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$A$ is a matrix in $\mathbb{R}^{m \times n}$ and $x \in \mathbb{R}^n$.

I want to prove that: $$\langle Ax, x \rangle = \langle x , A^T x \rangle $$

Having $A^T$ as the transpose of the matrix $A$.

Just realized I use this property so much, yet I don't known how to prove it.

Bernard
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Figurinha
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    Since $Ax \in \mathbb{R}^m$, that statement only really makes sense if $m=n$. What's true more generally: if $x \in \mathbb{R}^n$ and $y \in \mathbb{R}^m$, then $\langle Ax, y \rangle = \langle x, A^T y \rangle$. – Daniel Schepler Aug 26 '20 at 21:01
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    Also: https://math.stackexchange.com/q/2411517/42969, https://math.stackexchange.com/q/353147/42969 – Martin R Aug 26 '20 at 21:05

1 Answers1

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We have that by definition

$$\langle Ax, x \rangle = (x^TA^T)x=x^T(A^Tx)=\langle x , A^T x \rangle$$

with $A\in \mathbb{R}^{n \times n}$, otherwise

$$\langle Ax, y \rangle = (x^TA^T)y=x^T(A^Ty)=\langle x , A^T y \rangle$$

with $y \in \mathbb{R}^m$.

user
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