$T$-invariance of $U$ is equivalent to $T^{*}$-invariance of $U^{\perp}$

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Is the Following argument correct?

Suppose $T\in\mathcal{L}(V)$ and $U$ is a subspace of $V$. Prove that $U$ is invariant under $T$ if and only if $U^{\perp}$ is invariant under $T^*$.

Proof. Given that $U$ is invariant under $T$ then $\langle w,Tu\rangle = 0,\forall (u,w)\in U\times U^{\perp}$ equivalently $\langle w,(T^*)^*u\rangle = 0 ,\forall (u,w)\in U\times U^{\perp}$ this when taken together with the definition of the adjoint of $T$ is equivalent to $\langle T^*w,u\rangle = 0,\forall (u,w)\in U\times U^{\perp}$ thus $U^{\perp}$ is invariant under $T^{*}$.

$\blacksquare$