Riemann surface
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TASK. 1.
Compute the singular value decomposition (SVD) of the matrix
$$
B=\left[\begin{array}{ccc}
3 & 1 & 1 \
-1 & 3 & 1
\end{array}\right] .
$$
.
SOLUTION OF THE TASK 2.
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 $z_1, z_2, z_3 \in \mathbb{R}$.
SOLUTION OF THE TASK 3.
TASK. 5.
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 $A^{\dagger}$.
SOLUTION OF THE TASK 5.
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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