Questions 11 - 16

Q11

Find $ABAB$ and $BABA$ if ${\displaystyle AA=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]}$ and ${\displaystyle BB=\left[\begin{array}{cc}5& 6\\ 7& 8\end{array}\right]}$ . Do these matrices commute and if not what is $[ABAB]$?

Q12

(a) Find $A{A}^2,B{B}^2,ABAB,BABA,A{A}^{2}BB$,and $B{B}^{2}AA$ if ${\displaystyle AA=\left[\begin{array}{cc}1& 1\\ 0& 1\end{array}\right]}$ and ${\displaystyle BB=\left[\begin{array}{cc}1& 0\\ 1& 1\end{array}\right]}$.

(b) Do $A{A}^{2}BB$ and $B{B}^{2}AA$ commute ?

Q13

If $xx=[x_1,x_2],yy=[y_1,y_2{]}^T$ ,and ${\displaystyle QQ=\left[\begin{array}{cc}1& -4\\ -9& 6\end{array}\right]}$, find $xQyxQy$ and $yQxyQx$.

Q14

(a) Explain why if $j$ is a constant factor that $|jMM|=j^n|MM|$ for an $n\times n$ matrix.

(b) Confirm equation 4 if ${\displaystyle AA=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]}$ by calculating $A{A}^{-1}$ and $|A{A}^{-1}|$.

(c) Using Sympy (or by hand ) find $A{A}^{-1}$ and $|A{A}^{-1}|$ for the $3\times 3$ matrix ${\displaystyle AA=\left[\begin{array}{ccc}0& b& c\\ b& 0& d\\ c& d& 0\end{array}\right]}$.

Strategy: See Section 4.14, but now the problem is algebraic not numerical.

Q15 Commuting operators

(a) Show that if $PP$ and $QQ$ are linear operators, not necessarily matrices, $[PP,QQ]=-[QQ,PP]$.

(b) (i) find $[d/dx,x]\sin (x)$ , (ii) $[d/dx,x]f(x)$, and (iii) $[d/dx,x]$ and

(c) Show that operators $d/dx$ and $x$ do not commute.

(d) Write a Python/Sympy commutator function and test your results.

(e) Do ${\displaystyle \frac{df(x)}{dx}}$ and ${\displaystyle {\int }_a^{b}f(x)dx}$ commute?

(f) Does the operator $df/dx$ and a displacement operator $\mathrm{\Delta }f=f(x+c)$ commute?

(g) Does the operator $xx$ which means multiply twice by $x$, commute with the inversion operator, $Inv(f(x))=f(-x)$?

Q16 Commuting matrices

$BB$ and $CC$ are commuting square matrices and the matrix $AA$ is defined as $AA\equiv e^{BB}=11+BB+B{B}^2/2!+\cdots$ show that:

(a) $e^{BB}{e}^{CC}=e^{B+CB+C}$,

(b) $A{A}^{-1}=e^{-BB}$,

(c) $e^{CBC{CBC}^{-1}}=CAC{CAC}^{-1}$.

Strategy: (a) Expand out the exponentials as shown in Section 5.5 and collect terms and try to reform an exponential series in $B+CB+C$. (b) Use the fact that for an inverse matrix $AA{AA}^{-1}=11$. (c) The expression $CAC{CAC}^{-1}$ is a similarity transform and is itself a square matrix. To prove this result, expand the exponential on both sides of the equation