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5.29 Adjoints and Unitary

Linear functionals and adjoints \(F: V \rightarrow \mathbb{F}\)

Theorem: \(V\) f.d. i.p.s. and \(F\) is a l.f. Then there exists a unique vector \(w\) in \(V\) s.t. \(F(v) = \langle v, w \rangle\).

Def: \(V\) f.d.i.p.s. and \(f: V \rightarrow V\) l.op. \(\exists!\) \(f^*: V \rightarrow V\) l.t. s.t. \(\langle f(v), w \rangle = \langle v, f^*(w) \rangle \quad \forall v, w \in V\).

Remark: Let \(B = \{v_1, \ldots, v_n\}\) be an ON basis for \(V\). \(f: V \rightarrow V\) a l.t.

\(f^*(w) = \sum_{i=1}^{n} \langle f^*(w), v_i \rangle v_i\) \(= \sum_{i=1}^{n} \overline{\langle v_i, f^*(w) \rangle} v_i\) \(= \sum_{i=1}^{n} \overline{\langle f(v_i), w \rangle} v_i\)
Since \(f^*(w) = \sum_{i=1}^{n} \overline{\langle f(v_i), w \rangle} v_i\), then \(([f^*]_B)_{ij} = \overline{\langle f(v_i), v_j \rangle}\) \(= (\overline{[f]_B})_{ji}\)

Thus \([f^*]_B = \overline{[f]_B}^t \quad \text{ON basis!}\)

Exercise: \(V = \mathbb{R}_3[x]\), \(\langle p, q \rangle = \int_{-1}^{1} p(x) q(x) dx\). \(f: V \rightarrow V\), \(p \mapsto f(p) = x \cdot p'\). Find \([f^*]_B\) with \(B\) is given by G-S and \(E = \{1, x, x^2, x^3\}\).

If \(E = \{1, x, x^2, x^3\} \xrightarrow{\text{Ex! G-S}} B = \left\{\frac{\sqrt{2}}{2}, \frac{\sqrt{6}}{2} x, \frac{3\sqrt{10}}{4} (x^2 - 1/3), \frac{5\sqrt{14}}{4} (x^3 - \frac{3}{5}x)\right\}\)
\(v_{1}\quad v_{2}\quad v_{3}\quad\quad\quad\quad\quad\quad\quad\quad v_{4}\)

Let compute \([f]_B\): \(f(\frac{\sqrt{2}}{2}) = x \cdot 0 = 0 = 0v_1 + 0v_2 + 0v_3 + 0v_4\)
\(f(\frac{\sqrt{6}}{2} x) = \frac{\sqrt{6}}{2} x = 0v_1 + 1v_2 + 0v_3 + 0v_4\)
\(f(\frac{3\sqrt{10}}{4}(x^{2} - 1/3)) = 2 \left(\frac{3\sqrt{10}}{4}x^{2}\right)=\sqrt5v_1+0v_2+2v_3+0v_4\)
\(f(\frac{5\sqrt{14}}{4}(x^{3}-\frac{3}{5}x))=\frac{5\sqrt{14}}{4}(3x^{3}-\frac{3}{5} x)=3v_{1}+\sqrt{21}v_{2}\)

Then \([f]_{B}= \begin{pmatrix} 0 & 0 & \sqrt{5} & 0 \\ 0 & 1 & 0 & \sqrt{21} \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \end{pmatrix}\), then \([f^{*}]_{B} = \overline{[f]_B}^{t} = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ \sqrt{5} & 0 & 2 & 0 \\ 0 & 3 & 0 & 3 \end{pmatrix}\)

Ex: What if \(V\) is not a f.d. v.s.?
Consider \(V = C[0,1]\) with \(\langle f, g \rangle = \int_{0}^{1} f(x)g(x) dx\). And \(T: V \rightarrow V\) where \(f \mapsto T(f) = \frac{df}{dx}\)

\(\langle T(f), g \rangle = \langle f', g \rangle = \int_0^1 f'(x)g(x) \, dx = \int_0^1 (f(x)g(x))' - f(x)g'(x) \, dx\) \(= f(1)g(1) - f(0)g(0) - \int_{0}^{1} f(x)g'(x) \, dx=f(1)g(1) - f(0)g(0) -\langle f, g' \rangle\)
\(\therefore\langle f^{\prime},g\rangle=(f(1)g(1)-f(0)g(0))-\langle f,g^{\prime}\rangle =\langle f,T^{*}(g)\rangle\quad\forall f,g\in C[0,1]\)

If \(\exists T^*\) then \(\langle f, T^*(g) \rangle = f(1)g(1) - f(0)g(0) - \langle f, g' \rangle\).
Then \(\langle f,T^{*}(g)+g^{\prime}\rangle=f(1)g(1)-f(0)g(0)\quad\forall f,g\in C[0,1]\).

For a given g: \(F_g(f) = f(1)g(1) - f(0)g(0)\) is a l.f. (linear functional)
If \(\exists h \in C[0,1]\) s.t. \(F_g(f) = \langle f, h \rangle \forall f\). then \(\int_0^1 f(x)h(x) dx = f(1)g(1) - f(0)g(0) \quad \forall f\).

If \(f(x)=1 \quad \forall x \in [0,1]\), then \(\int_0^1 h(x) dx = g(1) - g(0)\).

If \(f\) is s.t. \(f(1) = f(0) = 0\), \(\int_0^1 h(x)f(x) dx = 0\). \((\equiv f \perp h)\).

In particular if \(f(x) = x(x-1)t(x)\) with \(t \in C[0,1]\), then \(\langle f, h \rangle = 0=\int_{0}^{1} h(x)x(x-1)t(x) \, dx\)

Let \(t(x) = x(x-1)h(x)\), then \(0 = \int_0^1 (h(x))^2 x^2(x-1)^2 \, dx\)
​\(\therefore h^2(x) = 0 \implies h(x)=0 \quad \forall x \in (0,1)\). \(h\) is cont\(\implies h(x)=0 \quad \forall x \in [0,1]\)

\(\Rightarrow F_g(f) = \langle f, h \rangle = \langle f, 0 \rangle = 0. \quad \forall f \in C[0,1]\).

But if \(g(x)=1\) and \(f(x)=x\) then \(F_g(f) = f(1)g(1) - f(0)g(0)\) \(= 1 \neq 0\)
Thus no adjoint!

Def: \(V\) f.d.i.p.s. and \(f: V \to V\) a l.o. (linear operator)

\(f\) is self-adjoint if \(f = f^*\) (or Hermitian)\(=\) \(\langle f(u), v \rangle = \langle u, f(v) \rangle \quad \forall u, v \in V\).

\(A \in \mathbb{R}^{n \times n}\): \(A\) is self-adjoint if \(A^T = A\) (\(A^T = A\))
​\(A \in \mathbb{C}^{n \times n}\): \(A\) is self-adjoint if \(\overline{A^{T}} = A\) (\(A^* = A\))

Prop: \(V\) f.d.i.p.s. \(f\) a l.o. on \(V\)

\(f\) is self-adjoint iff \(\forall\) o.n. basis \(B\) for \(V\), \([f]_B\) is Hermitian iff \(\exists\) o.n. basis \(B\) for \(V\), \([f]_B\) is Hermitian.

\(f\) is self-adjoint \(\implies \chi_f(x)\) has all its roots in \(\mathbb{R}\)​\(\implies f\) is diagonalizable.

Remark: \(A \in \mathbb{C}^{n \times n}\) Hermitian, consider \(\mathbb{C}^n\) with the s.t. inner product and \(f_A: \mathbb{C}^n \to \mathbb{C}^n\) by \(f_A(x) = Ax\).

Thus \(f_A\) is self-adjoint and \([f_{A}]_{E}= A\) \(\Rightarrow \exists\) o.n. basis \(B\) s.t. \([f_A]_B = D \in \mathbb{R}^{n \times n}\)

\(A = C(E,B)^{-1} D C(E,B) = C(E,B) A C(E,B)^{-1} = D\)

\(B = \{v_1, v_2, ..., v_n\}\), \(C(B,E) = [[v_1]_E | [v_2]_E | ... | [v_n]_E]\)

\((C(E,B)^{-1})_{ij} = C(B,E)_{ij} = \langle v_j, e_i \rangle\) \(= \overline{\langle e_i, v_j \rangle}\) \(= \overline{(C(E,B))_{ji}} = (C(E,B)^*)_{ij}\)

\(\therefore C(E,B)^{-1} = C(B,E)^*\)​

We say that a matrix \(U\) is unitary if \(U^{-1}=U^*\) with \(U \in \mathbb{C}^{n \times n}\). Orthogonal if \(U^{-1}=U^t\) with \(U \in \mathbb{R}^{n \times n}\).

Ex 3: Find a matrix \(P\) in \(\mathbb{R}^{3 \times 3}\) s.t. \(A = PDP^t\) where \(D\) is diagonal and \(A = \begin{bmatrix} 5 & -2 & 1 \\ -2 & 5 & -1 \\ 1 & -1 & 8 \end{bmatrix}\)

Remark: \(A=A^t\), thus \(A\) is Hermitian (self-adjoint) and each eigenvalue is in \(\mathbb{R}\) and \(A\) is diag.
\(\Rightarrow \exists B\) orthonormal basis for \(\mathbb{R}^3\) s.t. \([f_A]_B\) is Hermitian.

\(\chi_A(x) = (x-3)(x-6)(x-9)\), then for \(\lambda=3 \Rightarrow E_3 = \text{span } \{e_1+e_2\}\).
\(\lambda=6 \Rightarrow E_6 = \text{span } \{-e_1+e_2+e_3\}\). \(\lambda=9 \Rightarrow E_9 = \text{span } \{e_1-e_2+2e_3\}\).

Then \(A = \begin{bmatrix} 1 & -1 & 1 \\ 1 & 1 & -1 \\ 0 & 1 & 2 \end{bmatrix} \begin{bmatrix} 3 & & \\ & 6 & \\ & & 9 \end{bmatrix} \begin{bmatrix} 1 & -1 & 1 \\ 1 & 1 & -1 \\ 0 & 1 & 2 \end{bmatrix}^{-1}\), but we need ON basis

Let \(B = \left\{ \frac{(1,1,0)}{\sqrt{2}}, \frac{(-1,1,1)}{\sqrt{3}}, \frac{(1,-1,2)}{\sqrt{6}} \right\}\) (ON basis for \(\mathbb{R}^3\)), \(A = \underbrace{\begin{bmatrix} 1/\sqrt{2} & -1/\sqrt{3} & 1/\sqrt{6} \\ 1/\sqrt{2} & 1/\sqrt{3} & -1/\sqrt{6} \\ 0 & 1/\sqrt{3} & 2/\sqrt{6} \end{bmatrix}}_{P} \begin{bmatrix} 3 & & \\ & 6 & \\ & & 9 \end{bmatrix} P^t\)

Unitary Operators. \(V\) f.d.i.p.s. \(\mathbb{F} = \mathbb{C} \text{ or } \mathbb{R}\).

Def: \(f:V \to V\) a l.o. \(f\) is a unitary operator if \(f\) is an isomorphism of inner product spaces. \(||f(v)||=||v||\).

Theorem: \(f\) is unitary iff \(f\) preserves inner products i.e. \(\langle f(v), f(w) \rangle = \langle v, w \rangle\).
iff \(f\) is s.t. \(ff^* = f^*f = id\).
iff \(\forall\) ON basis \(B\), \(f(B)\) is an ON basis for \(V\).
iff \(\exists\) ON basis \(B\), \(f(B)\) is an ON basis for \(V\).
iff \(\forall\) ON basis \(B\), \([f]_B\) is a unitary matrix.
iff \(\exists\) ON basis \(B\), \([f]_B\) is a unitary matrix.

Ex 4: Consider \(\mathbb{R}^4\) with the standard inner product and let \(B=\{v_1, v_2, v_3, v_4\}\) be an orthogonal basis for \(\mathbb{R}^4\) s.t. \(||v_1|| = 2||v_2||\) and \(||v_3|| = 2||v_4||\).

Let \(f:\mathbb{R}^4 \to \mathbb{R}^4\) be a l.op s.t. \(f(v_1)=4v_2\), \(f(v_2)=v_1\), \(f(v_3)=2v_4\) and \(f(v_4)=-\frac{1}{2}v_3\).

a) Find a non-trivial \(f\)-invariant subspace \(S\) for \(\mathbb{R}^4\) s.t. \(f|_S\) is unitary.

Ans: \([f]_B = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 4 & 0 & 0 & 0 \\ 0 & 0 & 0 & 2 \\ 0 & 0 & -1/2 & 0 \end{bmatrix}\).

Let \(U = \text{span}\{v_1, v_2\}\), \(U' = \text{span}\{v_3, v_4\}\). Notice that \(U\) and \(U'\) are \(f\)-invariant.

\(B_U = \left\{ \frac{v_1}{||v_1||}, \frac{v_2}{||v_2||} \right\}\), \(B_{U'} = \left\{ \frac{v_3}{||v_3||}, \frac{v_4}{||v_4||} \right\}\).

\(f\left(\frac{v_1}{||v_1||}\right) = \frac{4v_2}{||v_1||} = \frac{4v_2}{2||v_2||} = \frac{2v_2}{||v_2||}\). \(f\left(\frac{v_2}{||v_2||}\right) = \frac{v_1}{||v_2||} = \frac{2v_1}{||v_1||}\).

\(f\left(\frac{v_3}{||v_3||}\right) = \frac{2v_4}{||v_3||} = \frac{2v_4}{2||v_4||} = \frac{v_4}{||v_4||}\). \(f\left(\frac{v_4}{||v_4||}\right) = -\frac{1}{2} \frac{v_3}{||v_4||} = -\frac{1}{2} \frac{2v_3}{||v_3||} = -\frac{v_3}{||v_3||}\).

\(|| f\left(\frac{v_1}{||v_1||}\right) || = \left|\left| \frac{2v_2}{||v_1||} \right|\right| = 2\). But \(\left|\left| \frac{v_1}{||v_1||} \right|\right| = 1\). Since they are not equal, then \(\left. f \right|_U\) is not unitary!

\(\left. f \right|_{U'}\) is unitary since \(f\left(\left\{ \frac{v_3}{||v_3||}, \frac{v_4}{||v_4||} \right\}\right) = \left\{ \frac{v_4}{||v_4||}, -\frac{v_3}{||v_3||} \right\}\)​\(\leftarrow\) ON basis for \(U'\)!
Define \(S = U'\)

b) Find a non-trivial \(f\)-invariant subspace \(W\) for \(\mathbb{R}^4\) s.t. \(f|_W\) is self-adjoint.

Take \(B_{U}\)(ON basis for \(U\)), \(\left[ f \right]_{B_U} = \begin{bmatrix} 0 & 2 \\ 2 & 0 \end{bmatrix}\) this is Hermitian \(\Rightarrow \left. f \right|_U\) is self-adjoint.

Thus let \(W = U\)