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Consider two vectors $x\in\mathbb{R}^n$ and $y\in\mathbb{R}^n$. Obviously, $x^\top y = 0$ implies that $x$ and $y$ are orthogonal. Now, let $A\in\mathbb{R}^{n \times n}$ be a matrix. What is the meaning of $$x^\top A y = 0$$? Does this imply orthogonality of $x$ and $y$ with respect to something?

Arthur
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    Note: if $A$ is the zero matrix, $x^TAy=0$ for all $x,y\in\mathbb R^n$ – J. W. Tanner Feb 03 '20 at 02:39
  • Obviously, this means that $x$ is orthogonal to the image of $y$ under $A$. I was wondering if there are more meanings behind it. – Arthur Feb 03 '20 at 02:39
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    Yes, as you said in the previous comment, $x^tAy = 0$ means that $x$ is orthogonal to some linear combination of the columns of $A$, but also we that $0 = x^tAy = (A^tx)^ty$, that is, $y$ is orthogonal to some combination of the rows of $A$. – azif00 Feb 03 '20 at 02:42
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    If $x^TAy = 0$, then we can say that $x,y$ are orthogonal relative to the bilinear form induced by $A$. – Ben Grossmann Feb 03 '20 at 10:03
  • Possible duplicate of https://math.stackexchange.com/q/1364346/293177? – Al.G. Jun 22 '21 at 11:46

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