Homework 10

(20 p.) Consider a finite-dimensional inner product space \(V\) and let \(f\) be a linear operator on \(V\). Show that the range of \(f^*\) is the orthogonal complement of the null space of \(f\). (i.e., show that \(\text{Im}(f^*) = \text{ker}(f)^\perp\))

NTP: \(\text{Range}(f^*) = (\text{Null}(f))^\perp\), \(f^*: V \to V\) and \(f: V \to V\)

\(\subseteq\)). Take \(w \in \text{Range}(f^*): \exists v \in V: f^*(v) = w\). NTP \(w \in (\text{Null}(f))^\perp\)\(\implies \langle w, t \rangle = 0, \forall t \in \text{Null}(f)\)
\(\implies\left<w,t\right>=0,\forall t:f(t)=0\implies\overline{\left<t,w\right>}=0 ,\forall t:f(t)=0\)
\(\overline{\left< t, w \right>}= \overline{\left< t, f^*(v) \right>}= \overline{\left< f(t), v \right>} = \overline{\left< 0, v \right>} = 0, \forall t: f(t)=0. \checkmark\)

\(\supseteq\)) Since \(\text{range } f^* \subseteq \text{Null } f^\perp\), NTP \(\dim (\text{range } f^*) = \dim ((\text{Null } f)^\perp)\).

We know \(\dim V = \dim (\text{Null } f) + \dim ((\text{Null } f)^\perp)\).
\(\dim V = \dim (\text{Null } f) + \dim (\text{range } f) \implies \dim ((\text{Null } f)^\perp) = \dim (\text{range } f)\).

Then we know \([f^{*}]_{\mathcal{B}}= ([f]_{\mathcal{B}})^{*} = \overline{[f]_{\mathcal{B}}}^{t}\). Then \(\dim (\text{Null } f) = \dim (\text{Null } f^*)\) and \(\dim (\text{range } f) = \dim (\text{range } f^*)\). Thus \(\dim ((\text{Null } f)^\perp) = \dim (\text{range } f^*)\).

Thus \(V = \text{Range}(f^*) = (\text{Null}(f))^\perp\).
(30 p.) Find the adjoint of the linear operator \(f: V \to V\) in the following cases.
(a) \(V = \mathbb{C}^{n \times n}\) with the inner product \(\langle A, B \rangle = \mathrm{tr}(AB^*)\), and \(f(A) = P^{-1}AP\) with \(P\) a fixed invertible matrix in \(V\).
(b) \(V = \mathbb{R}_3[x]\) with the inner product \(\langle p, q \rangle = \int_0^1 p(x)q(x) dx\), and \(f(p) = p'\).

(a) \(\left\langle f(A), B \right\rangle = \text{tr}(f(A) \cdot B^*) = \text{tr}(P^{-1} A P B^*) = \text{tr}(A \cdot P B^* P^{-1})\)
\(\left\langle A, f^*(B) \right\rangle = \text{tr}(A \cdot (f^*(B))^*)\)

Then since \(\left\langle f(A), B \right\rangle = \left\langle A, f^*(B) \right\rangle\), then \((f^*(B))^* = P^{-1} B^* P\)

\(\left[ (f^*(B))^* \right]^* = \overline{P^{-1}}^t \cdot B \cdot \overline{P}^t = (P^*)^{-1} \cdot B \cdot P^*\). Thus \(f^*(A) = (P^*)^{-1} \cdot A \cdot P^*\)

(b) \(\langle f(p), q \rangle = \int_{0}^{1} f(p) \cdot q \, dx = \int_{0}^{1} p'( x)q(x) \, dx = p(x)q(x)\Big|_{0}^{1} - \int_{0}^{1} p(x)q'(x)\, dx\)

\(\langle p, f^*(q) \rangle = \int_0^1 p(x)f^*(q)\, dx\). Since \(\langle f(p), q \rangle = \langle p, f^*(q) \rangle\).

Then \(\int_0^1 p(x)(f^*(q) + q'(x))\, dx = p(1)q(1) - p(0)q(0)\)

Let \(f^*(q) + q'(x) = g(x)\). Then we want \(g(x): \int_0^1 p(x)g(x)\, dx = p(1)q(1) - p(0)q(0)\)

Let \(\{1, x, x^2, x^3\}\) be basis of \(\mathbb{R}_3[X]\). Then \(\{Z_1, Z_2, Z_3, Z_4\}\) is orthogonal basis

\(Z_1 = 1\), \(Z_2 = x - \frac{\langle x, 1 \rangle}{||1||^2} \cdot 1 = x - \int_0^1 x \, dx = x - \frac{1}{2}\), \(Z_3 = x^2 - \frac{\langle x^2, 1 \rangle}{||1||^2} \cdot 1 - \frac{\langle x^2, x - \frac{1}{2} \rangle}{||x - \frac{1}{2}||^2} (x - \frac{1}{2}) = x^2 - x + \frac{1}{6}\)

\(Z_4 = x^3 - \frac{\langle x^3, 1 \rangle}{||1||^2} \cdot 1 - \frac{\langle x^3, x - \frac{1}{2} \rangle}{||x - \frac{1}{2}||^2} (x - \frac{1}{2}) - \frac{\langle x^3, x^2 - x + \frac{1}{6} \rangle}{||x^2 - x + \frac{1}{6}||^2} (x^2 - x + \frac{1}{6})\) \(= x^3 - \frac{3}{2}x^2 + \frac{3}{5}x - \frac{1}{20}\)

Then \(\{1, 2x-1, 6x^2-6x+1, 20x^3-30x^2+12x-1\}\) is orthogonal basis

Then \(g(x)=\sum_{i=1}^{4}\frac{\langle g(x),Z^{\prime}_{i}\rangle}{||Z^{\prime}_{i}||^{2}} Z^{\prime}_{i}\), \(||Z^{\prime}_{1}||^{2}=1\), \(||Z'_{2}||^{2}=\frac{1}{3}\), \(||Z'_{3}||^{2}=\frac{1}{5}\), \(||Z'_{4}||^{2}=\frac{1}{7}\)

\(\langle g(x), Z'_{i} \rangle = \int_{0}^{1} Z'_{i}(x)g(x)\, dx = Z'_{i}(1)q(1) - Z'_{i} (0)q(0)\)

Then \(g(x) = (q(1)-q(0)) \cdot 1 + 3(q(1)+q(0))(2x-1) + 5(q(1)-q(0))(6x^2-6x+1) + 7(q(1)+q(0))(20x^3-30x^2+12x-1)\)

Thus \(f^*(q) = -q'(x) + q(1)(140x^3-180x^2+60x-4) + q(0)(140x^3-240x^2+120x-16)\)
(30 p.) Show that the following operators \(f: V \to V\) are self-adjoint.
(a) \(V = \mathbb{R}^{3 \times 3}\), \(\langle A, B \rangle = \text{tr}(B^t A)\), \(f(A) = A + A^t\).
(b) \(V = \mathbb{R}_4[x]\), \(\langle p, q \rangle = \int_0^1 p(x)q(x) dx\), \(f(p)(x) = p(1-x)\).

(a)
\(\langle f(A), B \rangle = tr(B^t \cdot f(A)) = tr(B^t \cdot (A+A^t)) = tr(B^t A + B^t A^t)\)
\(\langle A, f(B) \rangle = tr(f(B)^t \cdot A) = tr((B+B^t)^t \cdot A) = tr((B^t+B) \cdot A) = tr(B^t A + BA)\)
Then since \(tr(B^t A + B^t A^t) = tr(B^t A + BA)\) then \(\langle f(A), B \rangle = \langle A, f(B) \rangle\)
Thus \(f\) is self adjoint

(b)
\(\langle f(p), q \rangle = \int_0^1 f(p)(x)q(x)dx = \int_0^1 p(1-x)q(x)dx\)
\(\langle p, f(q) \rangle = \int_0^1 p(x)f(q)(x)dx = \int_0^1 p(x)q(1-x)dx\)
Let \(1-x=t\). Then \(x=1-t\) and \(dx=-dt\).
When \(x=0, t=1\). When \(x=1, t=0\).
\(= \int_1^0 p(1-t)q(t)(-dt)\) \(= -\int_1^0 p(1-t)q(t)dt = \int_0^1 p(1-t)q(t)dt\)
Thus \(\langle f(p), q \rangle = \langle p, f(q) \rangle\) is self adjoint
(20 p.) Find an isomorphism of inner product spaces between \(\mathbb{R}^3\) equipped with the standard inner product and \(\mathbb{R}_2[x]\) equipped with the inner product defined by \(\langle p, q \rangle = \int_0^1 p(x)q(x)x dx.\)

Let \(\{e_1, e_2, e_3\}\) be an orthonormal basis of \(\mathbb{R}^3\).
Let \(\{1, x, x^2\}\) be a basis of \(\mathbb{R}_2[x]\), then let's find an orthonormal basis.
\(\mathcal{Z}_1 = 1\), \(\|\mathcal{Z}_{1}\|^{2} = \langle \mathcal{Z}_{1}, \mathcal{Z}_{1} \rangle = \int _{0}^{1} x dx = \frac{1}{2}\), then \(\mathcal{Z}_1' = \sqrt{2}\).
\(\mathcal{Z}_2 = x-\frac{2}{3}\), \(\|\mathcal{Z}_{2}\|^{2} = \int_{0}^{1} (x-\frac{2}{3})^{2} x dx = \frac{1}{36}\), then \(\mathcal{Z}_2' = 6x-4\).
\(\mathcal{Z}_3 = x^2-\frac{6}{5}x+\frac{3}{10}\), \(\|\mathcal{Z}_3\|^2 = \int_0^1 (x^2-\frac{6}{5}x+\frac{3}{10})^2 x dx = \frac{1}{600}\), then \(\mathcal{Z}_3' = \sqrt{6}(10x^2-12x+3)\).

Then by the first theorem of algebra, there exists a unique \(T\):
\(T(1,0,0) = \sqrt{2}\), \(T(0,1,0) = 6x-4\), \(T(0,0,1) = \sqrt{6}(10x^2-12x+3)\)

Then \(T(x,y,z)=x\sqrt{2}+y(6x-4)+\sqrt{6}z(10x^2-12x+3)\) which is an isomorphism, since it carries an orthonormal basis to another orthonormal basis.