Question

Prove for 2x2 matrices that tr(AB)=tr(A)tr(B) if and only if det(A+B)=det(A)+det(B).

Answer 1

*be careful, this isn't a typo

Answer 2

hmmm let me see

Answer 3

\[
C:=AB\\
C=\begin{pmatrix}
a_{11}b_{11}+a_{12}b_{21}& a_{11}b_{12}+a_{12}b_{22}\\
a_{21}b_{11}+a_{22}b_{21}& a_{21}b_{12}+a_{22}b_{22}
\end{pmatrix}\\
\operatorname{tr}(C)=a_{11}b_{11}+a_{12}b_{21}+a_{21}b_{12}+a_{22}b_{22}\\
\]
\[
\begin{align*}
&\phantom{{}={}}\operatorname{tr}(A)\operatorname{tr}(B)\\
&=(a_{11}+a_{22})(b_{11}+b_{22})\\
&=a_{11}b_{11}+a_{11}b_{22}+a_{22}b_{11}+a_{22}b_{22}\\
\end{align*}
\]
\[
\begin{align*}
\operatorname{tr}(A)\operatorname{tr}(B)&=\operatorname{tr}(C)=\operatorname{tr}(AB)\\
a_{11}b_{11}+a_{11}b_{22}+a_{22}b_{11}+a_{22}b_{22}&=a_{11}b_{11}+a_{12}b_{21}+a_{21}b_{12}+a_{22}b_{22}\\
a_{11}b_{22}+a_{22}b_{11}&=a_{12}b_{21}+a_{21}b_{12}
\end{align*}
\]
\[
D:=A+B\\
D=\begin{pmatrix}
a_{11}+b_{11} & a_{12}+b_{12}\\
a_{21}+b_{21} & a_{22}+b_{22}
\end{pmatrix}\\
\det(D)=(a_{11}+b_{11})(a_{22}+b_{22})-(a_{12}+b_{12})(a_{21}+b_{21})\\
\det(A)=a_{11}a_{22}-a_{12}a_{21}\\
\det(B)=b_{11}b_{22}-b_{12}b_{21}
\]
\[
\begin{align*}
&\phantom{{}={}}\det(D)\\
&=(a_{11}+b_{11})(a_{22}+b_{22})-(a_{12}+b_{12})(a_{21}+b_{21})\\
&=\operatorname{tr}(A)\operatorname{tr}(B)-(a_{12}a_{21}+a_{12}b_{21}+a_{21}b_{12}+b_{12}b_{21})\\
&=\operatorname{tr}(A)\operatorname{tr}(B)-(a_{11}a_{22}-\det(A)+a_{12}b_{21}+a_{21}b_{12}+b_{11}b_{12}-\det(B))\\
&=\operatorname{tr}(A)\operatorname{tr}(B)-(a_{11}a_{22}-\det(A)+a_{11}b_{22}+a_{22}b_{11}+a_{22}b_{11}-\det(B))\\
&=\operatorname{tr}(A)\operatorname{tr}(B)-(2a_{11}a_{22}+2a_{22}b_{11}-\det(A)-\det(B))
\end{align*}
\]
Smells interesting.

Answer 4

If \(\det(A+B)=\det(A)+\det(B)\) is true, then \(a_{11}b_{22}+a_{22}b_{11}=-a_{12}a_{21}-b_{12}b_{21}\), which is exactly the reverse of what happens when \(\operatorname{tr}(AB)=\operatorname{tr}(A)\operatorname{tr}(B)\)

Answer 5

\[
\begin{align*}
&\phantom{{}={}}(a_{12}+b_{12})(a_{21}+b_{21})\\
&=a_{12}a_{21}+a_{12}b_{21}+a_{21}b_{12}+b_{12}b_{21}
\end{align*}
\]
\[
\begin{align*}
\det(A)&=a_{11}a_{22}-a_{12}a_{21}\\
a_{12}a_{21}&=a_{11}a_{22}-\det(A)\\
\det(B)&=b_{11}b_{22}-b_{12}b_{21}\\
b_{12}b_{21}&=b_{11}b_{22}-\det(B)
\end{align*}
\]
Continuing:
\[
\begin{align*}
&\phantom{{}={}}a_{12}a_{21}+a_{12}b_{21}+a_{21}b_{12}+b_{12}b_{21}\\
&=a_{11}a_{22}+a_{12}b_{21}+a_{21}b_{12}+b_{11}b_{22}-\det(A)-\det(B)\\
&=a_{11}a_{22}+a_{11}b_{22}+a_{22}b_{11}+b_{11}b_{22}-\det(A)-\det(B) \quad\text{By }\operatorname{tr}(A)\operatorname{tr}(B)=\operatorname{tr}(C)\\
&=(a_{11}+b_{11})(a_{22}+b_{22})-\det(A)-\det(B)
\end{align*}
\]
\[
\begin{align*}
&\phantom{{}={}}\det(D)\\
&=(a_{11}+b_{11})(a_{22}+b_{22})-(a_{12}+b_{12})(a_{21}+b_{21})\\
&=(a_{11}+b_{11})(a_{22}+b_{22})-\left((a_{11}+b_{11})(a_{22}+b_{22})-\det(A)-\det(B)\right)\\
&=\det(A)+\det(B)
\end{align*}
\]
YAY!!!!

Answer 6

\[
\begin{align*}
\det(D)&=\det(A)+\det(B)\\
(a_{11}+b_{11})(a_{22}+b_{22})-(a_{12}+b_{12})(a_{21}+b_{21})&=a_{11}a_{22}-a_{12}a_{21}
+b_{11}b_{22}-b_{12}b_{21}\\
a_{11}b_{22}+a_{22}b_{11}-(a_{12}+b_{12})(a_{21}+b_{21})&=-a_{12}a_{21}-b_{12}b_{21}\\
a_{11}b_{22}+a_{22}b_{11}-a_{12}b_{21}-a_{21}b_{12}&=0\\
a_{11}b_{22}+a_{22}b_{11}&=a_{12}b_{21}+a_{21}b_{12}\\
a_{11}b_{22}+a_{22}b_{11}+a_{11}b_{11}+a_{22}b_{22}&=a_{12}b_{21}+a_{21}b_{12}+a_{11}b_{11}+a_{22}b_{22}\\
\operatorname{tr}(A)\operatorname{tr}(B)&=\operatorname{tr}(C)
\end{align*}
\]

Answer 7

Great success!

Answer 8

Hahaha all I can do is give you a medal, fun and weird huh? xD