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A complex representation of $G$ is a complex vector space $V$ together with a homomorphismIt is faithful when $\ker \rho =\{e\}$, equivalently when $\rho$ is injective. Every finite group has a faithful complex representation: in the regular representation, $G$ permutes the basis $\{e_h:h\in G\}$ byand a group element fixing every basis vector must be the identity.

If $g$ is conjugate to $g^k$, then $\rho (g)$ is similar to $\rho (g^k)=\rho (g{)}^k$. Thus the two matrices have the same spectrum, while an eigenvalue $\lambda$ of $\rho (g)$ gives the eigenvalue ${\lambda }^k$ of $\rho (g{)}^k$. Hence

Now take a $p$-cycle $g\in S_p$. It is conjugate to $g^k$ for every $1\le k<p$. Faithfulness makes $\rho (g)$ have order $p$, so its spectrum contains a nontrivial $p$th root of unity $\lambda$. The preceding implication puts all the $p-1$ distinct valuesin the spectrum. Therefore $\dim V\ge p-1$. This is the spectrum orbit bound for a faithful symmetric-group representation.

Finally let $G$ be the group of all permutations of $\mathbb{N}$. It has no faithful finite-dimensional complex representation. For every $n$, $G$ contains a subgroup $(C_2{)}^n$ generated by $n$ disjoint transpositions. In a $d$-dimensional complex representation, commuting involutions are simultaneously diagonalizable and hence map into the group of diagonal sign matrices, which has order $2^d$. A faithful restriction would require $2^n\le 2^d$ for every $n$, an impossibility.

The extension $L/K$ is finite when $L$ is finite-dimensional as a vector space over $K$, and its degree is $[L:K]={\dim }_{K}L$.

A complex matrix is diagonalisable when it is similar to a diagonal matrix, equivalently when the vector space has a basis of its eigenvectors.

The dual space $V^{\ast }$ is the vector space of linear maps $V\to \mathbb{F}$. If $B=(v_1,\dots ,v_n)$, its dual family satisfies $v_i^{\ast }(v_j)={\delta }_{ij}$. Every functional $f$ obeysso the family spans; evaluation on each $v_j$ proves linear independence. Thus it is a basis without any prior dimension argument.

For $p=a+bt+c{t}^2$,The basis dual to $(f_0,f_1,f_2)$ isas direct substitution gives $f_i(p_j)={\delta }_{ij}$. Hence the three functionals form a basis of $P_2^{\ast }$.

Conversely, assume condition (b), and let $M,N$ be nonzero. The stated property givesThese are nonzero vector spaces over the residue field $k=R/\mathfrak{m}$, so their tensor product over $k$ is nonzero. Associativity and base change giveTherefore $M{\otimes }_{R}N\ne 0$. This proves the reverse implication and the local tensor nonvanishing criterion.

The dual $X^{\ast }$ is the vector space of bounded linear maps $f:X\to \mathbb{F}$, with normIf $(f_n)$ is Cauchy in this norm, then $(f_n(x))$ is Cauchy in the complete scalar field for every $x$. Define $f(x)={\lim }_n{f}_n(x)$. Pointwise passage to the limit preserves linearity. Given $\epsilon >0$, choose $N$ such that $\Vert f_n-f_m\Vert <\epsilon$ for $m,n\ge N$; sending $m\to \infty$ givesThus $f$ is bounded and $\Vert f_n-f\Vert \le \epsilon$. Hence $X^{\ast }$ is Banach, even if $X$ was only normed.

Let $A,B,C$ represent $a,b,c$. Since $c$ is central, $C$ commutes with $A$ and $B$. Over the complex number field $\mathbb{C}$, $C$ has an eigenvalue $\gamma$, and its corresponding eigenspace is invariant under all three operators. The irreducibility of $V$ therefore makes this eigenspace all of $V$, so $C=\gamma I$. Taking the trace ofgives $(\dim V)\gamma =0$ by the cyclic property of the trace; hence $C=0$.

The remaining operators $A$ and $B$ commute. Two commuting operators on a nonzero finite-dimensional complex vector space have a common eigenvector, whose span is invariant. Irreducibility therefore forces $\dim V=1$. Conversely, every pair $(\alpha ,\beta )\in {\mathbb{C}}^2$ defines a one-dimensional irreducible representation byThese are all the finite-dimensional irreducible representations.

Let $V$ be the real vector space with basis $(e_i{)}_{i\in I}$. The Coxeter Gram matrix defines the symmetric bilinear formIts diagonal entries are $2$. The Geometric representation of a Coxeter group is generated by the reflectionsEach has square one, and on $\mathrm{span}\{e_i,e_j\}$ the product $\sigma (x_i)\sigma (x_j)$ has order $m_{ij}$. The reflections therefore satisfy the Coxeter relations and define a group representation $\sigma :W\to GL(V)$.

On an $n$-sample, each coordinate partition has at most $(n+1{)}^{m-1}$ placements, hence all $p$ partitions have at most $(n+1{)}^{p(m-1)}$ configurations. For each fixed configuration, the sign class from a vector space of dimension $1+p(m-1)$ has at most $(n+1{)}^{1+p(m-1)}$ patterns. Multiplication gives

For a finite set $S=\{s_1,\dots ,s_k\}$ of vectors in a real vector space, its span is the set of all linear combinations:The span of the empty set is $\{0\}$. The set $S$ is linearly independent if ${\sum }_i{a}_i{s}_i=0$ implies $a_i=0$ for every $i$. It is a spanning set for $V$ if $\mathrm{Sp}(S)=V$. A basis is both a linearly independent set and a spanning set.

For basis extension, start with the given linearly independent set. If it does not span $V$, choose $v\notin \mathrm{Sp}(S)$ and adjoin it. The enlarged set remains linearly independent: a relation $av+{\sum }_i{a}_i{s}_i=0$ with $a\ne 0$ would express $v$ as a linear combination of $S$, a contradiction; if $a=0$, independence of $S$ forces all other coefficients to be zero.

Repeat whenever the current span is proper. Each step increases the set's size by one, while the allowed bound says its size cannot exceed $\dim V$. Thus the process terminates with a linearly independent set spanning $V$, hence a basis containing the original set. This also covers an initially empty set and the zero-dimensional vector space.

Use $f,g$ from i. Their inner products with $x^2$ areThus $h=x^2-2/3$. Its values at $-1,0,1$ are $1/3,-2/3,1/3$, soBy ii the three polynomials are orthogonal vectors, and normalizing $h$ givesThis orthonormal set is a basis of the three-dimensional vector space $V$. It is the result of the Gram-Schmidt process applied to $1,x,x^2$.

The Steinitz exchange lemma states that if $u_1,\dots ,u_r$ are linearly independent and $w_1,\dots ,w_s$ span a vector space $V$, then $r\le s$ and some $r$ of the $w_j$ can be replaced by $u_1,\dots ,u_r$ while preserving the spanning set property.

Here is the exchange argument. Suppose that after $k$ replacements, a spanning set iswhere the remaining $w_j$ have been relabelled if necessary. Express $u_{k+1}$ as a linear combination of this family. At least one remaining $w_j$ has a nonzero coefficient; otherwise $u_{k+1}$ belongs to the span of $u_1,\dots ,u_k$, contradicting linear independence. Relabel this entry $w_{k+1}$. Solving for $w_{k+1}$ shows that it lies in the span ofThe new family therefore still spans $V$. Starting at $k=0$ repeats this step until all $r$ independent vectors have been inserted. If $r>s$, after $s$ steps the next $u_{s+1}$ would lie in the span of $u_1,\dots ,u_s$, a contradiction. Thus $r\le s$, and the construction proves the replacement assertion as well.

The definition of an eigenvalue and the kernel criterion for a singular square matrix giveFor the second equivalence, a linear map from a finite-dimensional vector space to itself has zero kernel exactly when its rank of a linear map is the full dimension, by the rank-nullity theorem, and therefore exactly when it is invertible. This proves both directions.

The transpose is a linear map, so $T(A)=A+A^T$ is a linear map on the $n^2$-dimensional vector space $M_{n\times n}(F)$. Let $E_{ij}$ denote a matrix unit. The answer depends on the characteristic of a field.

If $\mathrm{char}F\ne 2$, then $T(A)=0$ means $A^T=-A$. Such skew-symmetric matrices have zero diagonal, and their entries above the diagonal can be chosen freely. A basis of the kernel of a linear map is $E_{ij}-E_{ji}$ for $i<j$, giving nullity $n(n-1)/2$. The image of a linear map consists of all symmetric matrices: $T(A)$ is symmetric, and every symmetric matrix $S$ is $T(S/2)$. Its basis comprises $E_{ii}$ and $E_{ij}+E_{ji}$ for $i<j$, giving rank of a linear map $n(n+1)/2$.

If $\mathrm{char}F=2$, then $-A=A$, so $\ker T$ consists of all symmetric matrices, with dimension $n(n+1)/2$. Every $T(A)$ is symmetric with zero diagonal, and every symmetric zero-diagonal matrix $S$ arises by taking $A_{ij}=S_{ij}$ for $i<j$ and all other entries zero. The image of a linear map therefore has basis $E_{ij}+E_{ji}$ for $i<j$, and dimension $n(n-1)/2$. This is the symmetrization map in characteristic two.

In both cases the rank-nullity theorem is verified explicitly:

Take any finite selection of distinct indices $i_1,\dots ,i_r$ and a relationThe coordinate at $i_j$ on the left is exactly ${\alpha }_j$, while the zero vector has coordinate zero. Hence ${\alpha }_j=0$ for each $j$. Every finite subset of $E$ is therefore linearly independent, which is the definition of linear independence for an infinite set. Thus $E$ is linearly independent in the vector space of finitely supported sequences.

A map $T:V\to W$ between vector spaces over the same field $\mathbb{F}$ is a linear map precisely when it preserves vector addition and scalar multiplication:Equivalently, $T(au+bv)=aT(u)+bT(v)$ for all $u,v\in V$ and $a,b\in \mathbb{F}$.

A vector space over a field $\mathbb{F}$ is a set $V$ with vector addition $V\times V\to V$ and scalar multiplication $\mathbb{F}\times V\to V$ satisfying the following vector space axioms. Under vector addition, $V$ is an abelian group: addition is associative and commutative, there is a zero vector $0_V$, and each $v$ has an additive inverse $-v$. For all $a,b\in \mathbb{F}$ and $u,v\in V$,The operations are defined for every indicated pair and take their values in $V$.

Use basis extension to choose a basis $(v,v_2,\dots ,v_n)$ of the vector space $V$ over the finite field ${\mathbb{F}}_p={\mathbb{Z}}_p$. The condition $T(v)=0$ fixes the first value of a linear functional; its remaining $n-1$ values are arbitrary. There are $p^{n-1}$ such linear maps, exactly one of which is zero. Hence the required number isFor $n=1$ this is zero, as it should be: vanishing at a basis vector then forces the entire linear functional to vanish.

Yes. Consider the following $n^2$ matrices:Each has only row $i$ nonzero and has $i$th diagonal entry $1$, so part (c)(ii) proves that it is a diagonalisable matrix. The collection is a spanning set, because it contains every $E_{ii}$ andThus it spans the matrix unit basis. A spanning set of $n^2$ vectors in an $n^2$-dimensional vector space is a basis. Alternatively, in a vanishing linear combination of $\mathcal{B}$, each off-diagonal entry first forces its own coefficient to be zero, and then each diagonal entry forces the remaining coefficient to be zero.

The position vectors end at the midpoints of $BC$, $CA$ and $AB$, respectively. The drawing places $E$, $F$ and $G$ on those edges and shows the arrows from the origin. The volume of their parallelepiped is the absolute scalar triple product:Equivalently, this is the absolute determinant of the matrix whose columns are the three vectors. Since it is nonzero, the vectors are linearly independent. Three linearly independent vectors in the three-dimensional vector space ${\mathbb{R}}^3$ form a basis.