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I got this question in one of my classes and i'm really lost. So i'm given to matrices representing the mean and the covariance. The mean is m=\begin{bmatrix}10\\0\end{bmatrix} and the covariance is cov=\begin{bmatrix}16&-12\\-12&34\end{bmatrix}. Now after I do some computation, i get the eigenvector matrix eig=\begin{bmatrix}2&-1\\1&2\end{bmatrix}

Now i'm to compute the transformation matrix trs=eigt *cov *eig. which gives \begin{bmatrix}50&0\\0&200\end{bmatrix}. Now i'm asked to explain what happened to the covariance matrix after it was transformed and why.

Thanks in advance.!

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  • $\begingroup$ The trace of a matrix is invariant under similarity transformations. The trace of the original matrix is equal to $50$, but the trace of the transformed matrix is $250$, so you’ve made an error somewhere along the way. Since your formula has a matrix transpose instead of an inverse, it looks like the eigenvector matrix is meant to be orthogonal. $\endgroup$ – amd Dec 14 '17 at 8:18
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I am getting different Eigenvalues.

But intuitively what does it mean when you diagonalize the covariance matrix?

You have done what is called "principle component analysis."

You have derived independent factors that drive the volatility in your dependent variable. Each factor drives variance proportional to the size of the associated eigenvalue.

It is not always intuitive what your factors, in fact, represent in "real world" terms.

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  • $\begingroup$ Ill review my computations. Thanks a lot for the answer. Question is a lot more clearier now! $\endgroup$ – Jacques Dec 14 '17 at 3:14

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