Linear Algebra with Applications, 10th Edition

7.4

1. 1. 

   1. (a) $\Vert A{\Vert }_F=\sqrt{2},\Vert A{\Vert }_{\infty }=1,\Vert A{\Vert }_1=1;$

   2. (b) $\Vert A{\Vert }_F=5,\Vert A{\Vert }_{\infty }=5,\Vert A{\Vert }_1=6;$

   3. (c) $\Vert A{\Vert }_F=\Vert A{\Vert }_{\infty }=\Vert A{\Vert }_1=1;$

   4. (d) $\Vert A{\Vert }_F=7,\Vert A{\Vert }_{\infty }=6,\Vert A{\Vert }_1=10;$

   5. (e) $\Vert A{\Vert }_F=9,\Vert A{\Vert }_{\infty }=10,\Vert A{\Vert }_1=12;$

2. 2. 2

3. 3. Hint: $\parallel A{\parallel }_2$ is 1 since A has singular values ${\sigma }_1=1\text{and}{\sigma }_2=0$. To show that the $\parallel A{\parallel }_2=1$ using the definition of the 2-norm, we must show that

   $$\underset{\mathbf{\text{x}}\ne 0}{\text{max}}\frac{\parallel A_{\mathbf{\text{x}}}\parallel }{\parallel \mathbf{\text{x}}\parallel }=\underset{\mathbf{\text{x}}\ne 0}{\text{max}}\frac{\sqrt{x_1^2}}{\sqrt{x_1^2+x_2^2}}=1$$

   To do this, consider the 2 cases: (i) $x_2\ne 0$ and (ii) $x_2=0$.

4. 4. $\Vert I{\Vert }_1=\Vert I{\Vert }_{\infty }=1,\Vert I{\Vert }_F=\sqrt{n};$

5. 5. Hint: If

   $$\left|d_{kk}\right|=\underset{1\le i\le n}{\text{max}}\left(\right|d_{ii}\left|\right)$$

   you can establish the result by either using the definition and showing

   $$\underset{\mathbf{\text{x}}\ne 0}{\text{max}}\frac{\parallel D_{\mathbf{\text{x}}}\parallel }{\parallel \mathbf{\text{x}}\parallel }=\left|d_{kk}\right|$$

   or you can make use of Theorem7.4.3 and show that the largest singular value of $D\text{is}\left|d_{kk}\right|$.

6. 6. 

   1. (a) 10;

   2. (b) $(-1,1,-1{)}^T$

7. 10. 

   1. (a) Hint: If x in ${ℝ}^1$, then x is just a scalar and hence

      $$\parallel \mathbf{\text{x}}{\parallel }_2=\left|x\right|\text{and }\parallel Y\mathbf{\text{x}}{\parallel }_2=\parallel x\mathbf{\text{y}}{\parallel }_2$$

      Alternatively, ...