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If A is an $n\times n$ matrix and u and v are $n\times 1$ matrices, then prove that $\boldsymbol{A u} \cdot \mathbf{v}=\mathbf{u} \cdot \boldsymbol{A}^{T} \mathbf{v}$
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Suppose that $$A=\left[\begin{array}{rrr} 1 & -2 & 3 \\ 2 & 4 & 1 \\ -1 & 0 & 1 \end{array}\right], \quad \mathbf{u}=\left[\begin{array}{r} -1 \\ 2 \\ 4 \end{array}\right], \quad \mathbf{v}=\left[\begin{array}{r} -2 \\ 0 \\ 5 \end{array}\right]$$ Then $$\begin{array}{r} A \mathbf{u}=\left[\begin{array}{rrr} 1 & -2 & 3 \\ 2 & 4 & 1 \\ -1 & 0 & 1 \end{array}\right]\left[\begin{array}{r} -1 \\ 2 \\ 4 \end{array}\right]=\left[\begin{array}{r} 7 \\ 10 \\ 5 \end{array}\right] \\ A^{T} \mathbf{v}=\left[\begin{array}{rrr} 1 & 2 & -1 \\ -2 & 4 & 0 \\ 3 & 1 & 1 \end{array}\right]\left[\begin{array}{r} -2 \\ 0 \\ 5 \end{array}\right]=\left[\begin{array}{r} -7 \\ 4 \\ -1 \end{array}\right] \end{array}$$ from which we obtain \begin{aligned} A \mathbf{u} \cdot \mathbf{v} &=7(-2)+10(0)+5(5)=11 \\ \mathbf{u} \cdot A^{T} \mathbf{v} &=(-1)(-7)+2(4)+4(-1)=11 \end{aligned}
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