Csdrhrt

My textbook says the following:

Given a vector $mathrm{mathbf{x}}$ of random variables $x_i$ for $i = 1, dots, N,$ with mean $bar{mathrm{mathbf{x}}} = E[mathrm{mathbf{x}}]$, where $E[cdot]$ represents the expected, and $Delta mathrm{mathbf{x}} = mathrm{mathbf{x}} - bar{mathrm{mathbf{x}}}$, the covariance matrix $Sigma$ is an $N times N$ matrix given by

$$Sigma = E[Delta mathrm{mathbf{x}} Delta mathrm{mathbf{x}}^T]$$

so that $Sigma_{i j} = E[ Delta x_i Delta x_j]$. The diagonal entries of the matrix $Sigma$ are the variances of the individual variables $x_i$, whereas the off-diagonal entries are the cross-covariance values.

The variables $x_i$ are said to conform to a joint Gaussian distribution, if the probability distribution of $mathrm{mathbf{x}}$ is of the form

$$P(bar{mathrm{mathbf{x}}} + Delta mathrm{mathbf{x}}) = (2 pi) ^{-N/2} det(Sigma^{-1})^{1/2} exp(-(Delta mathrm{mathbf{x}})^T Sigma^{-1} (Delta mathrm{mathbf{x}})/2) tag{A2.1}$$

for some positive-semidefinite matrix $Sigma^{-1}$.

$vdots$

Change of coordinates. Since $Sigma$ is symmetric and positive-definite, it may be written as $Sigma = U^TDU$, where $U$ is an orthogonal matrix and $D = (sigma_1^2, sigma_2^2, dots, sigma_N^2)$ is diagonal. Writing $mathrm{mathbf{x}}' = U mathrm{mathbf{x}}$ and $bar{mathrm{mathbf{x}}}' = U bar{mathrm{mathbf{x}}}$, and substituting in (A2.1), leads to

$$ begin{align*}exp(-(mathrm{mathbf{x}} - bar{mathrm{mathbf{x}}})^T Sigma^{-1} (mathrm{mathbf{x}} - bar{mathrm{mathbf{x}}})/2) &= exp(-(mathrm{mathbf{x}}' - bar{mathrm{mathbf{x}}}')^T U Sigma^{-1} U^T (mathrm{mathbf{x}}' - bar{mathrm{mathbf{x}}}')/2) \ &= exp(-(mathrm{mathbf{x}}' - bar{mathrm{mathbf{x}}}')^T D^{-1} (mathrm{mathbf{x}}' - bar{mathrm{mathbf{x}}}')/2) end{align*}$$

Thus, the orthogonal change of coordinates from $mathrm{mathbf{x}}$ to $mathrm{mathbf{x}}' = U mathrm{mathbf{x}}$ transforms a general Gaussian PDF into one with diagonal covariance matrix. A further scaling by $sigma_i$ in each coordinate direction may be applied to transform it to an isotropic Gaussian distribution. Equivalently stated, a change of coordinates may be applied to transform Mahalanobis distance to ordinary Euclidean distance.

I don't understand how the author derived these expressions:

$$ begin{align*}exp(-(mathrm{mathbf{x}} - bar{mathrm{mathbf{x}}})^T Sigma^{-1} (mathrm{mathbf{x}} - bar{mathrm{mathbf{x}}})/2) &= exp(-(mathrm{mathbf{x}}' - bar{mathrm{mathbf{x}}}')^T U Sigma^{-1} U^T (mathrm{mathbf{x}}' - bar{mathrm{mathbf{x}}}')/2) \ &= exp(-(mathrm{mathbf{x}}' - bar{mathrm{mathbf{x}}}')^T D^{-1} (mathrm{mathbf{x}}' - bar{mathrm{mathbf{x}}}')/2) end{align*}$$

My attempt was as follows:

$$ begin{align*}exp(-(mathrm{mathbf{x}} - bar{mathrm{mathbf{x}}})^T Sigma^{-1} (mathrm{mathbf{x}} - bar{mathrm{mathbf{x}}})/2) &= exp(-(U^{-1}mathrm{mathbf{x}}' - U^{-1} bar{mathrm{mathbf{x}}}')^T Sigma^{-1}(U^{-1} mathrm{mathbf{x}} - U^{-1} bar{mathrm{mathbf{x}}})/2 ) \ &= exp(-(U^{-1}(mathrm{mathbf{x}}' - bar{mathrm{mathbf{x}}}'))^T Sigma^{-1} (U^{-1}(mathrm{mathbf{x}}' - bar{mathrm{mathbf{x}}}'))/2) \ &= exp(-((mathrm{mathbf{x}}' - bar{mathrm{mathbf{x}}}')^T U) Sigma^{-1} dfrac{U^T}{2} (mathrm{mathbf{x}}' - bar{mathrm{mathbf{x}}}')/2) tag{*} end{align*}$$

(*) Since $(AB)^T = B^T A^T$ and $U^T = U^{-1}$ ($U$ is orthogonal).

As you can see, I can't figure out how to derive the expressions that the author outlines. In fact, based on my work, as shown above, I can't see how such a derivation is possible?

I would greatly appreciate it if people could please take the time to demonstrate this.

You are only missing the implication $Sigma = U^{T}DU Rightarrow Sigma^{-1} = U^{T}D^{-1}U$. Now, by definition we can write $Sigma = U^{T}DU$. For invertible matrices $A,B$ it holds that $(AB)^{-1} = B^{-1}A^{-1}$. Therefore
$$
Sigma^{-1} = (U^{T}DU)^{-1} = U^{-1}(U^{T}D)^{-1} = U^{-1}D^{-1}(U^{T})^{-1} = U^TD^{-1}U
$$
where we used the fact that $U^{-1} = U^T$.