Exercise of symmetric matrices

Asked Nov 30 '17 at 03:52

Active Nov 30 '17 at 03:52

Viewed 28 times

How can I solve this exercise?

Suppose that $A,B,C,D$ are $n\times n$ matrices with entries in a field $F$, satisfying the conditions that $AB^{t}$ and $CD^{t}$ are symmetric and $AD^{t}-BC^{t}=I$. Here $I$ is the $n\times n$ identity matrix, $M^{t}$ is the transpose of $M$. Prove that $A^{t}D-C^{t}B=I$

Thanks for your help

asked Nov 30 '17 at 03:52

Victor Huuanca Sullca

https://math.stackexchange.com/questions/463791/prove-that-atd-ctb-i?rq=1 – user284331 Nov 30 '17 at 03:56
Use $(AB)^t = B^tA^t$ and $(A+B)^t=A^t + B^t$ with $I^t=I$ and the result follows by taking transposes on both sides of the expression. – CyclotomicField Nov 30 '17 at 04:14

Exercise of symmetric matrices

0 Answers0