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$A$ mapping $\displaystyle *$ from $\displaystyle F_{n}$ to $\displaystyle F_{n}$ is called adjoint if

$\displaystyle \begin{array}{{>{\displaystyle}l}} ( A+B)^{*} =A^{*} +B^{*}\\ ( AB)^{*} =B^{*} A^{*}\\ \left( A^{*}\right)^{*} =A;\\ \text{for all } A,B\in F_{n} \end{array}$

if $\displaystyle \lambda $ is any scalar matrix in $\displaystyle F_{n}$ then prove that $\displaystyle \lambda ^{*}$ must also be a scalar matrix.

For Hermitian adjoint, I know it is true. But for general adjoint, how can I prove this?

Kenta S
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Tanya Rastogi
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1 Answers1

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Recall that a matrix $A \in M_n(F)$ is a scalar matrix if and only if $AB = BA$ for all $B \in M_n(F)$.

Consider the scalar matrix $\lambda I$ for some $\lambda \in F$. For any matrix $B \in M_n(F)$ we have $$(\lambda I) B = B(\lambda I).$$ Applying $*$ on this equality gives $$B^*(\lambda I)^*= (\lambda I)^*B^*, \quad \forall B \in M_n(F).$$ Now, $*$ is bijective since $(B^{*})^* = B$ so $$B(\lambda I)^*= (\lambda I)^*B, \quad \forall B \in M_n(F)$$ which implies that $(\lambda I)^*$ is again a scalar matrix.

mechanodroid
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