0

Let $B \in M_{n×n}(\Bbb F)$. Define the functions $L_B$ and $R_B$ by: $L_{B}(A) = BA$ and $R_{B}(A) = AB$. Prove that $\det(L_B) = \big(\det(B)\big)^n$, $\det(R_B) = \big(\det(B)\big)^n$.

$L_B$ and $R_B$ are linear transformations, thus can be represented by matrices. So the determinant of those functions exists.

gebruiker
  • 6,154
Yonatan Izutskiver
  • 440
  • 1
  • 6
  • 13
  • Is $F^{n\times n}$ the set of the $n\times n$ matrices ? Also when you say "$\det(L_B)=$..." this doesn't make sense to me because $L_B$ is only a function and only $L_B(X)$ is a matrix . –  Jan 09 '16 at 14:28
  • yes, $F^{n*n}$ is the set of the $n×n$ matrices. – Yonatan Izutskiver Jan 09 '16 at 14:31
  • 6
    This question was answered here: http://math.stackexchange.com/questions/704928/trace-and-determinant-of-composition-of-a-left-multiplication-and-a-right-multip – J.Cork Jan 09 '16 at 14:52
  • The maps $L_B,R_B$ are linear on vector space $\mathbb{F}^{n\times n}$, which accounts for the strange notion that $\det L_B = \det R_B = (\det B)^n$. – hardmath Jan 09 '16 at 17:38

0 Answers0