91Ó°ÊÓ

A function T from $R^m$to $R^n$is called a linear transformation if there exists an$n×m$ matrix A such that

$T\left(\stackrel{→}{x}\right)=A\stackrel{→}{x}$

For all$\stackrel{→}{x}$ in the vector space

The image of a function consists of all the values the function takes in its target space.If f is a function from X to Y ,then:

Image(f)=$\{f\left(x\right):x\text{ }in\text{ }X\}$

=$\{b\text{ â¶Ä‰}in\text{ â¶Ä‰}Y:b=f\left(x\right)\text{ â¶Ä‰}for\text{ â¶Ä‰}some\text{ â¶Ä‰}x\text{ â¶Ä‰}in\text{ â¶Ä‰}X\}$

The kernel of a linear transformation$T\left(\stackrel{→}{x}\right)=A\stackrel{→}{x}$ from $R^m$ to $R^n$consists of all the zeros of the transformation

i.e.,the solutions of the equation$T\left(\stackrel{→}{x}\right)=A\stackrel{→}{x}=0$

We denote the kernel of T by ker(T) or ker(A).

Consider a linear transformation $L\left(\stackrel{→}{x}\right)=A\stackrel{→}{x}$from$R^n$ to$R^m$ where rank(A)=m.

The pseudoinverse $L^{+}$of L is the transformation from $R^m$ to $R^n$given by

$L^{+}\left(\stackrel{→}{y}\right)$=(the minimum solution of the system$L^{+}\left(\stackrel{→}{x}\right)=\stackrel{→}{y}$ )

By the definition,the pseudoinverse$L^{+}$ is linear.

Thus the proof.

Consider a linear transformation $L\left(\stackrel{→}{x}\right)=A\stackrel{→}{x}$from $R^n$to $R^m$where rank(A)=m.

The pseudoinverse$L^{+}$ of L is the transformation from$R^m$to $R^n$given by:

$L^{+}\left(\stackrel{→}{y}\right)$=(the minimum solution of the system$L^{+}\left(\stackrel{→}{x}\right)=\stackrel{→}{y}$)

Hence the pseudoinverse of the transformation$L\left(L^{+}\left(\stackrel{→}{y}\right)\right)$for $\stackrel{→}{y}$in$R^m$is$L\left(L^{+}\left(\stackrel{→}{y}\right)\right)=pro{j}_W\stackrel{→}{y}$and also the pseudoinverse of the transformation$L^{+}\left(L\left(\stackrel{→}{x}\right)\right)$for $\stackrel{→}{x}$in $R^n$is .$L^{+}\left(L\left(\stackrel{→}{x}\right)\right)=pro{j}_V\stackrel{→}{x}$

Where,

$\begin{array}{l}V={\left(\ker A\right)}^{âŠ\yen }\\ \text{ â¶Ä‰â€‰â¶Ä‰â€‰}=im\left(A^T\right)\end{array}$

And also W is:

$\begin{array}{l}W={\left(\ker A\right)}^{âŠ\yen }\\ \text{ â¶Ä‰â€‰â¶Ä‰â€‰}=im\left(A^T\right)\end{array}$

Hence the solution.

By the definition,we obtain that the image of the pseudoinverse $L^{+}$of L is the transformation from $R^m$ to$R^n$ becomes:

$im\left(L^{+}\right)=im\left(A^T\right)$

And also the kernel of the pseudoinverse $L^{+}$of L is the transformation from$R^m$ to$R^n$ becomes:

$\ker \left(L^{+}\right)=\left(A^T\right)$

Hence the solution.

Consider a linear transformation$L\left(\stackrel{→}{x}\right)=A\stackrel{→}{x}$from $R^n$to$R^m$where rank(A)=m.

The pseudoinverse $L^{+}$of L is the transformation from $R^m$to$R^n$given by:

$L^{+}\left(\stackrel{→}{y}\right)$=(the minimum solution of the system$L^{+}\left(\stackrel{→}{x}\right)=\stackrel{→}{y}$)

Thus the matrix is in transpose form,then the matrix in pseudo inverse becomes:

$L^{+}\left(\stackrel{→}{y}\right)=\left[\begin{array}{l}\frac{1}{2}\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}0\\ 0\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}0\\ 0\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}0\end{array}\right]\stackrel{→}{y}$

Thus the solution.