1. Determine the nullity of the linear map L:\mathbb{R}_n[x]\rightarrow\mathbb{R} given by a_0+a_1x+\cdots+a_nx^n\mapsto\int_{-1}^0(a_0+a_1x+\cdots+a_nx^n)\mathrm{d}x, where a_i\in\mathbb{R}.

  • Lx=0\Rightarrow\sum_{i=0}^{n}\frac{a_i}{i+1}x^{i+1}|_{-1}^0=\sum{i=0}^n\frac{(-1)^i}{i+1}a_i=0\Rightarrow\dim\mathrm{Ker}L=n.

2. Let L:\mathbb{R}_2[x]\rightarrow\mathbb{R}_3[x] be defined by L(f(x))=xf(x)-2f'(x) for f\in\mathbb{R}_2[x]. Show that L is a linear map, and find bases for both \mathcal{N}(L) and \mathcal{R}(L). Then determine whether L is injective or surjective.

  • \forall f(x),g(x)\in\mathbb{R}_2[x],\lambda\in R,L((f+\lambda g)(x))=x(f(x)+\lambda g(x))-2(f'(x)+\lambda g'(x))=L(f(x))+\lambda L(g(x)), so L is a linear map.
  • L(f)=0\Rightarrow xf=2f'\Rightarrow f(x)=0, so the bases for \mathcal{N}(L) is \emptyset. Therefore \mathrm{rank}L=3, means that \dim\mathcal{R}(L)=\mathrm{rank}L=3=\dim\mathbb{R}_2[x]. Notice that 1,x,x^2 is a base for \mathbb{R}_2[x], so x,x^2-2,x^3-4x is a base for \mathcal{R}(L).
  • As \dim\mathcal{R}(L)=3<4=\dim\mathbb{R}_3[x] and \dim\mathcal{N}(L)=0, L is injective but not surjective.

3. Let V be a non-zero finite dimensional real vector space. Show that there are no L,S\in\mathcal{L}(V) with LS-SL=\mathrm{id}_V. Does this hold if V is an infinite dimensional vector space?

  • Let n=\dim V, consider \sigma:V\rightarrow \mathbb{R}^n defined by maping a base for V to a base for \mathbb{R}^n, \varphi:\mathcal{L}(V)\rightarrow M_n(\mathbb{R}) defined by \varphi(L)=\sigma\circ L\circ\sigma^{-1}, which implied that V\cong\mathbb{R}^n,\mathcal{L}(V)\cong M_n(\mathbb{R}).
    Notice that \mathrm{tr}(AB-BA)=0\neq \mathrm{tr}(I), therefore AB-BA=I has no solutions in M_n(\mathbb{R}). Apply \sigma^{-1},\sigma on the two sides of the equation, we get \sigma^{-1}\circ(AB-BA)\circ\sigma=\sigma^{-1}\circ AB\circ\sigma-\sigma^{-1}\circ BA\circ\sigma=\sigma^{-1}\circ A\circ\sigma\circ\sigma^{-1}\circ B\circ\sigma-\sigma^{-1}\circ B\circ\sigma\circ\sigma^{-1}\circ A\circ\sigma=\varphi^{-1}(A)\circ\varphi^{-1}(B)-\varphi^{-1}(B)\circ\varphi^{-1}(A)=LS-SL\neq\sigma^{-1}\circ I\circ\sigma=\sigma^{-1}\circ\sigma=\mathrm{id}_V,\forall L,S\in\mathcal{L}(V),A=\varphi(L),B=\varphi(S). This is exactly what we want to prove.
  • No. Let V=R^{\aleph_0}=\mathrm{Span}_\mathbb{R}\{e_i|i\in\mathbb{N}\};L:e_n\mapsto 0(n=0),ne_{n-1}(n>0);S:e_n\mapsto e_{n+1}, we have LS:e_n\mapsto(n+1)e_n;SL:e_n\mapsto0(n=0),ne_n(n>0), therefore LS-SL:e_n\mapsto e_n, which means LS-SL=\mathrm{id}_V.

4. Let F be a field. If f:M_n(F)\rightarrow F is a linear functional such that f(AB-BA)=0 for any A,B\in M_n(F), then f=c\cdot\mathrm{tr} for some c\in F.

  • Consider A=E_{a,b},B=E_{c,d}, we have AB-BA=0((a\neq d,b\neq c)\vee (a=d,b=c)),E_{a,d}(a\neq d,b=c),-E_{c,b}(a=d,b\neq c). Therefore f(E_{a,b})=0,\forall a\neq b;f(E_{i,i}-E_{j,j})=0\Rightarrow f(E_{i,i})=f(E_{j,j}),\forall i,j. And let f(E_{i,i})=c, we get (f-c\cdot \mathrm{tr})(E_{i,j})=f(E_{i,j})-c\cdot \mathrm{tr}(E_{i,j})=0,\forall i,j. Notice that \bigcup\{E_{i,j}\} is a base for M_n(F), f must be c\cdot \mathrm{tr}.

5.Suppose V is finaite dimensional and S,T\in\mathcal{L}(V). Prove that ST is invertible \Leftrightarrow S and T are invertible. Is it true for infinite dimensional V?

  • ST is invertible \Leftrightarrow \dim \mathrm{Im}ST=\dim V\xRightarrow{\dim\mathrm{Im}ST\leq{\dim\mathrm{Im}T,\dim\mathrm{Im}S}\leq\dim V}\dim\mathrm{Im}S=\dim\mathrm{Im}T=\dim V\Rightarrow S and T are invertible \Rightarrow STT^{-1}S^{-1}=\mathrm{id}_V\Rightarrow ST is invertible.
  • No. Let V=R^{\aleph_0}=\mathrm{Span}<em>\mathbb{R}{e_i|i\in\mathbb{N}};S:e_n\mapsto 0(n=0),e</em>{n-1}(n>0);T:e_n\mapsto e_{n+1}, we have ST=\mathrm{id}_V;e_0\notin\mathrm{Im}T, therefore ST is invertible, but T is not surjectivity, which means T is not invertible.

6. Suppose q\in\mathbb{R}[x]. Prove that there exists a polynomial p(x)\in\mathbb{R}[x] such that q(x)=(x^2+x)p''(x)+2xp'(x)+p(3) for all x\in\mathbb{R}.

  • Let N=\deg(q(x))\neq0;p(x)=(1,x,\cdots,x^N)\alpha;q(x)=(1,x,\cdots,x^N)\beta;p'(x)=(1,x,\cdots,x^N)D\alpha;D_{i,j}=0(i\neq j-1),i(i=j-1). Notice that \deg((x^2+x)p''(x)+2xp'(x)+p(3))=N\Rightarrow\deg(p(x))\leq N.
    So, let U_{i,j}=0(i\neq j+1),1(i=j+1);P_{i,j}=3^{j-1}(i=1),0(i\neq 1), we have xp(x)=(1,x,\cdots,x^N)U\alpha,\forall\deg(p(x))<N;p(3)=(1,x,\cdots,x^N)P\alpha. In this way, the equation turns to \beta=(U^2D^2+UD^2+2UD+P)\alpha:=A\alpha.
    Notice that A_{i,j}=0(i>j),A_{i,i}\neq0,\forall i, therefore \mathrm{rank}A=N, which means the equation has one unique solution.

7. Suppose m is a positive integer. Show that the dual basis of the basis 1,x,\cdots,x^m of \mathbb{R}_m[x] is \varphi_0,\cdots,\varphi_m, where \varphi_k(f)=\frac{f^{(k)}(0)}{k!}.

  • Notice that \varphi_i(x^j)=\frac{0}{i!}(j<i),1(j=i),\frac{j!x^{j-i}|_{x=0}}{(j-i)!i!}(j>i)=\delta_{ij}, so \varphi_0,\cdots,\varphi_m is the dual basis of the basis 1,x,\cdots,x^m of \mathbb{R}_m[x].

8. Suppose m is a positive integer.

(a) Show that 1,x-5,\cdots,(x-5)^m is a basis of \mathbb{R}_m[x].

  • Notice that (x-5)^k=\sum_{i=0}^k C^i_k(-5)^{k-i}x^i, we have (1,x-5,\cdots,(x-5)^m)=(1,x,\cdots,x^m)T,T_{i,j}=C^i_j(-5)^{j-i}(i\leq j),0(i>j). Because |T|\neq0\Rightarrow \mathrm{rank}T=m, 1,x-5,\cdots,(x-5)^m is a basis of \mathbb{R}_m[x].

(b) What is the dual basis of the basis in (a)?

  • Notice that if we define \varphi_k(f)=\frac{f^{(k)}(5)}{k!}, we can get \varphi_i((x-5)^j)=\delta_{ij}. So, \varphi_0,\cdots,\varphi_m is the dual basis of the basis in (a).

9. State and prove three isomorphism theorems on vector spaces.

(a) (First isomorphism theorem). Let V be a vector space and T:V\rightarrow W is a linear transformation, we have \mathrm{Im}T\cong V/\mathrm{Ker}T.

  • Consider \tau:V/\mathrm{Ker}T\rightarrow\mathrm{Im}T,v+\mathrm{Ker}T\mapsto T(v). As v+\mathrm{Ker}T=v'+\mathrm{Ker}T\xRightarrow{v'-v\in\mathrm{Ker}T} T(v')=T((v'-v)+v)=T(v'-v)+T(v)=T(v), so \tau is well-defined. Notice that \tau((v+\mathrm{Ker}T)+\lambda(v'+\mathrm{Ker}T))=T(v+\lambda v')=T(v)+\lambda T(v')=\tau(v+\mathrm{Ker}T)+\lambda\tau(v'+\mathrm{Ker}T), so \tau is linear. Additionally, \tau(v+\mathrm{Ker}T)=T(v)=0\Rightarrow v\in\mathrm{Ker}T\Rightarrow v+\mathrm{Ker}T=0;\forall\alpha\in\mathrm{Im}T\exists v\in V\ni T(v)=\alpha\Rightarrow\alpha=\tau(v+\mathrm{Ker}T)\in\mathrm{Im}\tau, so \tau is an isomorphism. Therefore, \mathrm{Im}T\cong V/\mathrm{Ker}T.

(b) (Second isomorphism theorem). Let V be a vector space and U,W\subseteq V two subspaces, we have U/(U\cap W)\cong (U+W)/W.

  • Consider \tau:U/(U\cap W)\rightarrow(U+W)/W,u+(U\cap W)\mapsto u+W. As u+(U\cap W)=u'+(U\cap W)\xRightarrow{u'-u\in U\cap W\subseteq W} u'+W=(u'-u+W)+(u+W)=u+W, so \tau is well-defined. Notice that \tau((u+(U\cap W))+\lambda(u'+(U\cap W)))=u+\lambda u'+W=u+W+\lambda (u'+W)=\tau(u+(U\cap W))+\lambda\tau(u'+(U\cap W)), so \tau is linear. Additionally, \tau(u+(U\cap W))=u+W=0\Rightarrow u\in W\Rightarrow u\in(U\cap W)\Rightarrow u+(U\cap W)=0;\forall\alpha\in(U+W)/W\exists u'=u+w,u\in U,w\in W\ni u'+W=\alpha\Rightarrow\alpha=u+w+W=u+W=\tau(u+(U\cap W))\in\mathrm{Im}\tau, so \tau is an isomorphism. Therefore, U/(U\cap W)\cong (U+W)/W.

(c) (Third isomorphism theorem). Let V be a vector space, W a subspace of V, and U a subspace of W, we have (V/U)/(W/U)\cong V/W.

  • Consider \tau:(V/U)/(W/U)\rightarrow V/W,(v+U)+W/U\mapsto v+W. As (v+U)+W/U=(v'+U)+W/U\xRightarrow{v'-v+U\in W/U\Rightarrow v'-v\in W} v'+W=(v'-v+W)+(v+W)=v+W, so \tau is well-defined. Notice that \tau(((v+U)+W/U)+\lambda((v'+U)+W/U))=v+\lambda v'+W=v+W+\lambda (v'+W)=\tau((v+U)+W/U)+\lambda\tau((v'+U)+W/U), so \tau is linear. Additionally, \tau((v+U)+W/U)=v+W=0\Rightarrow v\in W\Rightarrow v+U\in W/U\Rightarrow (v+U)+W/U=0;\forall\alpha\in V/W\exists v\in V\ni v+W=\alpha\Rightarrow\alpha=v+W=\tau((v+U)+W/U)\in\mathrm{Im}\tau, so \tau is an isomorphism. Therefore, (V/U)/(W/U)\cong V/W.