Perform elementary row operations to

\left[A\mid \overrightarrow{b^{*} }\mid I_{3} \right] = \left[\begin{array}{cc:c:ccc}  1 & 0 & b_{1} & 1 & 0 & 0 \\  1 & 1 & b_{2} & 0 & 1 & 0 \\ 1 & -1  & b_{3} & 0 & 0 & 1 \end{array} \right]

\underset{\begin{matrix} E_{\left(2\right)-\left(1\right) } \\ E_{\left(3\right)-\left(1\right)} \end{matrix} }{\longrightarrow} \left[\begin{array}{cc:c:ccc}  1 & 0 & b_{1} & 1 & 0 & 0 \\ 0 & 1  & b_{2}-b_{1} & -1 & 1 & 0 \\ 0 & -1  & b_{3}-b_{1} & -1 & 0 & 1 \end{array} \right]

\underset{\text{}E_{\left(3\right)+\left(2\right) } }{\longrightarrow} \left[\begin{array}{cc:c:ccc}  1 & 0 & b_{1} & 1 & 0 & 0 \\ 0 & 1  & b_{2}-b_{1} & -1 & 1 & 0 \\ 0 & 0  & b_{2}+b_{3}-2b_{1} & -2 & 1 & 1 \end{array} \right].\left(*_{18} \right)

From \left(*_{18} \right)

A \overrightarrow{x^{*}}=\overrightarrow{b^{*}} has a solution \overrightarrow{x}=\left(x_{1},x_{2}\right) \in R^{2}.

\Leftrightarrow b_{2} +b_{3}-2b_{1}=0

⇒ The solution is  x_{1}=b_{1}, x_{2}=b_{2}-b_{1}.

The constrained condition b_{2}+b_{3}-2b_{1}=0 can also be seen by eliminating x_{1},x_{2},x_{3} from the set of equations x_{1}=b_{1}, x_{1}+x_{2}=b_{2},x_{1}-x_{2}=b_{3}.

\left(*_{18} \right) also indicates that

BA=I_{2},  where B = \left[\begin{matrix} 1 & 0 & 0 \\ -1 & 1 & 0  \end{matrix} \right]

i.e. B is a left inverse of A. In general, let B=\left[\begin{matrix} \overrightarrow{v_{1}} \\ \overrightarrow{v_{2}} \end{matrix} \right]_{2\times3}. Then

BA=\left[\begin{matrix} \overrightarrow{v_{1}} \\ \overrightarrow{v_{2}} \end{matrix} \right] A =\left[\begin{matrix} \overrightarrow{v_{1}}A \\ \overrightarrow{v_{2}}A \end{matrix} \right]=I_{2}

\Leftrightarrow \overrightarrow{v_{1}}A=\overrightarrow{e_{1}}

\overrightarrow{v_{2}}A=\overrightarrow{e_{2}}.

Suppose \overrightarrow{v_{1} }=\left(x_{1},x_{2},x_{3}\right), Then

\overrightarrow{v_{1}}A=\overrightarrow{e_{1}}

\Leftrightarrow \left\{\begin{matrix} x_{1}+x_{2}+x_{3}=1 \\ x_{2}-x_{3}=0  \end{matrix} \right.

\Leftrightarrow\overrightarrow{v_{1}}=\left(1,0,0 \right)+t_{1}\left(-2,1,1 \right), t_{1}\in R.

Similarly, let \overrightarrow{v_{2} }=\left(x_{1},x_{2},x_{3}\right). Then

\overrightarrow{v_{2}}A=\overrightarrow{e_{2}}

\Leftrightarrow \left\{\begin{matrix} x_{1}+x_{2}+x_{3}=0 \\ x_{2}-x_{3}=1   \end{matrix} \right.

\Leftrightarrow\overrightarrow{v_{2}}=\left(-1,1,0 \right)+t_{2}\left(-2,1,1 \right), t_{2}\in R.

Thus, the left inverses of A are

B=\left [ \begin{matrix} 1-2t_{1}  & t_{1} & t_{1}  \\ -1-2t_{2} & 1+t_{2} & t_{2} \end{matrix} \right ] _{2\times 3}   for t_{1},t_{2}\in R.  \left(*_{19} \right)

On the other hand, \left(*_{18} \right) says

E_{\left(3\right)+\left(2\right) }E_{\left(3\right)-\left(1\right)}E_{\left(2\right)-\left(1\right) }A=PA= \left [ \begin{matrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{matrix} \right ]

(row-reduced echelon matrix and
normal form of A),

P=E_{\left(3\right)+\left(2\right) }E_{\left(3\right)-\left(1\right) }E_{\left(2\right)-\left(1\right) }=\left [ \begin{matrix} 1 & 0 & 0 \\ -1 & 1 & 0  \\ -2 & 1 & 1  \end{matrix} \right ]

\Rightarrow A= E^{-1}_{\left(2\right)-\left(1\right) }E^{-1}_{\left(3\right)-\left(1\right) }E^{-1}_{\left(3\right)+\left(2\right) }\left [ \begin{matrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{matrix} \right ]=P^{-1}\left [ \begin{matrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{matrix} \right ]

=\left [ \begin{matrix} 1 & 0 & 0 \\ 1 & 1 & 0  \\ 1 & -1 & 1  \end{matrix} \right ]\left [ \begin{matrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{matrix} \right ] (LU-decomposition).

Refer to (1) in (2.7.70).

To investigate A^{*} and A^{*}A, consider \overrightarrow{x}A=\overrightarrow{b} for \overrightarrow{x}=\left(x_{1},x_{2},x_{3}\right)\in R^{3} and \overrightarrow{b}=\left(b_{1},b_{2}\right)\in R^{2}.

By simple computation or by \left(*_{18} \right)

Ker\left(A \right)= \ll \left(-2,1,1\right) \gg = Im \left(A^{*} \right) ^{\bot },

Im\left(A \right)=R^{2}=Ker\left(A ^{*}\right)^{\bot },

Ker\left(A^{*} \right)=\left\{0\right\} = Im \left(A\right) ^{\bot },

Im\left(A^{*} \right)=\left\{\left(x_{1},x_{2},x_{3} \right)\in R^{3} \mid 2x_{1}-x_{2}-x_{3}=0 \right\}=\ll \left(1,2,0\right),\left(1,0,2\right) \gg

=\ll \left(1,1,1\right),\left(0,1,-1\right) \gg=Ker\left(A \right)^{\bot },

and

\left(A^{*} A\right)=\left[\begin{matrix} 3 & 0  \\ 0 & 2 \end{matrix} \right ],

\left(A^{*} A\right)^{-1} = \frac{1}{6} \left[\begin{matrix} 2 & 0  \\ 0& 3  \end{matrix} \right ].

For any fixed \overrightarrow{b} \in R^{2}, \overrightarrow{x}A= \overrightarrow{b} always has a particular solution \overrightarrow{b}\left(A^{*} A\right)^{-1}A^{*} and the solution set is

\overrightarrow{b}\left(A^{*} A\right)^{-1}A^{*} + Ker\left(A \right),      \left(*_{20} \right)

which is a one-dimensional affine subspace of R³. Among so many solutions, it is \overrightarrow{b}\left(A^{*} A\right)^{-1}A^{*} that has the shortest distance to the origin\overrightarrow{0} (see Ex. <B> 7 of Sec. 3.7.3). For simplicity, let

A^{+}=\left(A^{*} A\right)^{-1}A^{*}

= \frac{1}{6} \left[\begin{matrix} 2 & 0  \\ 0& 3  \end{matrix} \right ]\left[\begin{matrix} 1 & 1 & 1  \\ 0 & -1 & -1  \end{matrix} \right ]=\frac{1}{6}\left[\begin{matrix} 2 & 2 & 2  \\ 0 & 3 & -3  \end{matrix} \right ]_{2 \times 3}      \left(*_{21} \right)

and can be considered as a linear transformation from R² into R³ with the range space Im\left(A^{*} \right). Since

A^{+}A=I_{2},

A^{+} is a left inverse of A.

Therefore, it is reasonable to expect that one of the left inverses shown in \left(*_{19} \right) should be A^{+}. Since the range of A^{+}. is Im\left(A^{*} \right), then

B in \left(*_{19} \right) is A^{+}.

\Leftrightarrow The range space of B =Im\left(A^{*} \right)

\Leftrightarrow \left(1-2t_{1} ,t_{1},t_{1}\right) and \left(-1-2t_{2} ,1+t_{2},t_{2}\right) are in Im\left(A^{*} \right).

\Leftrightarrow \left\{\begin{matrix} 2 \left(1-2t_{1}\right)-t_{1}-t_{1}=0 \\ 2 \left(-1-2t_{2}\right)- \left(1+t_{2}\right)-t_{2}=0 \end{matrix} \right.

\Rightarrow t_{1}=\frac{1}{3} and t_{2}=-\frac{1}{2}.

In this case, B in \left(*_{19} \right) is indeed equal to A^{+}. This A^{+} is called the generalized inverse of A.

How about AA^{*}?

AA^{*}= \left [ \begin{matrix} 1 & 0 \\ 1 & 1 \\ 1 & -1 \end{matrix} \right ] \left [ \begin{matrix} 1 & 1 & 1 \\ 0 & 1 & -1 \end{matrix} \right ] = \left [ \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 0 \\ 1 & 0 & 2 \end{matrix} \right ] \left(*_{22} \right)

is a linear operator on R³ with the range space Im\left(A^{*} \right). Actual computation shows that \left(AA^{*} \right)^{2}\neq AA^{*} . Therefore, AA^{*} is not a projection of R³ onto Im\left(A^{*} \right) along Ker\left(A\right). Also

indicates that AA^{*} is not a projection (see (3.7.34)). Notice that

QAA^{*}Q^{-1}=\left [ \begin{matrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 0 \end{matrix} \right ],  where  Q= \left[\begin{matrix} \overrightarrow{v_{1} } \\ \overrightarrow{v_{2} } \\ \overrightarrow{v_{3} } \end{matrix} \right] is orthogonal.

\Rightarrow QAA^{*}Q^{*}=\left [ \begin{matrix} \sqrt{2} & 0 & 0 \\ 0 & \sqrt{3} & 0 \\ 0 & 0 & 1 \end{matrix} \right ] \left [ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{matrix} \right ]\left [ \begin{matrix} \sqrt{2} & 0 & 0 \\ 0 & \sqrt{3} & 0 \\ 0 & 0 & 1 \end{matrix} \right ]

\Rightarrow R\left(AA^{*} \right)R^{*}=\left[\begin{matrix} I_{2} & 0 \\ 0 & 0 \end{matrix} \right] _{3\times 3},

where

R= \left [ \begin{matrix} \frac{1}{\sqrt{2}} & 0 & 0 \\ 0 & \frac{1}{\sqrt{3}} & 0 \\ 0 & 0 & 1 \end{matrix} \right ] \left [ \begin{matrix} 0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \\ -\frac{2}{\sqrt{6}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{6}} \end{matrix} \right ] = \left [ \begin{matrix} 0 & \frac{1}{2} & -\frac{1}{2} \\ \\  \frac{1}{3} & \frac{1}{3} & \frac{1}{3} \\ \\  -\frac{2}{\sqrt{6}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{6}} \end{matrix} \right ].

Thus, the index of AA^{*} is 2 and the signature is equal to 2. By the way, we pose the question: What is the preimage of the unit circle (or disk) y^{2}_{1} +y^{2}_{2}=1 (or ≤1) under A? Let \overrightarrow{y}=\left(y_{1},y_{2}\right)=\overrightarrow{x}A. Then

y^{2}_{1} +y^{2}_{2}=\overrightarrow{y}\overrightarrow{y^{*} } =1

\Leftrightarrow \left(\overrightarrow{x}A\right)\left(\overrightarrow{x}A\right)^{*}=\overrightarrow{x}AA^{*}\overrightarrow{x^{*} }

=x^{2}_{1} +2x^{2}_{2}+2x^{2}_{3}+2x_{1}x_{2}+2x_{1}x_{3}, in the natural basis for R³

= 2x^{\prime2}_{1}+3x^{\prime2}_{2}, in the basis \left\{\overrightarrow{v_{1}},\overrightarrow{v_{2}},\overrightarrow{v_{3}}\right\}=B

= x^{\prime \prime 2}_{1}+x^{\prime \prime 2}_{2}, in the basis \left\{R_{1^{*} }, R_{2^{*} },R_{3^{*} }\right\}=C

= 1,    \left(*_{23} \right)

where \left(x^{\prime}_{1},x^{\prime}_{2},x^{\prime}_{3}\right)= \left[\overrightarrow{x} \right] _{B} =\overrightarrow{x}Q^{-1} and \left(x^{\prime \prime}_{1},x^{\prime \prime}_{2},x^{\prime \prime}_{3}\right)=\left[\overrightarrow{x} \right] _{C} =\overrightarrow{x}R^{-1}. See Fig. 3.53.

Replace A^{*}  in  AA^{*}  by A^{+} and, in turn, consider

AA^{+} =\left [ \begin{matrix} 1 & 0 \\ 1 & 1 \\ 1 & -1 \end{matrix} \right ] \cdot \frac{1}{6}\left [ \begin{matrix} 2 & 2 & 2 \\ 0 & 3 & -3 \end{matrix} \right ] = \frac{1}{6}\left [ \begin{matrix} 2 & 2 & 2 \\ 2 & 5 & -1 \\ 2 & -1 & 5 \end{matrix} \right ].

AA^{+} is symmetric and

\left(AA^{+}\right)^{2}=AA^{+}AA^{+}=AI_{2}A^{+}=AA^{+}.

Therefore, AA^{+}: R^{3} \rightarrow Im \left(A^{*} \right)\subseteq R^{3} s the orthogonal projection of R³ onto Im \left(A^{*} \right) along Ker \left(A \right). Note that

Im \left(AA^{+} \right) = Im \left(A^{+} \right) = Ker \left(A\right) ^{\bot } = Ker\left(AA^{+} \right) ^{\bot },

Ker\left(AA^{+} \right) = Ker \left(A \right) = Im \left(A^{*} \right) ^{\bot } = Im \left(AA^{+} \right) ^{\bot }.

What is the reflection or symmetric point of a point \overrightarrow{x} in R³ with respect to the plane Im \left(A^{*} \right)? Notice that (see \left(*_{16} \right))

\overrightarrow{x} \in R^{3}

\rightarrow \overrightarrow{x} AA^{+}, the orthogonal projection of \overrightarrow{x} on Im \left(A^{*} \right)

\rightarrow \overrightarrow{x} AA^{+} + \left(\overrightarrow{x} AA^{+}-  \overrightarrow{x}\right)= \overrightarrow{x}\left(2AA^{+} -I_{3}\right), the reflection point. \left(*_{24} \right)

Thus, denote the linear operator

P_{A}= 2AA^{+} -I_{3} = 2 \cdot \frac{1}{6} \left [ \begin{matrix} 2 & 2 & 2 \\ 2 & 5 & -1 \\ 2 & -1 & 5 \end{matrix} \right ]-\left [ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right ]=\left [ \begin{matrix} -\frac{1}{3} & \frac{2}{3} & \frac{2}{3} \\ \\  \frac{2}{3} & \frac{2}{3} & -\frac{1}{3} \\ \\  \frac{2}{3} & -\frac{1}{3} & \frac{2}{3} \end{matrix} \right ].

P_{A} is symmetric and is orthogonal, i.e. P^{*}_{A}=P^{-1}_{A} and is called the reflection of R³ with respect to Im \left(A^{*} \right).

A simple calculation shows that

D=\left\{\overrightarrow{u_{1} } ,\overrightarrow{u_{2} } ,\overrightarrow{u_{3} } \right\} is an orthonormal basis for R³. In D,

\left[AA^{+}\right] _{D}=S\left(AA^{+}\right)S^{-1}=\left [ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right ] ,   where S = \left[\begin{matrix} \overrightarrow{u_{1} } \\ \overrightarrow{u_{2} } \\ \overrightarrow{u_{3} } \end{matrix} \right] is orthogonal;

\left[P_{A}\right] _{D}= SP_{A}S^{-1}=\left [ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{matrix} \right ].

Try to explain \left[AA^{+}\right] _{D} and \left[P_{A}\right] _{D} graphically.

As a counterpart of (3.7.40), we summarize in