024-2025)/Linear subspace/Intersection/Dimension estimate/Fact/Factreferencenumber

Corollary 9.8
# Linear algebra (Osnabrück 2024-2025)/Vector space/Finite dimensional/Linear subspace/Direct complement/Fact/Factreferencenumber

Lemma 9.12
# Linear algebra (Osnabrück 2024-2025)/Linear mapping/Linear combinations/Exercise/Exercisereferencenumber

Exercise 10.2
# Linear algebra (Osnabrück 2024-2025)/Linear mapping/Composition/Fact/Proof/Exercise/Exercisereferencenumber

Exercise 10.12
# Linear algebra (Osnabrück 2024-2025)/Linear mapping/Bijective/Inverse mapping linear/Fact/Proof/Exercise/Exercisereferencenumber

Exercise 10.13
# Linear algebra (Osnabrück 2024-2025)/Linear mapping/Determination on basis/Fact/Factreferencenumber

Theorem 10.10
# Linear algebra (Osnabrück 2024-2025)/Linear mapping/Matrix/Commutative diagram/Fact/Factreferencenumber

Lemma 10.14
# Linear algebra (Osnabrück 2024-2025)/Linear mapping/Matrix/Commutative diagram/Fact/Proof/Exercise/Exercisereferencenumber

Exercise 10.27
# Linear algebra (Osnabrück 2024-2025)/Linear mapping/Matrix to bases/Correspondence/Fact/Factreferencenumber

Theorem 10.15
# Linear algebra (Osnabrück 2024-2025)/Vector space/Isomorphic iff equal dimension/Fact/Proof/Exercise/Exercisereferencenumber

Exercise 10.32
# Linear algebra (Osnabrück 2024-2025)/Linear mapping/Determination on basis/Scaling/Proof/Exercise/Exercisereferencenumber

Exercise 10.21
# Vector space/Basis/Index partition/Direct sum/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.
# Linear mapping/One-dimensional/Proportionality and rule of three/Example

The easiest
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 fact, but this is also directly clear) by $\varphi (1)$, or by the value $\varphi (t)$ for a 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 mapping/Matrix to bases/Correspondence/Fact/Proof

We show that both compositions are the identity. We start with a matrix $M={\left(a_{ij}\right)}_{ij}$ and consider the matrix
    $M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi _{\mathfrak {w}}^{\mathfrak {v}}(M)).$
Two matrices are equal, when the entries coincide for every index pair $(i,j)$. We have
    ${\begin{aligned}(M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi _{\mathfrak {w}}^{\mathfrak {v}}(M)))_{ij}&=i-{\text{th coordinate of }}(\varphi _{\mathfrak {w}}^{\mathfrak {v}}(M))(v_{j})\\&=i-{\text{th coordinate of }}\sum _{i=1}^{m}a_{ij}w_{i}\\&=a_{ij}.\end{aligned}}$
Now, let $\varphi$ be a linear mapping, we consider
    $\varphi _{\mathfrak {w}}^{\mathfrak {v}}(M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )).$
Two linear mappings coincide, due to fact, when they have the same values on the basis $v_{1},\ldots ,v_{n}$. We have
    $(\varphi _{\mathfrak {w}}^{\mathfrak {v}}(M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi )))(v_{j})=\sum _{i=1}^{m}(M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi ))_{ij}\,w_{i}\,.$
Due to the definition, the coefficient $(M_{\mathfrak {w}}^{\mathfrak {v}}(\varphi ))_{ij}$ is the $i$-th coordinate of $\varphi (v_{j})$ with respect to the basis $w_{1},\ldots ,w_{m}$. Hence, this sum equals $\varphi (v_{j})$.
# Homomorphism space/Dimension/Fact

Let $K$ denote a field, and let $V$ and $W$ denote finite-dimensional $K$-vector spaces with dimensions $n$ and $m$.
    $\dim _{K}{\left(\operatorname {Hom} _{K}{\left(V,W\right)}\right)}=nm\,.$
# Linear inhomogeneous system/Elimination/2x+5y+2z-v is 3, 3x-4y+u+2v is 1, 4x -2z+2u is 7/Example

We want to solve the inhomogeneous linear system
    ${\begin{matrix}2x&+5y&+2z&&-v&=&3\\3x&-4y&&+u&+2v&=&1\\4x&&-2z&+2u&&=&7\,\end{matrix}}$
over $\mathbb {R}$ (or over $\mathbb {Q}$). Firstly, we eliminate $x$ by keeping the first row $I$, replacing the second row $II$ by $II-{\frac {3}{2}}I$, and replacing the third row $III$ by $III-2I$. This yields
    ${\begin{matrix}2x&+5y&+2z&&-v&=&3\\&-{\frac {23}{2}}y&-3z&+u&+{\frac {7}{2}}v&=&{\frac {-7}{2}}\\&-10y&-6z&+2u&+2v&=&1\,.\end{matrix}}$
Now, we can eliminate $y$ from the (new) third row, with the help of the second row. Because of the fractions, we rather eliminate $z$ (which eliminates also $u$). We leave the first and the second row as they are, and we replace the third row $III$ by $III-2II$. This yields the system, in a new ordering of the variables,
    ${\begin{matrix}2x&+2z&&+5y&-v&=&3\\&-3z&+u&-{\frac {23}{2}}y&+{\frac {7}{2}}v&=&{\frac {-7}{2}}\\&&&13y&-5v&=&8\,.\end{matrix}}$
Now we can choose an arbitrary (free) value for $v$. The third row determines $y$ uniquely, we must have
    $y={\frac {8}{13}}+{\frac {5}{13}}v\,.$
In the second equation, we can choose $u$ arbitrarily, this determines $z$ via
    ${\begin{aligned}z&=-{\frac {1}{3}}{\left(-{\frac {7}{2}}-u-{\frac {7}{2}}v+{\frac {23}{2}}{\left({\frac {8}{13}}+{\frac {5}{13}}v\right)}\right)}\\&=-{\frac {1}{3}}{\left(-{\frac {7}{2}}-u-{\frac {7}{2}}v+{\frac {92}{13}}+{\frac {115}{26}}v\right)}\\&=-{\frac {1}{3}}{\left({\frac {93}{26}}-u+{\frac {12}{13}}v\right)}\\&=-{\frac {31}{26}}+{\frac {1}{3}}u-{\frac {4}{13}}v.\end{aligned}}$
The first row determines $x$, namely
    ${\begin{aligned}x&={\frac {1}{2}}{\left(3-2z-5y+v\right)}\\&={\frac {1}{2}}{\left(3-2{\left(-{\frac {31}{26}}+{\frac {1}{3}}u-{\frac {4}{13}}v\right)}-5{\left({\frac {8}{13}}+{\frac {5}{13}}v\right)}+v\right)}\\&={\frac {1}{2}}{\left({\frac {30}{13}}-{\frac {2}{3}}u-{\frac {4}{13}}v\right)}\\&={\frac {15}{13}}-{\frac {1}{3}}u-{\frac {2}{13}}v.\end{aligned}}$
Hence, the solution set is
    ${\left\{{\left({\frac {15}{13}}-{\frac {1}{3}}u-{\frac {2}{13}}v,{\frac {8}{13}}+{\frac {5}{13}}v,-{\frac {31}{26}}+{\frac {1}{3}}u-{\frac {4}{13}}v,u,v\right)}\mid u,v\in \mathbb {R} \right\}}.$
A particularly simple solution is obtained by equating the free variables $u$ and $v$ with $0$. This yields the special solution
    $(x,y,z,u,v)=\left({\frac {15}{13}},\,{\frac {8}{13}},\,-{\frac {31}{26}},\,0,\,0\right)\,.$
The general solution set can also be written as
    ${\left\{{\left({\frac {15}{13}},{\frac {8}{13}},-{\frac {31}{26}},0,0\right)}+u{\left(-{\frac {1}{3}},0,{\frac {1}{3}},1,0\right)}+v{\left(-{\frac {2}{13}},{\frac {5}{13}},-{\frac {4}{13}},0,1\right)}\mid u,v\in \mathbb {R} \right\}}.$
Here,
    ${\left\{u{\left(-{\frac {1}{3}},0,{\frac {1}{3}},1,0\right)}+v{\left(-{\frac {2}{13}},{\frac {5}{13}},-{\frac {4}{13}},0,1\right)}\mid u,v\in \mathbb {R} \right\}}$
is a description of the general solution of the corresponding homogeneous linear system.
# Linear inhomogeneous system/Elimination/Echelon form and triangular form/Fact

Every (inhomogeneous) system of linear equations over a field $K$
fact, to an equivalent linear system of the form
    ${\begin{matrix}b_{1s_{1}}x_{s_{1}}&+b_{1s_{1}+1}x_{s_{1}+1}&\ldots &\ldots &\ldots &\ldots &\ldots &+b_{1n}x_{n}&=&d_{1}\\0&\ldots &0&b_{2s_{2}}x_{s_{2}}&\ldots &\ldots &\ldots &+b_{2n}x_{n}&=&d_{2}\\\vdots &\ddots &\ddots &\vdots &\vdots &\vdots &\vdots &\vdots &=&\vdots \\0&\ldots &\ldots &\ldots &0&b_{m{s_{m}}}x_{s_{m}}&\ldots &+b_{mn}x_{n}&=&d_{m}\\(0&\ldots &\ldots &\ldots &\ldots &\ldots &\ldots &0&=&d_{m+1}),\end{matrix}}$
$b_{1s_{1}},b_{2s_{2}},\ldots ,b_{ms_{m}}$
$0$
Here, either $d_{m+1}=0$, and the last row can be omitted, or $d_{m+1}=0$, and then the system has no solution at all.
With the help of renaming the variables, we get an equivalent system of the form
    ${\begin{matrix}c_{11}y_{1}&+c_{12}y_{2}&\ldots &+c_{1m}y_{m}&+c_{1m+1}y_{m+1}&\ldots &+c_{1n}y_{n}&=&d_{1}\\0&c_{22}y_{2}&\ldots &\ldots &\ldots &\ldots &+c_{2n}y_{n}&=&d_{2}\\\vdots &\ddots &\ddots &\vdots &\vdots &\vdots &\vdots &=&\vdots \\0&\ldots &0&c_{mm}y_{m}&+c_{mm+1}y_{m+1}&\ldots &+c_{mn}y_{n}&=&d_{m}\\(0&\ldots &\ldots &0&0&\ldots &0&=&d_{m+1})\end{matrix}}$
with diagonal elements $c_{ii}\neq 0$.
# Invertible matrix/No zero row/Exercise

Show that an invertible matrix $M$ neither has a zero row nor a zero column.
# Invertible matrices/K/Product/Exercise

Let $M$ and $N$ be invertible $n\times n$-matrices. Show that also $M\circ N$ is invertible, and that
    $(M\circ N)^{-1}=N^{-1}\circ M^{-1}\,$
holds.
# Inverse matrix/2/Linear system/1/Exercise

* Transform the matrix equation into a system of linear equations.
    ${\begin{pmatrix}3&7\\-4&5\end{pmatrix}}{\begin{pmatrix}x&y\\z&w\end{pmatrix}}={\begin{pmatrix}1&0\\0&1\end{pmatrix}}\,$
* Solve this linear system.
# Matrix/2/Left inverse/Right inverse/Exercise

Let $M={\begin{pmatrix}a&b\\c&d\end{pmatrix}}$ and $A={\begin{pmatrix}x&y\\z&w\end{pmatrix}}$ be matrices over a field $K$ such that
    $A\circ M={\begin{pmatrix}1&0\\0&1\end{pmatrix}}\,$
holds. Show directly that
    $M\circ A={\begin{pmatrix}1&0\\0&1\end{pmatrix}}\,$
holds as well.
# Inverse matrix/2 4 0 -1 0 3 0 1 1/Exercise

Determine the inverse matrix of
    ${\begin{pmatrix}2&4&0\\-1&0&3\\0&1&1\end{pmatrix}}.$
# Matrix/Inverting algorithm/Breaks down/1/Exercise

Perform, for the matrix
    ${\begin{pmatrix}3&-2&5\\1&7&-4\\9&17&-2\end{pmatrix}},$
the inverting algorithm, until it is obvious that the matrix is not invertible.
# Column rank/Product of matrices/Estimate/Exercise

Let $B$ be an $n\times p$-matrix, and let $A$ be an $m\times n$-matrix. Show that, for the column rank, the estimate
    $\operatorname {rk} \,A\circ B\leq {\min {\left(\operatorname {rk} \,A,\operatorname {rk} \,B,\right)}}\,$
holds.
# Column rank/Product with invertible matrix/Exercise

Let $M$ be an $m\times n$-matrix, and let $A$ be an invertible $m\times m$-matrix. Show that, for the column rank, the equality
    $\operatorname {rk} \,A\circ M=\operatorname {rk} \,M\,$
holds.
# Block matrix/Rank/Exercise

Let a block matrix of the form
    $M={\begin{pmatrix}A&0\\0&B\end{pmatrix}}\,$
be given. Show that the rank of $M$ equals the sum of the ranks of $A$ and of $B$.
# Matrix/Elementary matrices/Identity matrix/2/Exercise

Let
    $M={\begin{pmatrix}3&5&-1\\4&7&2\\2&-3&6\end{pmatrix}}\,.$
Find elementary matrices $E_{1},\ldots ,E_{k}$ such that $E_{k}\circ \cdots \circ E_{1}\circ M$ is the identity matrix.
# Invertible matrix/Find inverse matrix/C/5+8i 3-7i/2-9i 4-5i/Exercise

Determine the inverse matrix of the complex matrix
    $M={\begin{pmatrix}5+8{\mathrm {i} }&3-7{\mathrm {i} }\\2-9{\mathrm {i} }&4-5{\mathrm {i} }\end{pmatrix}}\,.$
# Block matrix/2x2/Definition

A block matrix is an $m\times n$-matrix of the form
    ${\begin{pmatrix}A&B\\C&D\end{pmatrix}},$
where $A$ is an $r\times s$-matrix, $B$ is an $r\times (n-s)$-matrix, $C$ is an $(m-r)\times s$-matrix and $D$ is an $(m-r)\times (n-s)$-matrix.
# Homomorphism space/Evaluation at a vector/Linear/Exercise

Let $K$ be a field, and let $V$ and $W$ be $K$-vector spaces. Let $\operatorname {Hom} _{K}{\left(V,W\right)}$ be the $K$-vector space of all linear mappings from $V$ to $W$, and let $v\in V$ denote a fixed vector. Show that the mapping
    $F\colon \operatorname {Hom} _{K}{\left(V,W\right)}\longrightarrow W,\varphi \longmapsto F(\varphi ):=\varphi (v),$
is $K$-linear.
# Matrix space/K^2/Multiplication with fixed matrix/Exercise

We consider the space of matrices $\operatorname {Mat} _{2}(K)$, and we fix the matrix $A={\begin{pmatrix}9&-8\\-7&6\end{pmatrix}}$.
* Show that the mapping is linear.
    $\varphi \colon \operatorname {Mat} _{2}(K)\longrightarrow \operatorname {Mat} _{2}(K),M\longmapsto A\circ M,$
* Determine the describing matrix of $\varphi$ with respect to a suitable basis.
* Determine the kernel and the image of $\varphi$.
# Endomorphism space/K^2/Squaring/Exercise

We consider the space of endomorphisms $\operatorname {End} _{\left(K^{2}\right)}=\operatorname {Hom} _{K}{\left(K^{2},K^{2}\right)}$. Is the mapping
    $\operatorname {End} _{\left(K^{2}\right)}\longrightarrow \operatorname {End} _{\left(K^{2}\right)},\varphi \longmapsto \varphi \circ \varphi ,$
a linear mapping?
# Linear algebra (Osnabrück 2024-2025)/Projection/Linear subspace/Idempotent/Fact/Factreferencenumber

Lemma 13.5
# Matrix/K/Linear dependence of powers/Exercise

Let $K$ be a field, and let $M$ denote an $n\times n$-matrix over $K$. Show that the first $n^{2}+1$ powers
    $M^{i},i=0,\ldots ,n^{2},$
are linearly independent in $\operatorname {Mat} _{n}(K)$.
# Homomorphism space/Functorial properties/Matrices/Exercise

Formulate fact for matrices with respect to given bases.
# Linear algebra (Osnabrück 2024-2025)/Homomorphism space/Functorial properties/Fact/Factreferencenumber

Lemma 13.8
# Linear surjective mapping/Linear section/As affine-linear space/Exercise

Let $K$ be a field, let $V$ and $W$ be finite-dimensional $K$-vector spaces, and let
    $\varphi \colon V\longrightarrow W$
be a linear mapping.
a) Show that $\varphi$ is surjective if and only if there exists a linear mapping
    $\psi \colon W\longrightarrow V$
such that
    $\varphi \circ \psi =\operatorname {Id} _{W}\,.$
b) Let now $\varphi$ be surjective, and set
    $S={\left\{\psi :W\rightarrow V\mid \psi {\text{ linear}},\,\varphi \circ \psi =\operatorname {Id} _{W}\right\}}\,,$
and let $\psi _{0}\in S$ be fixed. Define a bijection between $\operatorname {Hom} _{K}{\left(W,\operatorname {kern} \varphi \right)}$ and $S$, such that $0$ is mapped to $\psi _{0}$.
# Endomorphisms on vector space/Ring/Exercise

Let $K$ be a field, and let $V$ be a $K$-vector space. Show that
    $\operatorname {End} _{\left(V\right)}={\left\{\varphi :V\rightarrow V\mid \varphi {\text{ linear}}\right\}}\,$
is, with the natural addition and the composition of mappings, a ring.
# Endomorphism space/Conjugation/Isomorphism/Exercise

Let $V$ be a $K$-vector space, and let
    $g\colon V\longrightarrow V$
denote an isomorphism. Show that the mapping
    $\Psi _{g}\colon \operatorname {End} _{\left(V\right)}\longrightarrow \operatorname {End} _{\left(V\right)},f\longmapsto gfg^{-1},$
is an isomorphism of vector spaces, and that, moreover, the identities
    $\Psi _{g}(f_{1}\circ f_{2})=\Psi _{g}(f_{1})\circ \Psi _{g}(f_{2})\,$
and
    $\Psi _{g}(\operatorname {Id} _{V})=\operatorname {Id} _{V}\,$
hold.
# Endomorphism/Basis/To scalar multiples/Dimension/Exercise

Let $V$ be a $K$-vector space, and let $v_{1},\ldots ,v_{n}$ denote a basis of $V$. Determine the dimension of the space of all endomorphisms
    $\varphi \colon V\longrightarrow V$
satisfying
    $\varphi (v_{i})\in Kv_{i}\,$
for all $i$. How do the matrices of such a $\varphi$ with respect to this basis look like?
# Endomorphisms/Bijective/Line to same line/Scalar/Exercise

Let $V$ be a $K$-vector space, and let
    $\varphi ,\psi \colon V\longrightarrow V$
denote automorphisms, such that, for every linear subspace $U\subseteq V$, the identity $\varphi (U)=\psi (U)$ holds. Show that $\varphi =a\psi$ with some $a\in K$.
# Dual bases/Dependence on the coordinates of basis/Exercise

Show by an example of two bases $v,u$ and $v,w$ in $\mathbb {R} ^{2}$, that the coordinate function $v^{*}$ depend on the basis, and not only on $v$.
# Space of matrices/Standard basis/Dual basis/Trace/Exercise

Let $V=\operatorname {Mat} _{n}(K)$ be the space of the $n\times n$-matrices over the field $K$, with the standard basis $e_{ij}$. Describe the trace as a linear combination with respect to the dual basis $e_{ij}^{*}$.
# R^3/Dual basis to (4,-2,1),(5,3,2),(0,-1,7)/Standard dual basis/Exercise

Express the vectors $u_{1}^{*},u_{2}^{*},u_{3}^{*}$ of the dual basis of the basis $u_{1}={\begin{pmatrix}4\\-2\\1\end{pmatrix}},\,u_{2}={\begin{pmatrix}5\\3\\2\end{pmatrix}},\,u_{3}={\begin{pmatrix}0\\-1\\7\end{pmatrix}}$ in $\mathbb {R} ^{3}$ as linear combinations with respect to the standard dual basis $e_{1}^{*},e_{2}^{*},e_{3}^{*}$.
# R^2/Dual basis to (4,7),(6,-1)/Standard dual basis/Exercise

Express the vectors $u_{1}^{*},u_{2}^{*}$ of the dual basis of the basis $u_{1}={\begin{pmatrix}4\\7\end{pmatrix}},\,u_{2}={\begin{pmatrix}6\\-1\end{pmatrix}}$ in $\mathbb {R} ^{2}$ as linear combinations with respect to the standard dual basis $e_{1}^{*},e_{2}^{*}$.