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Use the Polynomial representation of the Heisenberg Lie algebra on the infinite-dimensional polynomial ring $\mathbb{C}[x]$:The product rule gives $[d/dx,x]=1$, so this is a Lie algebra representation. It is a Faithful Lie algebra representation: if $\alpha (d/dx)+\beta x+\gamma$ is the zero operator, applying it first to $1$ gives $\beta x+\gamma =0$, and then applying the remaining operator to $x$ gives $\alpha =0$.

To prove irreducibility, let $W$ be a nonzero invariant subspace and choose a nonzero polynomial in $W$ of least degree. If its degree were positive, repeated differentiation would produce a nonzero element of smaller degree, so $W$ contains a nonzero constant. Invariance under multiplication by $x$ then puts every monomial $x^n$ in $W$, and hence $W=\mathbb{C}[x]$.

The image of a linear map $\mathrm{im}(ST)=S(\mathrm{im}T)$ lies inside $\mathrm{im}S$. Therefore $r(ST)\le r(S)$. Also $S$ restricted to $\mathrm{im}T$ cannot have rank of a linear map greater than its domain's dimension, so $r(ST)\le \dim \mathrm{im}T=r(T)$. ThusEquivalently, apply the rank-nullity theorem to that restriction: $r(ST)=r(T)-\dim (\mathrm{im}T\cap \ker S)\le r(T)$. This proof keeps track of the intermediate vector subspace, rather than assuming an invertible factor.

Restrict $S$ to the vector subspace $W=\mathrm{im}T$. This restriction has kernel of a linear map $W\cap \ker S$ and image of a linear map $S(W)=\mathrm{im}(ST)$. The rank-nullity theorem on $W$ therefore givesApply the rank-nullity theorem to $T$ and $ST$ on $V$ to obtain the exact nullity of a composition identity:Since the intersection is a vector subspace of $\ker S$, its dimension is at most $n(S)$. Hence

For the linear map $T:V\to W$, its image of a linear map and kernel of a linear map areBoth contain the zero vector, since $T(0)=0$. If $y_1=T(v_1)$ and $y_2=T(v_2)$ belong to the image of a linear map, then for any real numbers $a,b$,Thus $\mathrm{im}T$ is a vector subspace of $W$. If $u_1,u_2\in \ker T$, thenso $\ker T$ is a vector subspace of $V$.

Yes. A surjective $T_A$ has $\mathrm{im}{T}_A={\mathbb{R}}^n$, so every $\mathbf{b}$ is attained. Its rank of a linear map is $n$, and the rank-nullity theorem givesThe solution set of a linear system is therefore a translate of a nonzero vector subspace, containing infinitely many solutions as in iii.

The preimage isIt contains $0$, and if $u,v$ belong to it and $a,b\in \mathbb{R}$, then $T(au+bv)=aT(u)+bT(v)\in X$, since $X$ is a vector subspace. Thus $T^{-1}(X)$ is a vector subspace. It contains the kernel of $T$ because $0\in X$.

To establish the dimension of the preimage of a vector subspace formula without assuming the theorem to be deduced, choose a basis $k_1,\dots ,k_{n_T}$ of $\ker T$ and a basis $y_1,\dots ,y_d$ of $X\cap \mathrm{im}T$, where $d=\dim (X\cap \mathrm{im}T)$. For each $j$, choose $u_j\in U$ with $T(u_j)=y_j$. These $u_j$ lie in $T^{-1}(X)$.

The family $k_1,\dots ,k_{n_T},u_1,\dots ,u_d$ is linearly independent: applying $T$ to a vanishing linear combination forces all $u_j$ coefficients to vanish by linear independence of the $y_j$, and then all remaining coefficients vanish by linear independence of the $k_i$. It also spans $T^{-1}(X)$: for $u$ in that preimage, write $T(u)={\sum }_j{b}_j{y}_j$. Then $u-{\sum }_j{b}_j{u}_j\in \ker T$, so it is a linear combination of the $k_i$. ConsequentlySet $X=V$. Then $T^{-1}(V)=U$ and $V\cap \mathrm{im}T=\mathrm{im}T$, giving the rank-nullity theorem:

For $u\in U$, the vector $Tu$ already belongs to $\mathrm{im}T$, soTherefore $\ker (ST)=T^{-1}(\ker S^{\prime })$. Since $\ker S^{\prime }=\ker S\cap \mathrm{im}T$, its intersection with $\mathrm{im}T$ is itself. Applying a with $X=\ker S^{\prime }$ yieldsThe inequality follows from the inclusion $\ker S^{\prime }\subseteq \ker S$ and monotonicity of dimension for vector subspaces.

A real number $\lambda$ is an eigenvalue of $T$ if $Tv=\lambda v$ for some nonzero vector $v\in V$. Its eigenspace isIt is a vector subspace because it contains $0$ and $T(av+bw)=aTv+bTw=\lambda (av+bw)$ whenever $v,w\in E_{\lambda }$ and $a,b\in \mathbb{R}$.

If $v\in E_{\lambda }\cap E_{\mu }$, then $\lambda v=Tv=\mu v$, so $(\lambda -\mu )v=0$. Since $\lambda \ne \mu$, this forces $v=0$. Therefore

Put $L=\mathrm{span}\{v\}$. Every two-dimensional vector subspace containing $v$ is $W=\mathrm{span}\{v,w\}$ for some $w\notin L$. There are $p^n-p$ choices for $w$ in $V$. Within any one such $W$, exactly $p^2-p$ choices of $w\in W\setminus L$ yield that same $W$: each forms a linearly independent pair with $v$, and that pair is a basis of $W$.

These sets of choices partition $V\setminus L$, because $w$ determines $\mathrm{span}\{v,w\}$ uniquely. Thus the number isEquivalently these vector subspaces correspond to the one-dimensional vector subspaces of the quotient vector space $V/L$. For $n=1$ there are none and the formula gives zero.

Let $S$ be the vector subspace of symmetric matrices. A basis consists of $E_{ii}$ for $1\le i\le n$ and $E_{ij}+E_{ji}$ for $i<j$, soThe dimension formula for a sum of subspaces now givesChoose $0\ne A\in V\cap S$. It belongs to $V$ and is a symmetric matrix, so part (a)(i) makes it a diagonalisable matrix. The preceding example shows that the strict threshold is sharp.

Take the vector subspace of skew-symmetric matrices,It is closed under vector addition and scalar multiplication, and a basis isIndeed, its diagonal entries vanish, and its entries below the diagonal are forced by those above it. Hence $\boxed{\dim K=\left(\genfrac{}{}{0ex}{}{n}{2}\right)}$. Part (a)(ii) proves that the only diagonalisable matrix in $K$ is $0_n$.

The homogeneous linear system has solution set $\ker A$, a vector subspace. If it contains only the zero vector, no two different solutions of the full system can exist: their difference would be a nonzero homogeneous solution.

A particular solution ${\mathbf{v}}_p$ to the full system exists exactly when $\mathbf{w}$ belongs to the image of a linear map $\mathrm{im}A$. If it exists, every solution isIndeed, subtracting $A{\mathbf{v}}_p=\mathbf{w}$ from $A\mathbf{v}=\mathbf{w}$ gives $A(\mathbf{v}-{\mathbf{v}}_p)=0$, and conversely adding a kernel vector preserves the right-hand side. Thus there are no solutions, one solution if the kernel is trivial, or infinitely many solutions if it is nontrivial. The last solution set is an affine subspace of dimension $\dim \ker A$, determined by the rank-nullity theorem. For a square nonsingular matrix there is one solution for every right-hand side.

No. Every vector in the plane can be written asThe two linearly independent vectors in part (ii) therefore form a vector space basis for this vector subspace. Its dimension is $2$, so any three vectors in it have linear dependence.

A vector subspace $W\subseteq {\mathbb{C}}^n$ is an invariant subspace for the group representation $D$ ifBecause $g^{-1}$ is also in the group and $D(g)$ is invertible, the inclusion is equivalently $D(g)W=W$ for every $g$.

For $n>0$, the representation is an irreducible representation when its only invariant vector subspaces areThus no nonzero proper subspace is preserved by all representing matrices.

The symmetric group $S_n$ consists of all bijections of $\{1,\dots ,n\}$, with composition as its operation; its group order is $n!$.

An $m$-dimensional complex group representation of $G$ is a group homomorphism $\rho :G\to GL(V)$, where $V$ is an $m$-dimensional complex vector space. After a choice of basis, each $\rho (g)$ is an invertible $m\times m$ complex matrix, with $\rho (gh)=\rho (g)\rho (h)$.

An irreducible representation has $V\ne 0$ and no invariant vector subspace other than $\{0\}$ and $V$. Two group representations $\rho$ on $V$ and $\sigma$ on $W$ are isomorphic representations if there exists an invertible linear map $S:V\to W$ such that $S\rho (g)=\sigma (g)S$ for every $g$. They are inequivalent if no such map exists; in equal-dimensional matrix form this says no fixed change of basis conjugates every $\rho (g)$ to $\sigma (g)$.

Use the Weyl algebra convention $[y_i,x_j]={\delta }_{ij}$, with $[x_i,x_j]=[y_i,y_j]=0$, where the brackets are commutators. Its order filtration of a Weyl algebra isThus $x_i$ has degree zero and $y_i$ degree one. By the ordered monomial basis of a Weyl algebra, the associated graded ring iswhere ${\xi }_i$ is the symbol of $y_i$.

The Rees ring of a filtered algebra is the graded ringwhere $t$ is a central indeterminate. Here it is generated by $x_i$, $Y_i=y_{i}t$, and $t$, with degrees $0,1,1$, respectively, and relationsIt interpolates between the associated graded ring and the original filtered ring:

A good filtration for the order filtration on $M$ is an increasing exhaustive family of complex vector subspaces $(M_j{)}_{j\in \mathbb{Z}}$, zero for sufficiently negative $j$, with $F_{i}R{M}_j\subseteq M_{i+j}$, such that the associated graded moduleis a finitely generated module over $A$. Equivalently, there are finitely many $m_1,\dots ,m_s$ and integer shifts $d_1,\dots ,d_s$ such that $M_j={\sum }_{\ell =1}^s(F_{j-d_{\ell }}R)m_{\ell }$. In particular, choose any finite generating family $m_1,\dots ,m_s$ of $M$ and putThis is a good filtration for the order filtration.

For $M\ne 0$, its dimension of a Weyl algebra module for the order filtration isHere support of a module, annihilator of a module, and Krull dimension are taken over the commutative polynomial ring $A$. The corresponding algebraic set is the order characteristic variety of a Weyl algebra module. This dimension is independent of the chosen good filtration for the order filtration; one may set $d(0)=-\infty$.

The pieces $F_{j}R$ generally have infinite complex dimension, since they contain all polynomials in the $x_i$. Thus a polynomial-growth definition based directly on ${\dim }_{\mathbb{C}}{M}_j$ would be inappropriate for this order filtration of a Weyl algebra.

Since $K$ is an $F$-vector subspace of the finite-dimensional vector space $L$, it has a finite $F$-basis $u_1,\dots ,u_m$. Any finite $F$-spanning set for $L$ also spans $L$ over $K$, so choose a finite $K$-basis $v_1,\dots ,v_r$ of $L$.

Every $\ell \in L$ can be expanded first in the $v_j$ and then in the $u_i$, showing that the products $u_i{v}_j$ span $L$ over $F$. If ${\sum }_{i,j}{c}_{ij}{u}_i{v}_j=0$ with $c_{ij}\in F$, independence of the $v_j$ over $K$ gives ${\sum }_i{c}_{ij}{u}_i=0$ for every $j$; independence of the $u_i$ then gives every $c_{ij}=0$. The products form an $F$-basis, proving the tower law for field extensions:

Use the same $4\times 4$ square grid with four centre-based triangles per small square, but identify boundary points byThis gives the real projective plane. To see the model explicitly, a radial homeomorphism from the square, centred at $(1/2,1/2)$, to the closed disc carries the boundary involution to the antipodal map. The map from that disc to the projective planeis continuous and surjective: every one-dimensional vector subspace of ${\mathbb{R}}^3$ has a unit vector representative in the closed upper hemisphere. Its only identifications are antipodal boundary points. The compact-to-Hausdorff continuous bijection theorem therefore identifies this disc quotient space with ${\mathbb{RP}}^2$.

The square triangulation descends to an abstract simplicial complex $K_2$. Small-square centres remain distinct interior vertices; each triangle has one centre and two adjacent grid vertices. Boundary pairings identify whole grid edges, while preserving distinct vertices in each triangle, so this is a finite triangulation of the real projective plane. Its counts are $33$ vertices, $96$ edges and $64$ triangles, giving Euler characteristic one.

Choose $\sigma$ to be the interior triangle with vertices $(1/4,1/4)$, $(1/2,1/4)$ and $(3/8,3/8)$. Removing its relative interior and cutting from $(0,0)$ to $(1/4,1/4)$ gives the same nine-sided cut disc as before. Parametrize it by a convex polygon, transporting the boundary identifications. Its boundary word is nowThe boundary involution pairs the two $a$ sides and the two $b$ sides in matching traversal directions; the $c$ pair restores the slit. Leave $d,e,f$ unpaired. The quotient space isthe real projective plane with an open disc removed. In particular it is a compact connected surface with boundary consisting of one three-edge circle.

Take the linear mapIts kernel of a linear map consists of all $(x,0)$, and its image of a linear map is the same line:Their set intersection is nonzero and their sum is only one-dimensional. In particular, $(0,1)$ cannot be written as a sum of one vector from each of these vector subspaces. The nonzero set intersection also prevents an internal direct sum of vector spaces. ThusThe matrix is a nonzero nilpotent matrix; its failure of diagonalizability explains why the preceding result does not apply.

By diagonalizability, $V$ has an eigenbasis $e_1,\dots ,e_m$ with $T{e}_j={\lambda }_j{e}_j$. Arrange the basis so that ${\lambda }_1=\cdots ={\lambda }_r=0$ and the remaining eigenvalues are nonzero. If there are no zero or no nonzero eigenvalues, the corresponding linear span below is simply the zero vector subspace.

For $v={\sum }_j{a}_j{e}_j$, the equation $Tv=0$ holds precisely when $a_j=0$ for every $j>r$. ThusEvery output is in $\mathrm{span}\{e_{r+1},\dots ,e_m\}$, and conversely $e_j=T({\lambda }_j^{-1}{e}_j)$ for each $j>r$. ThereforeThese two disjoint parts of a basis span $V$ and their linear spans intersect only at zero. This is exactly the direct sum of vector spaces assertion