91影视

First note that $\mathrm{n}鈮�\mathrm{m}$since rank$\left(\mathrm{A}\right)=\mathrm{m}$.

Let $\stackrel{藱}{\mathrm{x}}*={\mathrm{c}}_1{\stackrel{藱}{\mathrm{v}}}_1+{\mathrm{c}}_2{\stackrel{藱}{\mathrm{v}}}_2+\mathrm{L}+{\mathrm{c}}_{\mathrm{m}}{\stackrel{藱}{\mathrm{v}}}_{\mathrm{m}}$be the least square solution of $\mathrm{A}\stackrel{藱}{\mathrm{x}}=\stackrel{藱}{\mathrm{b}}$where $\left\{{\stackrel{藱}{v}}_1,...,{\stackrel{藱}{v}}_n\right\}$is an orthonormal basis of ${\mathrm{鈩�}}^m$.

We shall find the constants $c_k$'s .

Let $\stackrel{藱}{b}=b_1{\stackrel{藱}{u}}_1+...+b_m{\stackrel{藱}{u}}_m+...+b_n{\stackrel{藱}{u}}_n$where$\left\{{\stackrel{藱}{u}}_1,{\stackrel{藱}{u}}_2,...,{\stackrel{藱}{u}}_m,...,{\stackrel{藱}{u}}_n\right\}$ is an orthonormal basis of ${\mathrm{鈩�}}^n$.

Recall that $A{\stackrel{藱}{v}}_k={蟽}_k{\stackrel{藱}{u}}_k$by the property of the singular value decomposition.

localid="1660731980721" $$\begin{gathered} 鈭�A\stackrel{藱}{x}*-\stackrel{藱}{b}鈭�=鈭�A\left(c_1{\stackrel{藱}{v}}_1+c_2{\stackrel{藱}{v}}_2+...+c_m{\stackrel{藱}{v}}_m\right)-\left(b_1{\stackrel{藱}{u}}_1+b_2{\stackrel{藱}{u}}_2+...+b_m{\stackrel{藱}{u}}_m+...b_n{\stackrel{藱}{u}}_n\right)鈭� \\ =鈭�\left(c_{1}A{\stackrel{藱}{v}}_1+c_{2}A{\stackrel{藱}{v}}_2+...+c_{m}A{\stackrel{藱}{v}}_m\right)-\left(b_1{\stackrel{藱}{u}}_1+b_2{\stackrel{藱}{u}}_2+...+b_m{\stackrel{藱}{u}}_m+...b_n{\stackrel{藱}{u}}_n\right)鈭� \\ =鈭�\left(c_1{蟽}_1\stackrel{藱}{u}+c_2{蟽}_2\stackrel{藱}{u}+...+c_m{蟽}_m{\stackrel{藱}{u}}_m\right)-\left(b_1{\stackrel{藱}{u}}_1+b_2{\stackrel{藱}{u}}_2+...+b_m{\stackrel{藱}{u}}_m+...b_n{\stackrel{藱}{u}}_n\right)鈭� \\ =鈭�\left(\left(c_1{蟽}_1-b\right){\stackrel{藱}{u}}_1+\left(c_2{蟽}_2-b_2\right)\stackrel{藱}{u}+...+\left(c_m{蟽}_m-b_m\right){\stackrel{藱}{u}}_m-b_{m+1}{\stackrel{藱}{u}}_{m+1}-...-b_n{\stackrel{藱}{u}}_n\right)鈭� \\ =\sqrt{{\left(c_1{蟽}_1-b_1\right)}^2+{\left(c_2{蟽}_2-b_2\right)}^2+...+{\left(c_m{蟽}_m-b_m\right)}^2+b_{m+1}^2+...+b_n^2} \end{gathered}$$

This quantity is minimum if and only if $c_k{蟽}_k-b_k=0$for $1鈮�k鈮�m$if and

only if $c_k=\frac{b_k}{{蟽}_k}$for$a鈮�k鈮�m$

Note that $b_k=\stackrel{藱}{b}路{\stackrel{藱}{u}}_k$since $\left\{{\stackrel{藱}{u}}_1,{\stackrel{藱}{u}}_2,...,{\stackrel{藱}{u}}_m,...,{\stackrel{藱}{u}}_n\right\}$is an orthonormal basis of ${\mathrm{鈩�}}^n$

Thus, $鈭�A\stackrel{藱}{x}*-\stackrel{藱}{b}鈭�$is minimum if and only if $c_k=\frac{\stackrel{藱}{b}路{\stackrel{藱}{u}}_k}{{蟽}_k}$.

Hence, if $\stackrel{藱}{x}*$is the least square solution of $A\stackrel{藱}{x}=\stackrel{藱}{b}$, then

$\stackrel{藱}{x}*=\frac{\stackrel{藱}{b}路{\stackrel{藱}{u}}_1}{{蟽}_1}{\stackrel{藱}{v}}_1+\frac{\stackrel{藱}{b}路{\stackrel{藱}{u}}_2}{{蟽}_2}{\stackrel{藱}{v}}_2+...+\frac{\stackrel{藱}{b}路{\stackrel{藱}{u}}_m}{{蟽}_m}{\stackrel{藱}{v}}_m$

Thus the Set $\stackrel{藱}{\mathrm{x}}*={\mathrm{c}}_1{\stackrel{藱}{\mathrm{v}}}_1+{\mathrm{c}}_2{\stackrel{藱}{\mathrm{v}}}_2+\mathrm{L}+{\mathrm{c}}_{\mathrm{m}}{\stackrel{藱}{\mathrm{v}}}_{\mathrm{m}}$and find the values of $c_k\text{'}S$for which $鈭�A\stackrel{藱}{x}*-\stackrel{藱}{b}鈭�$is minimized.