Question

Find A+ and A+A and AA+ and x+ (shortest length least square solution) for this matrix A UVT (the SVD b: s given below)and these 48 .60

Answer 1

From the singular value decomposition (SVD) of a matrix $A$, we can calculate $A^{\dagger}$ as follows:

Let $A=U\Sigma V^{T}$ be SVD of the matrix $A$, then $A^{\dagger}=V\Sigma^{\dagger} U^{T}$ ,where $\Sigma^{\dagger}$ is calculated by replacing every nonzero entry by its reciprocal and transposing the resulting matrix.

Here,

$A=\begin{bmatrix} 3\\4\end{bmatrix}=\begin{bmatrix} .6&-.8\\.8&.6\end{bmatrix}\begin{bmatrix}5\\0 \end{bmatrix}\begin{bmatrix}1 \end{bmatrix}=U\Sigma V^{T}$, where $U=\begin{bmatrix} .6&-.8\\.8&.6\end{bmatrix},~\Sigma=\begin{bmatrix}5\\0 \end{bmatrix},~ V^{T}=\begin{bmatrix}1 \end{bmatrix}$ .

So,

$U^{T}=\begin{bmatrix} .6&.8\\-.8&.6\end{bmatrix},~\Sigma^{\dagger}=\begin{bmatrix}1/5\\0 \end{bmatrix}^{T}=\begin{bmatrix} 1/5&0\end{bmatrix},~ V=\begin{bmatrix}1 \end{bmatrix}$.

Thus,

$A^{\dagger}=V\Sigma^{\dagger}U^{T}\\ \Rightarrow A^{\dagger}=\begin{bmatrix}1 \end{bmatrix}\begin{bmatrix} 1/5&0\end{bmatrix}\begin{bmatrix} .6&.8\\-.8&.6\end{bmatrix}\\ \Rightarrow A^{\dagger}=\begin{bmatrix}1 \end{bmatrix}\begin{bmatrix} 0.2&0\end{bmatrix}\begin{bmatrix} .6&.8\\-.8&.6\end{bmatrix}\\ \Rightarrow A^{\dagger}=\begin{bmatrix}0.12&0.16 \end{bmatrix}$

Now,

$AA^{\dagger}=\begin{bmatrix}3\\4\end{bmatrix} \begin{bmatrix}0.12&0.16 \end{bmatrix}\\ \Rightarrow AA^{\dagger}=\begin{bmatrix}0.36&0.48\\0.48&0.64 \end{bmatrix}$

and

$A^{\dagger}A=\begin{bmatrix}0.12&0.16 \end{bmatrix}\begin{bmatrix}3\\4\end{bmatrix} \\ \Rightarrow A^{\dagger}A=\begin{bmatrix}0.36+0.64\end{bmatrix}\\ \Rightarrow A^{\dagger}A=\begin{bmatrix}1\end{bmatrix}$.

Shortest length least square solution to the system $Ax=b$ is $x=A^{\dagger}b$.

Here,

for $A=\begin{bmatrix}3\\4 \end{bmatrix}$ ,$A^{\dagger}=\begin{bmatrix}0.12&0.16 \end{bmatrix}$ and $b_{1}=\begin{bmatrix}3\\4 \end{bmatrix}$ , we have shortest length least square solution to the system $Ax=b_{1}$ is

$x=A^{\dagger}b_{1}\\ \Rightarrow x=\begin{bmatrix}0.12&0.16\end{bmatrix}\begin{bmatrix}3\\4 \end{bmatrix}\\ \Rightarrow x=\begin{bmatrix} 0.36+0.64\end{bmatrix}\\ \Rightarrow x=\begin{bmatrix} 1\end{bmatrix}$

So, $x^{\dagger}=\begin{bmatrix} 1\end{bmatrix}$ .

And for $A=\begin{bmatrix}3\\4 \end{bmatrix}$ , $A^{\dagger}=\begin{bmatrix}0.12&0.16 \end{bmatrix}$ and $b_{2}=\begin{bmatrix} -4\\3\end{bmatrix}$ , we have shortest length least square solution to the system $Ax=b_{2}$ is

$x=A^{\dagger}b_{2}\\ \Rightarrow x=\begin{bmatrix}0.12&0.16\end{bmatrix}\begin{bmatrix}-4\\3 \end{bmatrix}\\ \Rightarrow x=\begin{bmatrix} -0.48+0.48\end{bmatrix}\\ \Rightarrow x=\begin{bmatrix} 0\end{bmatrix}$

So, $x^{\dagger}=\begin{bmatrix} 0\end{bmatrix}$ .