Homework 11

Let \(f\) be a normal operator on a finite dimensional complex inner product space.
(a) Prove that \(\ker f = \ker f^*\).
(b) Show that \(f - \lambda id\) is a normal operator for each \(\lambda \in \mathbb{C}\).

(a) Take \(v \in \text{ker}f\): \(f(v) = 0 \iff v\) is an eigenvector corresponding to eigenvalue \(0\).
Since \(f\) is normal, then by theorem \(\iff v\) is an eigenvector corresponding to eigenvalue \(\bar{0}=0\) for \(f^* \iff f^*(v) = 0 \cdot v = 0\). Thus \(\text{ker}f = \text{ker}f^*\).

(b) \((f - \lambda I d)^* = f^* - \bar{\lambda} I d\)
Then \((f - \lambda I d)^* \circ (f - \lambda I d)(v) = (f^* - \bar{\lambda} I d) \circ (f - \lambda I d)(v)\)
\(= (f^* - \bar{\lambda} I d)(f(v) - \lambda v)\)

Since \(f\) is normal, then \(\exists B = \{v_1, \dots, v_n\}\): \(f(v_i) = \lambda_i v_i\)
Then \(v = a_1 v_1 + \dots + a_n v_n\), then \(= (f^* - \bar{\lambda} I d)(a_1 \lambda_1 v_1 + \dots + a_n \lambda_n v_n - \lambda (a_1 v_1 + \dots + a_n v_n))\)
\(= (f^* - \bar{\lambda} I d)(a_1 (\lambda_1 - \lambda) v_1 + \dots + a_n (\lambda_n - \lambda) v_n)\)
\(= a_1 (\bar{\lambda}_1 - \bar{\lambda})(\lambda_1 - \lambda) v_1 + \dots + a_n (\bar{\lambda}_n - \bar{\lambda})(\lambda_n - \lambda) v_n\)

\((f-\lambda Id) \circ (f-\lambda Id)^*(v) = (f-\lambda Id) (f^*(v)-\bar{\lambda}v) = (f-\lambda Id)\)\((\alpha_{1} \lambda_{1} v_{1} + \dots + \alpha_{n} \lambda_{n} v_{n} - \lambda ( \alpha_{1} v_{1} + \dots + \alpha_{n} v_{n}))\)
\(= (f-\lambda Id) (\alpha_{1} (\bar{\lambda_1} -\bar{\lambda}) v_{1} + \dots + \alpha_{n} (\bar{\lambda_n}-\bar{\lambda}) v_{n})\)
\(= \alpha_1 (\lambda_1-\lambda) (\bar{\lambda_1}-\bar{\lambda}) v_1 + \dots + \alpha_n (\lambda_n-\lambda) (\bar{\lambda_n}-\bar{\lambda}) v_n\)
\(\text{Thus } (f-\lambda Id)^* \circ (f-\lambda Id) (v) = (f-\lambda Id) \circ (f-\lambda Id)^*(v)\)
\(\text{Thus } f-\lambda Id \text{ is normal}\)
Let \(f\) be a linear operator on a finite dimensional inner product space \(V\). We say that \(f\) is positive-semidefinite if \(\langle f(v), v \rangle \geq 0\) for each \(v\) in \(V\). Show that:

(a) If \(f\) is positive semidefinite, then \(f\) is a normal operator.

(b) \(f\) is positive semidefinite if and only if its eigenvalues are non-negative real numbers.

(a) We know \(\langle f(v), v \rangle \geq 0\), then \(\langle f(v), v \rangle\) is real. Then by theorem in tutorial, \(f\) is self-adjoint.
Then \(f^* = f\), then \(f \cdot f^* = f \cdot f\). Then \(f\) is normal.

(2) \(\Rightarrow\)) Assume \(f(v)= \lambda v\). Then \(\langle f(v), v\rangle = \langle \lambda v, v\rangle = \lambda ||v||^{2} \ge 0\).
Then \(\lambda \ge 0\).

(\(\Leftarrow\))
- If \(f\) is self-adjoint, then \(\exists B = \{v_{1}, ..., v_{n}\}\) orthonormal and \(v_{i}\) are eigenvector.
  Then take \(v \in V\): \(v = a_{1} v_{1} + ... + a_{n} v_{n}\)
  Then \(\langle f(v), v\rangle = \langle a_{1} \lambda_{1} v_{1} + ... + a_{n} \lambda_{n} v_{n}, a_{1} v_{1} + ... + a_{n} v_{n}\rangle\) \(= \sum_{i=1}^{n} \lambda_{i} ||a_{i} v_{i}||^{2} = \sum_{i=1}^{n} \lambda_{i} || a_{i}||^{2}\).
  Since \(\lambda_i \ge 0\), then \(\langle f(v), v\rangle \ge 0\)
- If \(f\) is not self-adjoint, Counterexample: \([f]=\begin{pmatrix} 1&4\\0&1 \end{pmatrix}\)
Thus the statement holds only when \(f\) is self-adjoint
Let \(f\) be a normal operator on \(V\). Prove that if \(f\) is nilpotent, then \(f=0\).

Assume \(f^n=0\), \(f^i \ne 0\), \(\forall 1 \le i \le n-1\).
Since \(f\) is normal, then \(\exists B = \{v_1, \dots, v_n\}\) orthonormal and \(v_i\) are eigenvector.
Then claim: the only eigenvalue is \(0\). Suppose not, then \(\exists v_i: f(v_i) = \lambda(v_i)\), \(\lambda \ne 0\).
Then \(f^n(v_i) = \lambda^n(v_i) = 0 \Rightarrow \lambda = 0\). Contradiction.
Thus \(f(v_i) = 0\), \(\forall i\). Then \(f(v) = f(a_1 v_1 + \dots + a_n v_n) = a_1 f(v_1) + \dots + a_n f(v_n) = 0\). Thus \(f=0\).
Consider \(\mathbb{R}^3\) with the standard inner product and the orientation determined by the canonical basis. Let \(B = \{(1, 1, 0), (0, 1, 0), (0, 0, 1)\}\) be a basis of \(\mathbb{R}^3\), and let \(f\) be the linear operator on \(\mathbb{R}^3\) whose matrix associated in the basis \(B\) is \(\begin{bmatrix} \frac{3}{5} & 0 & -\frac{4}{5} \\ -\frac{8}{5} & -1 & \frac{4}{5} \\ \frac{4}{5} & 0 & \frac{3}{5} \end{bmatrix}\)

(a) Prove that \(f\) is an orthogonal operator.

(a). \(f(1,1,0) = \frac{3}{5}(1,1,0) - \frac{8}{5}(0,1,0) + \frac{4}{5}(0,0,1) = (\frac{3}{5}, -1, \frac{4}{5})\)
\(f(0,1,0) = -(0,1,0) = (0,-1,0)\)
\(f(0,0,1) = -\frac{4}{5}(1,1,0) + \frac{4}{5}(0,1,0) + \frac{3}{5}(0,0,1) = (-\frac{4}{5}, 0, \frac{3}{5})\)

Let \((1,1,0) = v_1\), \((0,1,0) = v_2\), \((0,0,1) = v_3\)

\(\langle f(v_1), f(v_1) \rangle = 2 = \langle v_1, v_1 \rangle\). \(\langle f(v_{1}),f(v_{3})\rangle=0=\langle v_{1},v_{3}\rangle\)
\(\langle f(v_1), f(v_2) \rangle = 1 = \langle v_1, v_2 \rangle\). \(\langle f(v_2), f(v_3) \rangle = 0 = \langle v_2, v_3 \rangle\)
\(\langle f(v_2), f(v_2) \rangle = 1 = \langle v_2, v_2 \rangle\). \(\langle f(v_3), f(v_3) \rangle = 1 = \langle v_3, v_3 \rangle\)

Thus, \(f\) preserves inner product, then \(f\) is orthogonal.

(b) Classify \(f\), indicating any symmetry subspaces and/or axis and angle of rotation.

We know \(f\) is orthogonal, then if \(f\) has eigenvalues, it is \(\pm 1\)

And from matrix \(f(0,1,0)=-(0,1,0)\), thus \(-1\) is an eigenvalue

\(f(1,0,0)=f((1,1,0)-(0,1,0))=\frac{3}{5}(1,1,0)-\frac{3}{5}(0,1,0)+\frac{4}{5}(0,0,1)=(\frac{3}{5},0,\frac{4}{5})\)

\(f(0,0,1)=-\frac{4}{5}(1,1,0)+\frac{4}{5}(0,1,0)+\frac{3}{5}(0,0,1)=(-\frac{4}{5},0,\frac{3}{5})\)

Then take \(\mathcal{B}'=\{(0,1,0)=z_1,(1,0,0)=z_2,(0,0,1)=z_3\}\)

Then \(f(z_{1})=-z_{1},f(z_2)=\frac{3}{5}z_2+\frac{4}{5}z_3,f(z_3)=-\frac{4}{5}z_2+\frac{3}{5}z_3\)

Thus \([f]_{\mathcal{B}'}=\begin{pmatrix} -1&&\\ &\frac{3}{5}&-\frac{4}{5}\\ &\frac{4}{5}&\frac{3}{5} \end{pmatrix}\), then \(\cos\theta =\frac{3}{5}\) and \(\sin\theta =\frac{4}{5}\)

Then \(f\) is symmetric respect to \(\lang (1,0,0),(0,0,1)\rang\) and a rotation in angle \(\arcsin\left(\frac{4}{5}\right)\approx 53.13\degree\)