Consider the matrix
\[ \pmb{A} = \begin{pmatrix} 2 & -1 \\ -1 & 2 \end{pmatrix} \]
\[ \begin{pmatrix}x &y\end{pmatrix}'\pmb{A} \begin{pmatrix}x \\y\end{pmatrix}=\\ \begin{pmatrix}x &y\end{pmatrix}'\begin{pmatrix} 2x-y \\ -x+2y \end{pmatrix}=\\ 2x^2-2xy+2y^2 = \\ x^2+y^2+(x-y)^2>0 \]
\[ \begin{aligned} &\vert\pmb{A-\lambda}\vert = (2-\lambda)^2-1 =\\ &\lambda^2-4\lambda+3 = (\lambda-3)(\lambda-1)=0 \end{aligned} \]
\[ \begin{aligned} &\pmb{(A-\lambda_1I)x} = 0\\ &x-y=0 \\ &x =y \\ &\sqrt{1^2+1^2}=\sqrt{2} \end{aligned} \]
yields the normalized eigenvector \(\begin{pmatrix}1 &1\end{pmatrix}/\sqrt2\)
\[ \begin{aligned} &\pmb{(A-\lambda_2I)x} = 0\\ &x+y=0 \\ &x =-y \\ &\sqrt{1^2+1^2}=\sqrt{2} \end{aligned} \]
yields the normalized eigenvector \(\begin{pmatrix}1 &-1\end{pmatrix}/\sqrt2\) So we find
\[ \begin{aligned} &\pmb{C'diag(\sqrt{\lambda})C} = \\ &\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}'/\sqrt 2 \begin{pmatrix} \sqrt 1 & 0 \\ 0 & \sqrt 3 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}/\sqrt 2= \\ &\frac12 \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ \sqrt 3 & -\sqrt 3 \end{pmatrix}=\\ &\frac12 \begin{pmatrix} 1+\sqrt 3 & 1-\sqrt 3 \\ 1-\sqrt 3 & 1+\sqrt 3 \end{pmatrix} \end{aligned} \]
Let
\[ \pmb{A} = \begin{pmatrix} 1 & 0 & x \\ 0 & 1 & 0 \\ x & 0 & 1 \\ \end{pmatrix} \]
Find \(\frac{\partial \log\vert\pmb{A}\vert}{\partial x}\)
\[ \vert \pmb{A} \vert = \begin{vmatrix} 1 & 0 & x \\ 0 & 1 & 0 \\ x & 0 & 1 \\ \end{vmatrix} = 1+x(-x)=1-x^2 \\ \log\vert \pmb{A} \vert=\log(1-x^2)\\ \frac{\partial \log\vert\pmb{A}\vert}{\partial x}=\frac{-2x}{1-x^2} \]
Note that
\[ \pmb{A}^{-1} = \frac1{1-x^2} \begin{pmatrix} 1 & 0 & -x \\ 0 & 1-x^2 & 0 \\ -x & 0 & 1\\ \end{pmatrix} \]
and so
\[ \begin{aligned} &\pmb{A}^{-1}\frac{\partial \pmb{A}}{\partial x} = \\ &\frac1{1-x^2} \begin{pmatrix} 1 & 0 & -x \\ 0 & 1-x^2 & 0 \\ -x & 0 & 1\\ \end{pmatrix} \begin{pmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \\ \end{pmatrix} = \\ &\frac1{1-x^2} \begin{pmatrix} -x & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & -x\\ \end{pmatrix} \\ &tr(\pmb{A}^{-1}\frac{\partial \pmb{A}}{\partial x}) = \frac1{1-x^2}(-2x) \end{aligned} \]
Say \(\pmb{Y} = \begin{pmatrix} Y_1& Y_2 \end{pmatrix}'\) with \(E[Y_1]=E[Y_2]=0\), \(var(Y_1)=1\), \(var(Y_2)=4\) and \(cov(Y_1,Y_2)=-2\). Let \(\pmb{A} = \begin{pmatrix} 2 & -1 \\ 0 & 3 \end{pmatrix}\).
Verify theorem (5.1.10) part i. directly.
Note that \(\pmb{\Sigma}=\begin{pmatrix} 1 & -2 \\ -2 & 4 \end{pmatrix}\), so
\[\pmb{A\Sigma A'} = \begin{pmatrix} 2 & -1 \\ 0 & 3 \end{pmatrix} \begin{pmatrix} 1 & -2 \\ -2 & 4 \end{pmatrix} \begin{pmatrix} 2 & 0 \\ -1 & 3 \end{pmatrix}= \\ \begin{pmatrix} 2 & -1 \\ 0 & 3 \end{pmatrix} \begin{pmatrix} 4 & -6 \\ -8 & 12 \end{pmatrix}= \begin{pmatrix} 16 & -24 \\ -24 & 36 \end{pmatrix}\]
and also
\[\pmb{AY} = \begin{pmatrix} 2 & -1 \\ 0 & 3 \end{pmatrix} \begin{pmatrix} Y_1\\ Y_2 \end{pmatrix}= \begin{pmatrix} 2Y_1-Y_2 \\ 3Y_2 \end{pmatrix}\]
\[ \begin{aligned} &var(2Y_1-Y_2) =4var(Y_1)+var(Y_2)+2cov(2Y_1,-Y_2)=\\ &4var(Y_1)+var(Y_2)-4cov(Y_1,Y_2)=\\ &4\times1+4-4\times2=16 \\ &var(3Y_2) =9var(Y_2)=9\times 4=36 \\ &cov(2Y_1-Y_2, 3Y_2) =\\ &cov(2Y_1, 3Y_2)-cov(Y_2, 3Y_2) = \\ &6cov(Y_1, Y_2)-3var(Y_2) = \\ &6(-2)-3\times4 =-24 \end{aligned} \]