$\DeclareMathOperator{\im}{Im}\DeclareMathOperator{\re}{Re}\DeclareMathOperator{\span}{Span}\DeclareMathOperator{\diag}{diag}\DeclareMathOperator{\tr}{tr}$

1. Let $W ⩽ V$ be finite-dimensional complex vector spaces and let $T: V → V$ be a linear map.
   1. Let $ℬ$ be a basis for $V$ containing a basis $ℬ_1$ for $W$, and let $ℬ_2:=ℬ∖ℬ_1$.
      1. Show that $q\left(ℬ_2\right)$ is a basis for $V / W$, where $q: V → V / W$ is given by $q(v)=v+W$.
      2. Suppose that $W$ is $T$-invariant. Prove that there exists a well-defined linear map $\bar{T}: V / W → V / W$ such that $\bar{T} ∘ q=q ∘ T$.
      3. Show that there exists a matrix $C$ such that$$\sideset{_ℬ}{_ℬ}{[T]}={\mmlToken{mo}(\begin{smallmatrix}\hskip5mu[{T|}_W]_{ℬ_1\kern-18.5mmℬ_1}&C\\0&\sideset{_{q(ℬ_2)}}{_{q(ℬ_2)}}{[\bar T]}\end{smallmatrix}\mmlToken{mo})}$$
      4. State and prove the Cayley-Hamilton Theorem.
   2. Now suppose that $V$ has basis $\{u, v\}$ and that $T(u)=i u$ and $T(v)=u+i v$.
      1. Prove that $\{u, i u, v, i v\}$ is a basis for $V$, viewed as an $ℝ$-vector space.
      2. Calculate the matrix $A$ of the $ℝ$-linear map $T: V → V$ with respect to $\{u, i u, v, i v\}$.
      3. Find the minimal polynomial of $A$, justifying your answer.
      4. Calculate the Jordan Normal Form of $A$, justifying your answer.
2. Let $V$ be a finite dimensional vector space over a field $𝔽$, and let $T: V → V$ be a linear map.
   1. Let $p(x), q(x) ∈ 𝔽[x]$ be coprime polynomials such that $\im p(T) ⊆ \ker q(T)$.
      1. Prove that $V=\ker p(T) ⊕ \ker q(T)$.
      2. Is the conclusion of (a)(i) true if $p(x)$ and $q(x)$ are not necessarily coprime?
      3. Suppose that the minimal polynomial of $T$ is equal to a product of distinct linear factors over $𝔽$. Prove that $T$ is diagonalisable.
      4. Does the converse to (a)(iii) hold?
   2. Suppose that $T^p=I$ for some prime number $p$. Assume that the characteristic of $𝔽$ is zero, and that there exists $ζ ∈ 𝔽$ such that $ζ^p=1$ but $ζ ≠ 1$. Let $ℰ$ denote the set of all $𝔽$-linear maps $S: V → V$, and for each integer $i$ let $$ ℰ(i):=\left\{S ∈ ℰ: T S=ζ^i S T\right\} $$
      1. Using part (a)(iii) applied to a suitable map defined on $ℰ$, prove that $$ ℰ=ℰ(0) ⊕ ℰ(1) ⊕ ⋯ ⊕ ℰ(p-1) $$ For each $S ∈ ℰ$ and any integer $i$, let $$ ϵ_i(S):=\sum_{j=0}^{p-1} ζ^{i j} T^j S T^{p-j} ∈ ℰ . $$
      2. Show that $ϵ_i(S) ∈ ℰ(-i)$ for any integer $i$.
      3. Deduce that $ϵ_i(S)^p$ commutes with $T$ for any integer $i$.
      4. Show that if $S ≠ 0$, then $ϵ_i(S) ≠ 0$ for at least one integer $i$. [Hint: you may find it helpful to consider the sum $ϵ_0(S)+⋯+ϵ_{p-1}(S)$.]
3. 1. 1. Define what is meant by a complex inner product space.
      2. Let $M_n(ℂ)$ be the vector space of $n × n$ complex matrices. Prove that $$ ⟨ x, y⟩:=\operatorname{tr}\left(\bar{x} y^T\right)   \text { for } x, y ∈ M_n(ℂ) $$ turns $M_n(ℂ)$ into a complex inner product space.
      3. Let $X:=\left\{x ∈ M_n(ℂ): x_{i j}=0\right.$ for all $\left.i>j\right\}$ be the subspace of upper triangular matrices. Find the orthogonal complement $X^{⟂}$ of $X$ in $M_n(ℂ)$.
   2. Let $U, V$ and $W$ be finite dimensional complex inner product spaces.
      1. Let $α: U → V$ be a linear map. Define the adjoint map $α^*: V → U$, and prove that it is unique. Show that $\left(α^*\right)^*=α$. [You may assume that $α^*: V → U$ always exists.]
      2. Show that $\imα∩\kerα^*=\{0\}$.
      3. Let $β: V → W$ be another linear map such that $\imα=\kerβ$, and let $$ γ:=α α^*+β^* β $$ Prove that $γ: V → V$ is invertible. MSE

Solution

1. 1. 1. Let $\{e_1, ⋯, e_k\}=ℬ_1,\{e_{k+1}, ⋯, e_n\}=ℬ_2$.
         To show $V/W=\spanℬ_2$. Take any $q(v)∈V/W$, since $V=\spanℬ$, $∃a_1, ⋯, a_n ∈ℂ:$ $$ v=a_1 e_1+⋯+a_k e_k+a_{k+1} e_{k+1}+⋯+a_n e_n $$Since $e_1,…,e_k∈\ker q$, \[q(v)=a_{k+1}\left(q(e_{k+1})\right)+⋯+a_n\left(q(e_n)\right)\] To show linear independence, assume for some $a_{k+1},…,a_n∈ℂ:$ $$ a_{k+1}\left(q(e_{k+1})\right)+⋯+a_n\left(q(e_n)\right)=0_W $$ Then $a_{k+1}e_{k+1}+⋯+a_ne_n∈\ker q=W$. Then $∃a_1,…,a_k∈ℂ:$ $$ a_1 e_1+⋯+a_k e_k=a_{k+1} e_{k+1}+⋯+a_n e_n $$Since ℬ is linearly independent, $a_{k+1}=⋯=a_n=0$.
      2. For any $q(v)∈V/W$, define $\bar{T}(q(v))=q(T(v))$.
         For any $w∈W$, $T(w)∈W$, so $\bar{T}(q(w))=q(T(w))=0$. $\bar T$ is well-defined Lemma 3.8
      3. Let $a_{ij}∈ℂ$ be the $(i,j)$-entry of $\sideset{_ℬ}{_ℬ}{[T]}$
         For $j≤k$, $e_j∈W$, so $T(e_j)∈W$
          so for $i>k,a_{ij}=0$; for $i≤k$, $a_{ij}$ is equal to the $(i,j)$-entry of $\sideset{_{ℬ_1}}{_{ℬ_1}}{[{T|}_W]}$
         For $j>k$, $q(T(e_j))=\bar T(q(e_j))$,
          so for $i>k$, $a_{ij}$ is equal to the $(i-k,j)$ entry of $\sideset{_{q(ℬ_2)}}{_{q(ℬ_2)}}{[\bar T]}$
      4. If $V$ is a finite-dimensional ℂ-vector space and $T:V→V$ is linear, then $χ_T(T)=0$.
         Proof: $∃P∈M_n(ℂ):A=P^{-1}TP$ is upper triangular. Suppose the theorem holds for $A$,\[χ_T(x)=\det(xI-T)=\det\left(P(xI-A)P^{-1}\right)=\det(xI-A)=χ_A(x)\]$⇒χ_T(T)=χ_A(T)=Pχ_A(A)P^{-1}=0$, the theorem holds for $T$.
         It remains to prove the theorem for any upper triangular matrix $A=\pmatrix{λ_1&*&*\\&⋱&*\\&&λ_n}$
         Let $e_1,…,e_n$ be the standard basis vectors for $ℂ^n$. Then for all $v∈ℂ^n$\begin{align*}(A-λ_nI)v&∈⟨e_1,…,e_{n-1}⟩\\ (A-λ_{n-1}I)(A-λ_nI)v&∈⟨e_1,…,e_{n-2}⟩\\ &⋮\\ (A-λ_1I)⋯(A-λ_{n-1}I)(A-λ_nI)v&∈\{0\} \end{align*}so $χ_A(A)=(A-λ_1I)⋯(A-λ_{n-1}I)(A-λ_nI)=0$ as required.
   2. 1. spanning: $∀x∈V,∃λ,μ∈ℂ:x=λu+μv$
         $⇒x=(\reλ)u+(\imλ)iu+(\reμ)v+(\imμ)iv∈⟨u,iu,v,iv⟩_ℝ$.
         linear independent: $a,b,c,d∈ℝ,au+biu+cv+div=0⇒(a+bi)u+(c+di)v=0$
         $\{u,v\}$ are linearly independent over ℂ$⇒a+bi=c+di=0⇒a=b=c=d=0$.
      2. $T(u)=iu,T(iu)=-u,T(v)=u+iv,T(iv)=iu-v⇒A=\pmatrix{&-1&1\\ 1&&&1\\ &&&-1\\ &&1}$
      3. By Cayley-Hamilton $m_A(x)|χ_A(x)=(x^2+1)^2$
         $T^2(v)=T(u)+T(iv)=2iu-v≠0⇒(A^2+I)v≠0$
         $⇒m_A(x)∤(x^2+1)⇒m_A(x)=(x^2+1)^2$
      4. Eigenvalues of $A$ are roots $±i$ of $m_A(x)$.
         By primary decomposition theorem,$V=\ker(A-iI)^2⊕\ker(A+iI)^2$
         Take $v_1∈\ker(A-iI)^2∖\ker(A-i)$, the matrix of $T$ over $\{(A-iI)v_1,v_1\}$ is $\pmatrix{i&1\\&i}$
         Take $v_2∈\ker(A+iI)^2∖\ker(A+iI)$, the matrix of $T$ over $\{(A+iI)v_2,v_2\}$ is $\pmatrix{-i&1\\&-i}$ so the Jordan Normal Form of $A$ is $\pmatrix{i&1\\&i\\&&-i&1\\&&&-i}$
2. 1. 1. Since $p(x),q(x)$ are coprime, $∃a(x),b(x)∈𝔽[x]:a(x)p(x)+b(x)q(x)=1$.
         $∀v∈V:v=\underbrace{a(T)p(T)v}_{∈\ker q(T)}+\underbrace{b(T)q(T)v}_{∈\ker p(T)}$
         And if $v∈\ker p(T)∩\ker q(T)$, $v=a(T)p(T)v+b(T)q(T)v=0+0$
      2. Set $𝔽=ℝ,\dim V=2,p(x)=q(x)=x$. We need $\im T⊆\ker T⇔T^2=0$
         Set $T=\pmatrix{0&1\\0&0}$ then $T^2=0$, $\ker T=\span\{\pmatrix{1\\0}\}$, $\ker p(T)∩\ker q(T)≠\{0\}$
      3. Induct on $\deg m_T$. For $\deg m_T=1$, $m_T(x)=x-λ_1⇒T=λ_1I$. Suppose it's true for $\deg m_T=n-1$.
         Let $m_T(x)=\prod_{i=1}^n(x-λ_i)$ for distinct $λ_i∈𝔽$.
         Let $p(x)=(x-λ_1)⋯(x-λ_{n-1}),q(x)=x-λ_n$.
         Let $a(x),b(x)$ be minimal polynomial of the restriction of $T$ to $\ker p(T),\ker q(T)$, then $a(x)|p(x),b(x)|q(x)$. Also $a(x)b(x)$ annihilates $T$ on $V$, we have $p(x)q(x)|a(x)b(x)$, so $a(x)=p(x),b(x)=q(x)$. So the minimal polynomial of ${T|}_{\ker p(T)}$ is $p(x)$, a product of $n-1$ distinct linear factors. By induction hypothesis $\ker p(T)=\bigoplus_{i=1}^{n-1}\ker(T-λ_iI)$.
         By (a.i) $V=\ker p(T)⊕\ker q(T)⇒V=\bigoplus_{i=1}^n\ker(T-λ_iI)$, let $0≠v_i∈\ker(T-λ_iI)$, $ℬ=\{v_1,…,v_n\}$ then $\sideset{_ℬ}{_ℬ}{[T]}=\diag(λ_1,…,λ_n)$.
      4. Yes. If $T$ is diagonalisable, ∃ a basis ℬ of $V$ such that $M=\sideset{_ℬ}{_ℬ}{[T]}$ is a diagonal matrix. Let $λ_1,…,λ_n$ be the distinct diagonal entries of $M$, and let $f(x)=(x-λ_1)⋯(x-λ_n)$.
         $∀v∈ℬ$, $∃k∈\{1,…,n\}:Tv=λ_kv⇒(T-λ_kI)v=0⇒f(T)v=(T-λ_1I)⋯(T-λ_kI)v=0$. Since ℬ spans $V$, $f(T)v=0∀v∈V$. Hence $m_T(x)$ divides $f(x)$, so $m_T(x)$ is a product of distinct linear factors.
   2. 1. Note that $T$ is invertible because $TT^{p-1}=T^{p-1}T=I$.
         Define $T$-conjugation map $C:ℰ→ℰ,S↦TST^{-1}$. Then $C$ is linear.
         $∀k≥1:C^k(S)=T^kST^{-k}$. Hence $C^p(S)=T^pST^{-p}=S$, so $C^p=\text{id}_ℰ$,
         $m_C(x)$ divides $x^p-1$ which factors over 𝔽 as $∏^{p-1}_{i=0}(x-ζ^i)$. Now $ζ$ has order dividing $p$ in $𝔽^×$. Since $ζ≠1$, this order is > 1. Since $p$ is prime, this order is $p$. Hence the elements $ζ^0,ζ^1,…,ζ^{p-1}$ are pairwise distinct. So $m_C(x)$ is a product of distinct linear factors, by (a)(iii) $C:ℰ→ℰ$ is diagonalisable. The only possible eigenvalues of $C$ are $ζ^i,i=0,…,p-1$, and $\ker(C-ζ^iI)=\{S∈ℰ:TST^{-1}=ζ^iS\}=ℰ(i)$.
         Hence $ℰ=ℰ(0) ⊕ ℰ(1) ⊕ ⋯ ⊕ ℰ(p-1)$.
      2. Since $T^p=I$, $ϵ_i(S)=\sum_{j=0}^{p-1} ζ^{i j} T^j S T^{-j}=\sum_{j=0}^{p-1} ζ^{i j} C^j(S)$\[C(ϵ_i(S))=\sum_{j=0}^{p-1} ζ^{i j} C^{j+1}(S)=\sum_{j=0}^{p-1} ζ^{i (j-1)} C^j(S)=ζ^{-i}ϵ_i(S)\] So $ϵ_i(S)∈ℰ(-i)$.
      3. If $U ∈ ℰ(i)$ and $V ∈ ℰ(j)$, then $C(U)=ζ^i U$ and $C(V)=ζ^j V$, so $C(U V)=T U V T^{-1}=T U T^{-1} T V T^{-1}=C(U) C(V)=ζ^{i+j} U V$. Hence $U V ∈ ℰ(i+j)$, which impies that $U^p ∈ ℰ(0)$ for any $U ∈ ℰ(i)$.
         In particular, $ϵ_i(S)^p ∈ ℰ(0)$ because $ϵ_i(S) ∈ ℰ(-i)$ by the above. But $ℰ(0)$ consists of precisely those $S ∈ ℰ$ that commute with $T$.
      4. Since $S ≠ 0$ and the characteristic of $𝔽$ is zero, it is enough to show that $ϵ_0(S)+ϵ_1(S)+⋯+ϵ_{p-1}(S)=p S$.
         The expression on the left hand side is equal to $$ \sum_{i=0}^{p-1}\left(\sum_{j=0}^{p-1} ζ^{i j} C^j(S)\right)=\sum_{j=0}^{p-1}\left(\sum_{i=0}^{p-1} ζ^{i j}\right) C^j(S) . $$ If $1 ⩽ j ⩽ p-1$, then because $p∤j$, $\left\{ζ^0, ζ^j, ζ^{2 j}, ⋯, ζ^{(p-1) j}\right\}=\left\{ζ^0, ζ^1, ⋯, ζ^{p-1}\right\}$.
         Hence $\sum_{i=0}^{p-1} ζ^{i j}=\sum_{i=0}^{p-1} ζ^i=0=\frac{ζ^p-1}{ζ-1}=0$. Hence $\sum_{i=0}^{p-1} ϵ_i(S)=p S$ as claimed.
3. 1. 1. A complex vector space $V$ with a sesquilinear, conjugate symmetric, positive definite form $⟨·,·⟩$ is called a complex inner product space.
         sesquilinear: $⟨u+v,w⟩=⟨u,w⟩+⟨v,w⟩,⟨\barλv,w⟩=λ⟨v,w⟩=⟨v,λw⟩$ for $v,w∈V$
         conjugate symmetric: $\overline{⟨v,w⟩}=⟨w,v⟩$ for $v,w∈V$
         positive definite: $⟨v,v⟩>0$ for $v≠0$
      2. $\tr(\overline{u+v}w)=\tr(\overline uw+\overline vw)=\tr(\overline uw)+\tr(\overline vw),\tr(\overline{\barλv}w)=λ\tr(\bar vw)=\tr(\bar vλw)$
         $\overline{\tr(\bar vw)}=\tr(\overline{\bar vw})=\tr(\bar wv)$
         $\tr(\bar vv)=\sum_{i,j=1}^n\bar v_{ij}v_{ij}=\sum_{i,j=1}^n\left|v_{ij}\right|^2>0$ for $v≠0$.
      3. Suppose $y∈X^⟂$, we'll show $y_{ij}=0$ for all $i≤j$.
         Let $x∈ M_n(ℂ)$ with $x_{ij}=\bar y_{ij}$ and all other entries 0, then$$x∈X⇒\tr(\bar xy)=\left|y_{ij}\right|^2=0⇒y_{ij}=0$$Conversely, if $y_{ij}=0$ for all $i≤j$, then $∀x∈X:$ \[\tr(\bar xy)=\sum_{i,j=0}^n\bar x_{ij}y_{ij}=\sum_{i≤j}\bar x_{ij}0+\sum_{i>j}0y_{ij}=0\]Hence $X^⊥$ is the space of strictly lower-triangular matrices.
   2. 1. $α^*$ is uniquely defined by $⟨v,α(w)⟩=⟨α^*(v),w⟩$
         Proof. Let $\tilde α$ be another map satisfying $⟨v,α(w)⟩=⟨\tilde α(v),w⟩$. Then for all $v,w∈V$ \begin{aligned}\left< α^*(v)-\tilde{α}(v),w\right>&=\left< α^*(v),w\right>-⟨\tilde{α}(v),w⟩\\&=⟨v,α(w)⟩-⟨v,α(w)⟩\\&=0\end{aligned}But $⟨·,·⟩$ is positive definite, hence $α^*(v)=\tilde α(v)∀v∈V$, so $α^*=\tilde α$.

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         $∀v,w∈V:⟨v,α(w)⟩=⟨α^*(v),w⟩=⟨v,(α^*)^*(w)⟩⇒α(w)=(α^*)^*(w)⇒α=α^*$
      2. $∀v∈\imα∩\kerα^*:v=α(x)$ for some $x$ and $α^*(v)=0$\[⟨v,v⟩=⟨v,α(x)⟩=⟨α^*(v),x⟩=0⇒v=0\]
      3. Lemma 1: $V=E⊕F$ where $E=\im(β^*)=\ker(α^*),F=\im(α)=\ker(β)$
         Proof. $∀v∈V,w∈\im(α)^⟂:0=⟨α(v),w⟩=⟨v,α^*(w)⟩⇒α^*(w)=0⇒w∈\ker(α^*)$Conversely $∀v∈V,w∈\ker(α^*):⟨α(v),w⟩=⟨v,α^*(w)⟩=0⇒w∈\im(α)^⟂$
         Therefore $\ker(α^*)=\im(α)^⟂⇒\ker(α^*)=\ker(β)^⊥$
         Similarly $\ker(β)=\im(β^*)^⊥⇒\im(α)=\im(β^*)^⊥$

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         Lemma 2: $\ker(αα^*)=\ker(α^*)$
         Proof. If $αα^*v=0$ then $0=⟨αα^*v,v⟩=⟨α^*v,α^*v⟩$, so $α^*v=0$
         Conversely if $α^*v=0$ then $αα^*v=α0=0$

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         Lemma 3: $αα^*:\im(α)→\im(α)$ is invertible.
         Proof. $\ker(αα^*)∩\im(α)=\ker(α^*)∩\im(α)=\{0\}$ by lemma 2 and (ii).
         By rank-nullity theorem, $αα^*:\im(α)→\im(α)$ is invertible.

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         Lemma 4: $β^*β$ restricted to $\im(β^*)$ is invertible.
         Proof. $\ker(β^*β)∩\im(β^*)=\ker(β)∩\im(β^*)=\{0\}$ by lemma 2 and (ii).
         By rank-nullity theorem, $β^*β:\im(β^*)→\im(β^*)$ is invertible.

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         $γ$ has matrix $\begin{pmatrix}αα^* & 0 \\ 0 & β^*β\end{pmatrix}$ with respect to the decomposition in Lemma 1, where both diagonal entries are invertible (as maps $E→E$ and $F→F$). Thus $γ$ is invertible.