Linear Algebra with Applications, 10th Edition

7.7

1. 1. 

   1. (a) $(\sqrt{2},0{)}^T;$

   2. (b) $(1-3\sqrt{2},3\sqrt{2},-\sqrt{2}{)}^T;$

   3. (c) $(1,0{)}^T;$

   4. (d) $(1-\sqrt{2},\sqrt{2}-\sqrt{2}{)}^T$

2. 2. $x_i=\frac{{\displaystyle d_i{b}_i+e_i{b}_{n+i}}}{{\displaystyle d_i^2+e_I^2}},i=1,\dots ,n$

3. 4. 

   1. (a) $\begin{array}{cc}Q=\left[\begin{array}{cc}\frac{1}{2}& \hfill -\frac{1}{6}\\ \frac{1}{2}& \hfill -\frac{1}{2}\\ \frac{1}{2}& \hfill \frac{5}{6}\\ \frac{1}{2}& \hfill -\frac{1}{6}\end{array}\right],& R=\left[\begin{array}{cc}2& 12\\ 0& 6\end{array}\right]\end{array}$

   2. (b) $\mathbf{x}=[\begin{array}{cc}0& \frac{1}{3}\end{array}{]}^T$

4. 5. Hint: $A^{+}$ will satisfy the four Penrose conditions if and only if ${\mathrm{\Sigma }}^{+}$ satisfies the four Penrose conditions.

5. 7. Hint: If $X=\frac{1}{{\parallel \mathbf{\text{x}}\parallel }_2^2}{\mathbf{\text{x}}}^T$, then

   $$X_{\mathbf{\text{x}}}=\frac{1}{{\parallel \mathbf{\text{x}}\parallel }_2^2}{\mathbf{\text{x}}}^T\mathbf{\text{x}}=1$$

   Using this it is easy to verify that x and X satisfy the four Penrose conditions.

6. 12. $A^{+}=\left[\begin{array}{ccc}\frac{1}{4}& \frac{1}{4}& 0\\ \frac{1}{4}& \frac{1}{4}& 0\end{array}\right]$

7. 13. Hint: Show first that the three conditions hold for the case $A=\mathrm{\Sigma }$

8. 15. $\parallel A_1-A_2{\parallel }_F=\rho ,\parallel A_1^{+}-A_2^{+}{\parallel }_F=1/\rho$. As $\rho \to 0$, $\parallel A_1-A_2{\parallel }_F\to 0$ and $\parallel A_1^{+}-A_2^{+}{\parallel }_F\to \infty$.