I'm trying to find the smallest euclidian norm of the general solution of an algebraic equation $Ax=y,$ where $A$ is an $n\times m$ matrix and the gral solution is:
$x=x_{p}+\alpha_{1}n_{1}+\alpha_{2}n_{2}=\begin{pmatrix}
0\\
-4\\
0\\
0
\end{pmatrix}+\alpha_{1}\begin{pmatrix}
1\\
1\\
-1\\
0
\end{pmatrix}+\alpha_{2}\begin{pmatrix}
0\\
2\\
0\\
-1
\end{pmatrix}$
So, $x$ is undetermined, $\alpha_{1}, \alpha_{2}$ can be any real number and $n_{1},n_{2}$ are null vectors of A.
My first approcah is the following:
$x=\begin{pmatrix}
\alpha_{1}\\
-4+\alpha_{1}+2\alpha_{2}\\
-\alpha_{1}\\
-\alpha_{2}
\end{pmatrix}$
and since $\left\lVert x \right\lVert_{2}=\sqrt{x^{T}x}\Rightarrow\left\lVert x \right\lVert^{2}_{2}=x^{T}x$, thus
$\left\lVert x \right\lVert^{2}_{2}=3\alpha^{2}_{1}+5\alpha^{2}_{2}-8\alpha_{1}-16\alpha_{2}+4\alpha_{1}\alpha_{2}+16$
And I know from the definition of a norm that
-
$\left\lVert x \right\lVert\geq 0$
-
$\left\lVert x_{1}+x_{2} \right\lVert\leq \left\lVert x_{1} \right\lVert+\left\lVert x_{2} \right\lVert$
So, if these contitions hold for $\left\lVert x \right\lVert$ then it should hold for $\left\lVert x \right\lVert^{2}$, but from here I'm not sure how to proceed, you guys have any advice?
update: I tried the Least-norm solution $x_{ln}=A^{T}(AA^{T})^{-1}y$, the issue here is that my matrix A is singular, since $det(A)=0$ and not invetible since $rank(A) \neq n$.
Best Answer
setting the gradient to the zero vector gives system $$ 6 \alpha_1 + 4 \alpha_2 = 8 \; , $$ $$ 4 \alpha_1 + 10 \alpha_2 = 16 \; . $$ Or augmented matrix $$ \left( \begin{array}{cc|c} 6 & 4 & 8 \\ 4 & 10 & 16 \end{array} \right) $$
$$ \left( \begin{array}{cc|c} 3 & 2 & 4 \\ 2 & 5 & 8 \end{array} \right) $$
$$ \left( \begin{array}{cc|c} 1 & -3 & -4 \\ 2 & 5 & 8 \end{array} \right) $$
$$ \left( \begin{array}{cc|c} 1 & -3 & -4 \\ 0 & 11 & 16 \end{array} \right) $$
$$ \left( \begin{array}{cc|c} 1 & -3 & -4 \\ 0 & 1 & \frac{16}{11} \end{array} \right) $$
$$ \left( \begin{array}{cc|c} 1 & 0 & \frac{4}{11} \\ 0 & 1 & \frac{16}{11} \end{array} \right) $$