Problem 1. Let \(U\) be a subspace of a vector space \(V\). Let \(V / U\) denote the quotient space and \(π : V→V / U\) the natural linear transformation. Prove that if \(W\) is a subspace of \(V / U\) then \(π^{- 1}W\) is a subspace of \(V\), and that if \(V\) is finite-dimensional then
\[\dim π^{- 1}W=\dim U + \dim W\]
Proof. Since \(W\) is a subspace of \(V / U\), \(\forall λ, \forall v_1, v_2 \in π^{- 1}W\),
\[π \left({v_1}\right), π (v_2) \in W \Rightarrow π (λ v_1 + v_2)=λ π \left({v_1}\right) + π (v_2) \in W \Rightarrow λ v_1+ v_2 \in π^{- 1}W\]So \(π^{- 1}W\) is a subspace of \(V\).
Image and kernel of \(π |_{π^{- 1}W}\) is \(W\) and \(U\), by rank-nullity theorem, \(\dim π^{- 1}W=\dim U + \dim W\).\(\Box\)
Problem 2. Suppose that \(V\) is a complex vector space of dimension \(n\) and that \(T : V→V\) is a linear transformation. Prove that there exist \(T\)-invariant subspaces \(V_1 \subseteq V_2 \subseteq \ldots \subseteq V_n\) of \(V\) with \(\dim V_i=i\) for each \(i=1, 2, \ldots, n\). Indicate briefly how to deduce a matrix version of this result.
Proof. Induct on \(n\). Since \(\mathbb{C}\) is algebraically closed, \(T\) has an eigenvalue λ. Let \(v\) be a λ-eigenvector, then \(V / \langle v \rangle\) is \(T\)-invariant and \(n - 1\) dimensional.
\[\dim V_i=\dim W_{i - 1}+ \dim \langle v \rangle=(i - 1) + 1=i\]
By induction hypothesis, there exist \(T\)-invariant subspaces \(W_1 \subseteq W_2 \subseteq \ldots \subseteq W_{n - 1}\) of \(\)\(V / \langle v \rangle\). Using Problem 1 for \(i=2, \ldots, n\) define \(V_i=π^{- 1}W_{i - 1}\), thenLet \(V_1=\langle v \rangle\), then \(\dim V_1=1\) and \(V_1 \subseteq V_2=π^{- 1}W_1\).\(\Box\)
A matrix version of this result: Wikipedia
The above argument can be slightly restated as follows: let λ be an eigenvalue of A, corresponding to some eigenspace Vλ. A induces an operator T on the quotient space Cn/Vλ. This operator is precisely the A22 submatrix from above. As before, T would have an eigenspace, say Wμ ⊂ Cn modulo Vλ. Notice the preimage of Wμ under the quotient map is an invariant subspace of A that contains Vλ. Continue this way until the resulting quotient space has dimension 0. Then the successive preimages of the eigenspaces found at each step form a flag that A stabilizes.
Problem 3. Suppose that \(A=(a_{ij})\) is a complex \(n \times n\) matrix with (not necessarily distinct) eigenvalues \(λ_1, λ_2, \ldots, λ_n\) Prove that
\[\sum_{i=1}^n \sum_{j=1}^n a_{ij}a_{ji}=\sum_{k=1}^n λ_k^2 \]
Proof. Because the eigenvalues of \(A^2\) are \(λ_k^2\),
\[\sum_{i=1}^n \sum_{j=1}^n a_{ij}a_{ji}=\operatorname{Tr}A^2=\sum_{k=1}^n λ_k^2 \]Be careful: the eigenvalues of \(A^{\dagger}A\) are not \(\)\(| λ_k |^2\),
\[\operatorname{Tr}A^{\dagger}A{\color{red}{\neq}}\sum_{k=1}^n | λ_k|^2\]See this thread\(\Box\)
Problem 1. Given a finite-dimensional real inner product space \(V\) with inner product ⟨,⟩ and a transformation \(T : V→V\), show that there is a unique linear transformation \(T^*: V→V\) satisfying
\[⟨T^*u, v⟩=⟨u, Tv⟩\quad \text{ for all }u, v \in V\]
Proof. (Uniqueness) See Lemma 8.14: Let \(\tilde{T}\) be another map satisfying (*). Then for all \(v, w \in V\)
\begin{align*}⟨T^*(v) - \tilde{T}(v), w⟩&=⟨T^*(v), w⟩-⟨\tilde{T}(v), w⟩\\ &=⟨v, T (w)⟩-⟨v, T (w)⟩\\ &=0.\end{align*}But ⟨,⟩ is non-degenerate and hence for all \(v \in V\)
\[ T^*(v) - \tilde{T}(v)=0,\]and so \(T^*=\tilde{T}\).
(Existence) See Theorem 8.15: Let \(v \in V\) and consider the map \(V→\mathbb{R}\) given by
\[ w \mapsto⟨v, T (w)⟩.\]Then \(⟨v, T (·)⟩\) is a linear functional as \(T\) is linear and as ⟨,⟩ is linear in the second coordinate. As \(V\) is finite dimensional, \(\phi : V→V'\) given by \(\phi (u)=⟨u,·⟩\) is an \(\mathbb{R}\)-linear isomorphism, and in particular a surjective map. Thus there exists \(u \in V\) such that
\[⟨v, T (·)⟩=⟨u,·⟩\]Defining \(T^*(v) :=u\) we therefore have
\[⟨v, T (·)⟩=⟨T^*(v),·⟩\text{, i.e., }⟨v, T (w)⟩=⟨T^*(v), w⟩\text{ for all }w \in V.\]To see that \(T^*\) is linear, note that for all \(v_1, v_2, w \in V, λ\in ℝ\),
\begin{align*}⟨T^*(v_1 + λv_2), w⟩&=⟨v_1 + λv_2, T (w)⟩\\ &=⟨v_1, T (w)⟩+ λ⟨v_2, T (w)⟩\\ &=⟨T^*(v_1), w⟩+ λ⟨T^*(v_2), w⟩\\ &=⟨T^*(v_1) + λT^*(v_2), w⟩.\end{align*}(These equalities have nothing to do with our actual definition of \(T^*\), but just follow from the fact that by construction it satisfies \(⟨v, T (w)⟩=⟨T^*(v), w⟩\) for all \(v, w \in V\).) As ⟨,⟩ is non-degenerate (equivalently, as \(\phi\) is injective)
\[ T^*(v_1 + λv_2)=T^*(v_1) + λT^*(v_2) .\] \(\Box\)
Problem 2. Now let \(V\) be the vector space of all \(n \times n\) real matrices with the usual addition and scalar multiplication. For \(A, B\) in \(V\), let
\[⟨A, B⟩=\operatorname{Tr}(A^t B),\]
where \(A^t\) denotes the transpose of \(A\). Show that this defines an inner product on \(V\). Let \(P\) be an invertible \(n \times n\) matrix and let \(θ: V→V\) be the linear transformation given by \(θ(A)=P^{-1}AP\). Find the adjoint \(θ^*\) of \(θ\).
Proof. Easy to show ⟨,⟩ is bilinear and symmetric.
\[⟨A, A⟩=\operatorname{Tr}(A^t A)=\sum_{i, j}A_{i, j}^2\geqslant 0 \text{ and }=0 \text{ iff }A=0\]So ⟨,⟩ is positive definite. So ⟨,⟩ is an inner product on \(V\).
\begin{align*}⟨θ^*A, B⟩&=⟨A,θB⟩\\ &=\operatorname{Tr}(A^t (P^{-1}BP))\\{\small \text{By cyclic property of Tr}}&=\operatorname{Tr}((PA^t P^{-1}) B)\\ &=\operatorname{Tr}(((P^t)^{-1}AP^t)^t B)\\ &=⟨(P^t)^{-1}AP^t, B⟩\end{align*}So \(θ^*(A)=(P^t)^{-1}AP^t\).\(\Box\)
Problem 3. Prove also that \(θ\) is self-adjoint if and only if \(P\) is either symmetric or skew-symmetric.
Proof. If \(P\) is symmetric, \(θ^*(A)=P^{-1}AP=(P^t)^{-1}AP^t=θ(A)\); if \(P\) is skew-symmetric, \(θ^*(A)=P^{-1}AP=(- P^t)^{-1}A (- P^t)=(P^t)^{-1}AP^t=θ(A)\).
\(θ^*=θ⇒∀A \in V : (P^t)^{-1}AP^t=P^{-1}AP⇒∀A \in V : A (P^t P^{-1})=(P^t P^{-1}) A\)
Since the center of \(\operatorname{GL}(n, \mathbb{R})\) is scalar matrices, \(P^t P^{-1}=λI⇒P^t=λP\) for some \(λ\in \mathbb{R}\). Taking det,
\[ λ^n=\det (λI)=\det (P^t P^{-1})=\det (P^t) \det (P^{-1})=\det (P) \det (P^{-1})=1\]So \(λ=\pm 1\). \(P\) is either symmetric or skew-symmetric.\(\Box\)