91影视

For a linear system $\mathit{A}\stackrel{\mathbf{鈬赌}}{\mathbf{x}}\mathbf{=}\mathit{b}$, a vector${\stackrel{\mathbf{鈬赌}}{\mathbf{x}}}^{\mathbf{+}}$in localid="1662034053845">鈩�nis considered as least-squares solution of the given linear system if $||\stackrel{鈬赌}{b}-A{\stackrel{鈬赌}{x}}^{+}||\mathbf{鈮�}||\stackrel{鈬赌}{b}-A\stackrel{鈬赌}{x}||$for every$\stackrel{\mathbf{鈬赌}}{\mathbf{x}}$in localid="1662034074963">鈩�m.

(a)

Given that $L\left(\stackrel{鈬赌}{x}\right)=A\stackrel{鈬赌}{x}$is a linear transformation from localid="1662034112751">鈩�n鈫�鈩�m with . Let L⁺ is pseudo inverse of L such that $L^{+}=\left(theleast-aquraessolutionofL\left(\stackrel{鈬赌}{x}\right)=\stackrel{鈬赌}{y}\right)$.

Now, the least-squares solution of$L\left(\stackrel{鈬赌}{x}\right)=\stackrel{鈬赌}{y}$if$ker\left(L\right)=\left\{\stackrel{鈬赌}{0}\right\}$is given by $\stackrel{鈬赌}{x}={\left(A^{T}A\right)}^{-1}{A}^T\stackrel{鈬赌}{y}$. The least-squares solution of$L\left(\stackrel{鈬赌}{x}\right)=\stackrel{鈬赌}{y}$is given as $L^{+}\left(\stackrel{鈬赌}{y}\right)$.

So, we have $L^{+}\left(\stackrel{鈬赌}{y}\right)={\left(A^{T}A\right)}^{-1}{A}^T\stackrel{鈬赌}{y}$.

Now, take $L^{+}\left(a\stackrel{鈬赌}{y}+b\stackrel{鈬赌}{z}\right)$as follows:

$\begin{array}{rcl}{L}^{+}\left(a\stackrel{鈬赌}{y}+b\stackrel{鈬赌}{z}\right)& =& \left(A^{T}A\right)A^T\left(a\stackrel{鈬赌}{y}+b\stackrel{鈬赌}{z}\right)\\ & =& a\left({\left(A^{T}A\right)}^{-1}{A}^T\right)\left(\stackrel{鈬赌}{y}\right)+b\left({\left(A^{T}A\right)}^{-1}{A}^T\right)\left(\stackrel{鈬赌}{z}\right)\\ & =& a{L}^{+}\left(\stackrel{鈬赌}{y}\right)+b{L}^{+}\left(\stackrel{鈬赌}{z}\right)\end{array}$

The above calculation shows that L⁺ is linear.

Now, we have $\begin{array}{rcl}{L}^{+}\left(\stackrel{鈬赌}{y}\right)& =& {\left(A^{T}A\right)}^{-1}{A}^T\stackrel{鈬赌}{y}\\ & =& \left({\left(A^{T}A\right)}^{-1}{A}^T\right)\stackrel{鈬赌}{y}\\ & =& A^{+}\stackrel{鈬赌}{Y}\end{array}$. This can be manipulated as follows:

Thus, the value of A⁺ is ${\left(A^{T}A\right)}^{-1}{A}^T$.

(b)

It is given in the question that L is invertible because $ker\left(L\right)=\left\{\stackrel{鈬赌}{0}\right\}$. Now, L⁺ is also invertible because L is invertible.

The above statement implies that if $L\left(\stackrel{鈬赌}{x}\right)=\stackrel{鈬赌}{y}$then $\stackrel{鈬赌}{x}=L^{-1}\left(\stackrel{鈬赌}{y}\right)$. From the calculation in part (a), we have $L^{+}\left(\stackrel{鈬赌}{y}\right)=\stackrel{鈬赌}{x}$.

Thus, the relationship between L⁺ and L^-1 is L⁺ = L^-1.

(c)

From part (a), we have $L^{+}\left(\stackrel{鈬赌}{y}\right)={\left(A^{T}A\right)}^{-1}{A}^T\stackrel{鈬赌}{y}$. So, the value of $L^{+}\left(L\left(\stackrel{鈬赌}{x}\right)\right)$can be computed as follows:

$\begin{array}{rcl}{L}^{+}\left(L\left(\stackrel{鈬赌}{x}\right)\right)& =& L^{+}\left(A\left(\stackrel{鈬赌}{x}\right)\right)\\ & =& {\left(A^{T}A\right)}^{-1}{A}^T\left(A\left(\stackrel{鈬赌}{x}\right)\right)\\ & =& {\left(A^{T}A\right)}^{-1}\left(A^{T}A\right)\left(\stackrel{鈬赌}{x}\right)\\ & =& \stackrel{鈬赌}{x}\end{array}$

Thus, the value of $\begin{array}{rcl}{L}^{+}\left(L\left(\stackrel{鈬赌}{x}\right)\right)& =& \stackrel{鈬赌}{x}\end{array}$.

(d)

From part (a), we have $L^{+}\left(\stackrel{鈬赌}{y}\right)={\left(A^{T}A\right)}^{-1}{A}^T\stackrel{鈬赌}{y}$. The value of$L\left(L^{+}\left(\stackrel{鈬赌}{y}\right)\right)$ can be obtained as follows:

$\begin{array}{rcl}L\left(L^{+}\left(y\right)\right)& =& L\left({\left(A^{T}A\right)}^{-1}{A}^T\stackrel{鈬赌}{y}\right)\\ & =& A\left({\left(A^{T}A\right)}^{-1}{A}^T\stackrel{鈬赌}{y}\right)\\ & =& A\left(A^{-1}{A}^{T^{-1}}{A}^T\stackrel{鈬赌}{y}\right)\\ & =& A{A}^{-1}{A}^T{A}^{T^{-1}}\stackrel{鈬赌}{y}\\ & =& \stackrel{鈬赌}{y}\end{array}$

Thus, the value of $\begin{array}{rcl}L\left(L^{+}\left(y\right)\right)& =& \stackrel{鈬赌}{y}\end{array}$.

(e)

The given linear transformation is $L\left(\stackrel{鈬赌}{x}\right)=\left[\begin{array}{cc}1& 0\\ 0& 1\\ 0& 0\end{array}\right]\stackrel{鈬赌}{x}$. From part (a), we have$L^{+}\left(\stackrel{鈬赌}{y}\right)=A^{+}\stackrel{鈬赌}{y}$ where $A^{+}={\left(A^{T}A\right)}^{-1}{A}^T$.

Now, it is given $A=\left[\begin{array}{cc}1& 0\\ 0& 1\\ 0& 0\end{array}\right]$. Find the value of A⁺ using $A^{+}={\left(A^{T}A\right)}^{-1}{A}^T$. First find $A^{T}A$:

$\begin{array}{rcl}{A}^{T}A& =& \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& 1\\ 0& 0\end{array}\right]\\ & =& \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\end{array}$

It is an identity matrix of 2x2 order. So, its inverse is also same which implies ${\left(A^{T}A\right)}^{-1}=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$.

Now, find the value of ${\left(A^{T}A\right)}^{-1}{A}^T$:

$\begin{array}{rcl}{\left(A^{T}A\right)}^{-1}{A}^T& =& \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\end{array}\right]\\ & =& \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\end{array}\right]\end{array}$

Thus, the required value is $L^{+}\left(\stackrel{鈬赌}{y}\right)\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & \left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\end{array}\right]\end{array}\begin{array}{rcl}& & \left(\stackrel{鈬赌}{y}\right)\end{array}$.