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Given $e = Y - XB$, where $ e = \begin{bmatrix} e_1 \\ \vdots \\ e_n \\ \end{bmatrix} $, $ Y= \begin{bmatrix} y_1 \\ \vdots \\ y_n \\ \end{bmatrix} $, $ X= \begin{bmatrix} 1 & x_1 & x^2_1 \\ \vdots & \vdots & \vdots \\ 1 & x_n & x^2_n \\ \end{bmatrix} $, and $ B = \begin{bmatrix} a \\ b \\ c \\ \end{bmatrix} $. Define the scalar sum of the squares as $$ \begin{equation} SSQ = e^T e \end{equation} $$ Substitute $Y - XB$ for $e$. $$ \begin{equation} SSQ = (Y-XB)^T (Y-XB) \tag{2} \end{equation} $$

I have seen the expansion of $(2)$ written as $$ \begin{equation} SSQ = Y^TY - 2B^TX^TY + X^TB^TXB \tag{3} \end{equation} $$

I don't completely understand the expansion. If the expansion follows the FOIL method, I understand $Y^TY$ and $X^TB^TXB$.

I don't understand how the middle term is created.

I would have expected $-X^TB^TY-Y^TXB$.

I guess I don't understand why one can substitute $Y^T$ for $Y$, $X^T$ for $X$, and $B^T$ for $B$, which would give $(3)$.

Thanks for helping this novice amateur mathematician.

KeithSmith
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1 Answers1

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The transpose is contravariant, meaning $(AB)^T=B^TA^T$ is to be used. So, we have $$Y^TY - B^TX^TY - Y^TXB +B^TX^TXB\,.$$ Now, $Y^TXB$ is a $1\times 1$ matrix, nonetheless, is the scalar product of vectors $Y$ and $XB$, but then it is symmetric as matrix, so we have $$Y^TXB=(Y^TXB)^T=B^TX^TY\,.$$

Berci
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