Answer to Question #118713 in Linear Algebra for jay

Question

A and X are the matrices
a b
c d !
and
x y
u v !
respectively, where b is not equal
to zero. Prove that if AX = XA then u = cy/b and v = x + (d − a)y/b. Hence prove
that if AX = XA then there are numbers p and q such that X = pA + qI, and find
p and q in terms of a, b, x, y.

Answer 1

$$A=\begin{pmatrix} a & b \\ c & d \end{pmatrix},X=\begin{pmatrix} x & y\\ u & v \end{pmatrix}$$

$$AX=\begin{pmatrix} a & b\\ c & d \end{pmatrix}\begin{pmatrix} x & y\\ u & v \end{pmatrix}=\begin{pmatrix} ax+bu & ay+bv\\ cx+du & cy+dv \end{pmatrix}$$

$$XA=\begin{pmatrix} x & y\\ u & v \end{pmatrix}\begin{pmatrix} a & b\\ c & d \end{pmatrix}=\begin{pmatrix} ax+cy & bx+dy\\ au+cv & bu+dv \end{pmatrix}$$

$$AX=XA=>$$

$$\begin{pmatrix} ax+bu & ay+bv\\ cx+du & cy+dv \end{pmatrix}=\begin{pmatrix} ax+cy & bx+dy\\ au+cv & bu+dv \end{pmatrix}$$

$$\begin{matrix} ax+bu=ax+cy \\ ay+bv=bx+dy \\ cx+du=au+cv \\ cy+dv=bu+dv \end{matrix}$$

Then

$$bu=cy=>u={cy\over b},b\not=0$$

$$bv=bx+(d-a)y=>v=x+{d-a\over b}y,b\not=0$$

$$X=pA+qI$$

$$X=\begin{pmatrix} x & y\\ u & v \end{pmatrix}=\begin{pmatrix} x & y\\ {cy \over b} &x+ {d-a \over b}y \end{pmatrix}=$$

$$=p\begin{pmatrix} a & b \\ c & d \end{pmatrix}+q\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$

$$\begin{matrix} pa+q=x \\ pb=y \\ pc={cy \over b} \\ pd+q=x+{d-a \over b}y \end{matrix}$$

$$p={y \over b}$$

$$q=x-{a \over b}y$$