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I've read that the results from multiple regression and canonical correlation analysis are the same, aside from scaling. No one online has shown how to prove this by hand, since most examples for canonical correlation and the related code require 2+ dependent variables, though it is possible using certain commands in software like R (namely the cancor() function in base R). Although cancor() only provides the canonical correlation and the final x and y coefficients, not the eigenvalues and eigenvectors used to construct them. I've asked several others for their assistance, but they were also confused as to how to find the correct eigenvalue($\lambda$) and eigenvectors (u).

I've listed all of the relevant equations as well as a miniature data set if you'd also like to try it out. If you have any suggestions or can see where I made an error, please let me know. I figure you all have worked with canonical correlation before and have an idea of how to translate it to multiple regression. Anyway, here's what I have so far:

Data: \begin{array}{|c|c|c|} \hline y& x1 &x2 \\ \hline 5 & 4& 3\\ \hline 6& 2&5 \\ \hline 3& 7& 6\\ \hline 4& 8& 4\\ \hline 9& 7& 7\\ \hline \end{array}

$M=R_{yy}^{-1}R_{xy}'R_{xx}^{-1}R_{xy}$

$M= \begin{bmatrix}1\end{bmatrix}\begin{bmatrix}-0.138 & 0.481\end{bmatrix}\begin{bmatrix}1.110 & -0.350\\-0.350 & 1.110\end{bmatrix}\begin{bmatrix}-0.138\\0.481\end{bmatrix} = 0.324$

$|M-\lambda I|= |0.324 - \lambda|=0,\\\lambda=0.324$

$(M-\lambda I)u = 0$, used to find eigenvectors u.

$\begin{bmatrix}0.324-0.324\end{bmatrix}\begin{bmatrix}u\end{bmatrix} = \begin{bmatrix}0\end{bmatrix}$

$0u=0$, indeterminate

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