How do I show ... if C and Q are nxn orthogonal matrices then so are CQ and (C^T)QC

Find the closest line to the points (-1, 0); (0, 1); (1, 2); (2, 4).

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- Nov 2nd 2009, 08:55 AMNoxideOrthogonality
How do I show ... if C and Q are nxn orthogonal matrices then so are CQ and (C^T)QC

Find the closest line to the points (-1, 0); (0, 1); (1, 2); (2, 4). - Nov 2nd 2009, 01:48 PMKrizalid
i'll do the first one since you can do the second in the same fashion.

since C and Q were given as orthogonal matrices, then $\displaystyle CC^t=C^tC=I$ and $\displaystyle QQ^t=Q^tQ=I,$ so $\displaystyle CQ(CQ)^t=CQQ^tC^t=I$ and $\displaystyle (CQ)^tCQ=Q^tC^tCQ=I$ so $\displaystyle CQ$ is orthogonal too. - Nov 2nd 2009, 03:41 PMmath2009
**Second problem :**

It's curve-fit. To suppose line equation : $\displaystyle y=ax+b,\vec{v}=\begin{bmatrix}a \\ b \end{bmatrix}, \begin{bmatrix}ax_1+b \\ax_2+b\\ax_3+b\\ax_4+b \end{bmatrix}=\begin{bmatrix}x_1&1 \\x_2&1\\x_3&1\\x_4&1 \end{bmatrix}\begin{bmatrix}a\\b \end{bmatrix}=A\vec{v}=\begin{bmatrix}0 \\1\\2\\4 \end{bmatrix}=\vec{y}$

Apply formulation $\displaystyle \vec{v}^*=(A^TA)^{-1}A^T\vec{y} =\frac{1}{10}\begin{bmatrix} 13 \\ 11 \end{bmatrix}$

$\displaystyle 10y=13x+11$