Linear Algebra with Applications, 10th Edition

6.4

1. 1. 

   1. (a) $\parallel \mathbf{z}\parallel =6,\parallel \mathbf{w}\parallel =3,\langle \mathbf{z},\mathbf{w}\rangle =-4+4i,\langle \mathbf{w},\mathbf{z}\rangle =-4-4i$;

   2. (b) $\parallel \mathbf{z}\parallel =4,\parallel \mathbf{w}\parallel =7,\langle \mathbf{z},\mathbf{w}\rangle =-4+10i,\langle \mathbf{w},\mathbf{z}\rangle =-4-10i$

2. 2. 

   1. (a) Hint: Show that

      $${\text{z}}_2^H{\text{z}}_1=0\text{and}{\text{z}}_1^H{\text{z}}_1={\text{z}}_2^H{\text{z}}_2=1$$

   2. (b) Hint: Note for a real inner product space V with an orthonormal basis $\{{\mathbf{\text{u}}}_1,\mathrm{...},{\mathbf{\text{u}}}_n\}$, we have by Theorem 5.5.2 that if

      $$\mathbf{\text{x}}=c_1{\mathbf{\text{u}}}_1+\mathrm{...}+c_n{\mathbf{\text{u}}}_n$$

      then

      $$c_j=\langle {\mathbf{\text{u}}}_j,\mathbf{\text{x}}\rangle =\langle \mathbf{\text{x,}}{\mathbf{\text{u}}}_j\rangle$$

      For the complex case if $\{{\mathbf{\text{u}}}_1,\mathrm{...},{\mathbf{\text{u}}}_n\}$ is an orthonormal basis for a complex inner product space and

      $$\mathrm{z}=c_1{\mathbf{u}}_1+\mathrm{...}+c_n{\mathbf{u}}_n$$

      then

      $$c_j=\langle \text{z,}{\mathbf{\text{u}}}_j\rangle \text{and}\overline{c_j}=\langle {\mathbf{\text{u}}}_j,\text{z}\rangle$$

3. 3. 

   1. (a) $\begin{array}{l}{\mathbf{u}}_1^H\mathbf{z}=4+2i,{\mathbf{z}}^H{\mathbf{u}}_1=4-2i,\\ {\mathbf{u}}_2^H\mathbf{z}=6-5i,{\mathbf{z}}^H{\mathbf{u}}_2=6+5i\end{array}$;

   2. (b) $\parallel \mathbf{z}\parallel =9$

4. 4. (b) and (f) are Hermitian while (b), (c), (e), and (f) are normal.

5. 9.  

   1. (c) Hint: Use the result from Exercise 20(b) of Section 6.1.

6. 10. Hint: Are AB and ABA Hermitian?

7. 12. Hint: $\langle \text{z},\alpha \mathbf{\text{x}}+\beta \mathbf{\text{y}}\rangle =\overline{\langle \alpha \mathbf{\text{x}}+\beta \mathbf{\text{y, z}}\rangle }$

8. 13. Hint: Show first that

   $$\langle z,w\rangle$$