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I begin with doing \begin{eqnarray*} \left \lVert Ax-y\right\rVert^{2} &=& (Ax-y)^{T}(Ax-y) \end{eqnarray*} \begin{eqnarray*} \left \lVert Ax-y\right\rVert^{2} &=& (x^{T}A^{T}-y^{T})(Ax-y) \end{eqnarray*}

\begin{eqnarray*} \left \lVert Ax-y\right\rVert^{2} &=& x^{T} A^{T} A x - y^{T} A x -x^{T} A^{T} y+ y^{T} y \end{eqnarray*}

So my doubt is how to show \begin{eqnarray*} \ y^{T} A x=x^{T} A^{T} y \end{eqnarray*}

because if it is true my proof will be completed.

Dhruv Joshi
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  • yes that is a well known fact about transpose. you can prove it by the definition of transpose and matrix mulitplication. – qwr Dec 14 '19 at 07:02
  • are you talking about (AB)'=B'A' – Dhruv Joshi Dec 14 '19 at 07:04
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    $y^T Ax$ is a 1x1 matrix, or simply a number, so taking the transpose will give you the same thing. But the transpose is precisely $x^TA^Ty$. Hence, they are equal – peek-a-boo Dec 14 '19 at 07:05
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    more generally https://math.stackexchange.com/questions/1440305/how-to-prove-abt-bt-at – qwr Dec 14 '19 at 07:06

1 Answers1

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$y^t Ax$ is just the dot product between $Ax$ and $y$, regarded as column vectors. Also, $x^tA^t y = (Ax)^t y$ is the dot product between $y$ and $Ax$.

azif00
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