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Suppose the $x$ -coordinates of the data $\left(x_{1}, y_{1}\right), \ldots,\left(x_{n}, y_{n}\right)$ are in mean deviation form, so that $\sum x_{i}=0 .$ Show that if $X$ is the design matrix for the least-squares line in this case, then $X^{T} X$ is a diagonal matrix.

$X^{T} X$ is a diagonal matrix.

Calculus 3

Chapter 6

Orthogonality and Least Square

Section 6

Applications to Linear Models

Vectors

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That's right, The medics of the coefficients Big Axe as the first column is all ones and the second column has the axes of the points X one Exxon. And we know that some of X, which is what I did Note X one plus plus Exxon and zero. So the point starting mean deviation form We want to show that Well, X transpose X is a dagger on metrics. So let me call the first column of X C one and the second column of Ecstasy too. Now X transpose x Well, that's see one transport see to transpose. But as Rose and then C wants you to thought, these are two by two matrix and well, we have C one transports you want you want transparency to and then see to transport C one and C to transport, see to All right, no c one Transparency One is just well, some of ones. And we have exactly in data points. So it's n them for C one trust. Beyonc? We see that that the some off ex multiply by one so there's just some of X and similarly c two transposed. The one is some of ex again and then see to transport see to it the sum of the squares. So some X squared. But now we're done because we know that some X zero. So we have an 00 sum of X squared, which means that X transpose x. He's a diamond.

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