Is it true that $ Av=\lambda v \implies v^{*T}Av = \lambda?$

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Is this generally true? I worked it out on paper and it seems to be the case but I'm not sure if I did everything correctly?

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Most of this was addressed in the comments, but here's the answer for the sake of having an answer:

I use $v^*$ instead of $v^{*T}$. Note that if $Av = \lambda v$ and $\|v\| = 1$, we have $$ v^*Av = v^*(\lambda v) = \lambda(v^*v) = \lambda \|v\|^2 $$ so, if $v$ is a unit eigenvector, $v^*Av = \lambda$ holds.