Riemann surface
matlab

Compute the singular value decomposition (SVD) of the matrix
$$B=\left[\begin{array}{ccc} 3 & 1 & 1 \ -1 & 3 & 1 \end{array}\right] .$$

.

a. The left upper $2 \times 2$ submatrix has determinant -1 and hence the first two columns are linearly independent. The third column is the sum of the first two and so $\operatorname{rank} A=2$.
b. We choose an invertible $2 \times 2$ submatrix of $A$, replace it with its transposed inverse, replace the other entries with $0 \mathrm{~s}$ and transpose the matrix obtained:
c. The candidate for the solution is
$$G b=\left[\begin{array}{c} 1 \ -1 \ 0 \end{array}\right] .$$
We can check that $A(G b)=b$ and hence $G b$ is really a solution of the system. All solutions are of the form
\begin{aligned} G b+(G A-I) z & =\left[\begin{array}{c} 1 \ -1 \ 0 \end{array}\right]+\left[\begin{array}{ccc} 0 & 0 & 1 \ 0 & 0 & 1 \ 0 & 0 & -1 \end{array}\right]\left[\begin{array}{c} z_1 \ z_2 \ z_3 \end{array}\right] \ & =\left[\begin{array}{c} 1+z_3 \ -1+z_3 \ -z_3 \end{array}\right], \end{aligned}
where $z_1, z_2, z_3 \in \mathbb{R}$.

Let
$$A=\left[\begin{array}{cc} 1 & -1 \ 1 & 1 \ -1 & 1 \end{array}\right] \quad \text { and } \quad b=\left[\begin{array}{c} 1 \ -1 \ 1 \end{array}\right]$$
a. Find the Moore-Penrose inverse $A^{\dagger}$ of the matrix $A$.
b. Does the system $A x=b$ have a solution?
c. If the system is solvable, find the solution closest to the origin. If the system does not have a solution, find the vector $x^{+}$such that the error $\left|A x^{+}-b\right|_2$ is the smallest possible.
d. Find the Moore-Penrose inverse $\left(A^{\dagger}\right)^{\dagger}$ of $A^{\dagger}$.

ن
a. Since rank $A=2$, the matrix $A^T A \in \mathbb{R}^{2 \times s}$ is invertible and hence $A^{\dagger}=\left(A^T A\right)^{-1} A^T$. So,
$$A^{\dagger}=\left[\begin{array}{cc} 3 & -1 \ -1 & 3 \end{array}\right]^{-1} \cdot A^T=\left[\begin{array}{cc} \frac{3}{8} & \frac{1}{8} \ \frac{1}{8} & \frac{3}{8} \end{array}\right]^{-1} \cdot A^T=\left[\begin{array}{ccc} \frac{1}{4} & \frac{1}{2} & -\frac{1}{4} \ -\frac{1}{4} & \frac{1}{2} & \frac{1}{4} \end{array}\right]$$
b. If the systema $A x=b$ has a solution, then one of the solutions is $x^{+}=A^{\dagger} b$. So, we have to compute $A x^{+}$and see, if the result is $b$.
$$x^{+}=A^{\dagger} b=\left[\begin{array}{c} -\frac{1}{2} \ \frac{1}{2} \end{array}\right] \quad \Rightarrow \quad A x^{+}=\left[\begin{array}{c} 0 \ -1 \ 0 \end{array}\right] \neq b .$$
So the system $A x=b$ does not have a solution.
c. The vector $x^{+}$such that the error $\left|A x^{+}-b\right|_2$ is the smallest possible, is
$$A^{\dagger} b=\left[\begin{array}{c} -\frac{1}{2} \ \frac{1}{2} \end{array}\right]$$
d. The Moore-Penrose inverse of $A^{\dagger}$ is always $A$ by the symmetry in $A$ and $A^{\dagger}$ in the conditions the MP inverse must satisfy.

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