Show that $(Au,Bv)=(u,A^tBv)$

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Let $ A, B $ be matrices of order $ n $, and $ \vec{u}, \vec{v} $ vectors from euclidean space $ \mathbb{R}^n $, then $ (Au,Bv) = (u,A^tBv) $

pd. $(\cdot ,\cdot)$ is my notation for inner product, and $ A^t $ is the transpose of $A$

I don't know how to demonstrate that, and I'm not sure if it's true.

There's my demo, I can't complete it. Maybe I'm not using the right definitions,

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