Let
\[ \pmb{A} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \]
Using (4.2.3) verify that \(\left(\pmb{A}^{-1}\right)^{-1}=\pmb{A}\).
\[ \pmb{A}^{-1} = \frac1{ad-bc} \begin{pmatrix} d& -b \\ -c& a \end{pmatrix}=\\ \begin{pmatrix} \frac{d}{ad-bc}& -\frac{b}{ad-bc} \\ -\frac{c}{ad-bc}& \frac{a}{ad-bc} \end{pmatrix}= \begin{pmatrix} s& u \\ v& t \end{pmatrix} \]
so \[ \left(\pmb{A}^{-1}\right)^{-1} = \frac1{st-uv} \begin{pmatrix} t& -u \\ -v& s \end{pmatrix}\\ \]
\[ st-uv = \frac{d}{ad-bc}\frac{a}{ad-bc}-(-\frac{b}{ad-bc})(-\frac{c}{ad-bc})=\\ \frac{ad-bc}{(ad-bc)^2}=\frac{1}{ad-bc}\\ \]
\[ \left(\pmb{A}^{-1}\right)^{-1} = ad-bc \begin{pmatrix} \frac{a}{ad-bc}& \frac{b}{ad-bc} \\ \frac{c}{ad-bc}& \frac{d}{ad-bc} \end{pmatrix}=\pmb{A} \]
Show that for any square \(n\times n\) matrix \(\pmb{A}\) the quadratic form can be written as
\[\pmb{y'Ay}=\pmb{y'}\left(\frac{\pmb{A}+\pmb{A'}}{2}\right)\pmb{y'}\]
\(\pmb{y'Ay}\) is a scalar, and therefore \((\pmb{y'Ay})'=\pmb{y'A'y}\). So
\[ \begin{aligned} &\pmb{y'}\left(\frac{\pmb{A}+\pmb{A'}}{2}\right)\pmb{y'} = \\ &\frac{\pmb{y'Ay}+\pmb{y'A'y}}{2} = \\ &\frac{\pmb{y'Ay}+\pmb{y'Ay}}{2} = \\ &\pmb{y'Ay}\\ \end{aligned} \]
Let
\[ \pmb{A} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \]
Show that \(\vert\pmb{A}^2\vert=\vert\pmb{A}\vert^2\)
\[ \begin{aligned} &\pmb{A}^2 = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} a & b \\ c & d \end{pmatrix}= \begin{pmatrix} a^2+bc & ab+bd \\ ac+dc & bc+d^2 \end{pmatrix}\\ &\vert\pmb{A}^2\vert= (a^2+bc)(d^2+bc)-bc(a+d)^2= \\ &a^2d^2+a^2bc+d^2bc+b^2c^2-a^2bc-2abcd+bcd^2 = \\ &a^2d^2-2abcd+b^2c^2=(ad-bc)^2\\ \end{aligned} \]
\[ \begin{aligned} &\vert\pmb{A}\vert^2 = \begin{vmatrix} a & b \\ c & d \end{vmatrix}^2 = (ad-bc)^2 \end{aligned} \]
Find all the solutions of the system of equations
\[ \begin{aligned} &2x+3y-z =1 \\ &x-y = 1 \end{aligned} \]
\[ \begin{aligned} &2x+3y =1+z \\ &x-y = 1\\ &\text{I-2II }\\ &5y=1+z-2=z-1\\ &x=1+y=1+(z-1)/5=(z+4)/5\\ &y=(z-1)/5 \end{aligned} \] and so for all z a solution is given by
\[\begin{pmatrix}z+4 \\z-1\\5z \end{pmatrix}/5\].
By (4.2.15) we can find a generalized inverse as follows:
\[ \pmb{A}_{11} = \begin{pmatrix} 2 & 3 \\ 1 & -1 \end{pmatrix}\\ \pmb{A}_{11}^{-1} = -\frac1{5} \begin{pmatrix} -1 & -3 \\ -1 & 2 \end{pmatrix}\\ \pmb{A}^{-} = \frac1{5} \begin{pmatrix} 1 & 3 \\ 1 &-2 \\ 0&0 \end{pmatrix} \]
and now
\[ \begin{aligned} &I-\pmb{A}^{-}\pmb{A} =\\ &\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0\\ 0 & 0& 1 \end{pmatrix}- \frac1{5} \begin{pmatrix} 1 & 3 \\ 1 &-2 \\ 0&0 \end{pmatrix} \begin{pmatrix} 2 & 3 & -1 \\ 1 & -1 & 0\\ \end{pmatrix}= \\ &\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0\\ 0 & 0& 1 \end{pmatrix}-\frac15 \begin{pmatrix} 5 & 0 & -1 \\ 0 & 5 & -1\\ 0 & 0& 0 \end{pmatrix} =\\ &\frac15\begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 1\\ 0 & 0& 5 \end{pmatrix} \end{aligned} \]
\[ \begin{aligned} &\pmb{A}^{-}c+(I-\pmb{A}^{-}\pmb{A})\pmb{h} = \\ &\frac1{5} \begin{pmatrix} 1 & 3 \\ 1 &-2 \\ 0&0 \end{pmatrix} \begin{pmatrix}1\\1\end{pmatrix}+ \frac15\begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 1\\ 0 & 0& 5 \end{pmatrix} \begin{pmatrix}x\\y\\z\end{pmatrix}= \\ &\frac15\begin{pmatrix} 4 \\ -1 \\ 0 \end{pmatrix} + \frac15\begin{pmatrix} z \\ z \\ 5z \end{pmatrix}= \\ &\frac15\begin{pmatrix} z+4 \\ z-1 \\ 5z \end{pmatrix} \end{aligned} \]
Say
\[ \pmb{A} = \begin{pmatrix} \frac12 & -\frac14 \\ -\frac14 & \frac12 \end{pmatrix} \]
Find \(\log(1+\pmb{A})\).
We have the power series expansion \(\log(1+x)=\sum_{n=1}^\infty (-1)^{n+1}\frac{x^n}{n}\), \(\vert x\vert\le 1\), so
\[ \begin{aligned} &\log(1+\pmb{A}) = \sum_{n=1}^\infty (-1)^{n+1}\frac{A^n}{n} = \\ &\sum_{n=1}^\infty (-1)^{n+1}\frac{\pmb{C'D}^n\pmb{C}}{n} = \\ &\pmb{C}'\left[\sum_{n=1}^\infty (-1)^{n+1}\frac{\pmb{D}^n}{n} \right]\pmb{C} \end{aligned} \] Next we need the eigenvalues and eigen vectors:
\[ \begin{aligned} &\vert\pmb{A-\lambda}\vert = (\frac12-\lambda)^2-\frac1{16} =\\ &\lambda^2-\lambda+\frac3{16} = (\lambda-\frac34)(\lambda-\frac14)=0 \end{aligned} \] so we have eigenvalues \(\frac14\) and \(\frac34\). Now
\[ \begin{aligned} &\pmb{(A-\lambda_1I)x} = 0\\ &x/4-y/4=0 \\ &x =y \\ &\sqrt{1^2+1^2}=\sqrt{2} \end{aligned} \]
yields the normalized eigenvector \(\begin{pmatrix}1 &1\end{pmatrix}/\sqrt2\)
\[ \begin{aligned} &\pmb{(A-\lambda_2I)x} = 0\\ &x/4+y/4=0 \\ &x =-y \\ &\sqrt{1^2+1^2}=\sqrt{2} \end{aligned} \] therefore
\[ \begin{aligned} &\sum_{n=1}^\infty (-1)^{n+1}\frac{\pmb{D}^n}{n} = \\ &\begin{pmatrix} \sum_{n=1}^\infty (-1)^{n+1}\frac{(1/4)^n}{n} & 0 \\ 0 & \sum_{n=1}^\infty (-1)^{n+1}\frac{(\pmb{D}(3/4)^n}{n} \end{pmatrix} = \\ &\begin{pmatrix} \log(1+1/4) & 0 \\ 0 & \log(1+3/4) \end{pmatrix} = \\ &\begin{pmatrix} \log(5/4) & 0 \\ 0 & \log(7/4) \end{pmatrix} \end{aligned} \]
\[ \begin{aligned} &\log(1+\pmb{A}) = \pmb{C'D^nC} = \\ &\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}'/\sqrt 2 \begin{pmatrix} \log(5/4) & 0 \\ 0 & \log(7/4) \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}/\sqrt 2= \\ &\frac12 \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} \log(5/4) & \log(5/4) \\ \log(7/4) & -\log(7/4) \end{pmatrix}=\\ &\frac12 \begin{pmatrix} \log(5/4)+\log(7/4) & \log(5/4)-\log(7/4) \\ \log(5/4)-\log(7/4) & \log(5/4)+\log(7/4) \end{pmatrix} \end{aligned} \]