Why linear transformation can make correlated random variables uncorrelated? A specific linear transformation can make a correlated covariance matrix uncorrelated. What is its physical meaning. For example weight and height are always correlated.What will it mean to make them uncorrelated, while we know physically it is not possible? But mathematically we can find a transformation which can make these uncorrelated.
 A: It means that we won't look at height and weight as our variables anymore. We're not saying that height and weight are decorrelated; we're saying that we are going to work with some new variables, linearly related to our old ones, that ARE decorrelated.
Suppose height is $H$ and weight is $W$ in some units. Together, these make my variable, $X$:
$$
X = \left[\begin{array}{c}
H \\
W
\end{array}
\right]
$$ 
Suppose I have a linear transformation, given by a matrix $A$.
$$
A = \left[\begin{array}{cc}
a_{11} & a_{12} \\
a_{21} & a_{22} \\
\end{array}
\right]
$$
And then I introduce new variables, $D,E$, given by appliing $A$ to $X$:
$$
\left[\begin{array}{c}
D \\
E
\end{array}
\right] = 
AX =
\left[\begin{array}{c}
a_{11}H + a_{12}W \\
a_{21}H + a_{22}W
\end{array}
\right]
$$
Now, $D$ and $E$ are my two variables. They are linear functions of height and weight, but they aren't height and weight; they might physically represent nothing. For example, ten times your weight in kilograms minus six times your height in inches does not have any easy physical meaning. However, you can possibly say that, as new variable, this value is uncorrelated with five times height + your weight divided by fourty. The linear transformation $A$, applied to the original data, can generate new variables that are uncorrelated.
A: We can always find an orthogonal matrix that transforms a covariance matrix $Q$ into a diagonal matrix whose (non-negative) entries consist of eigenvalues of $Q$. The eigenvalues correspond to the variance of the corresponding transformed variables. So, in some sense, the 'decorrelation transformation' is just a consequence of the fact that the covariance matrix is positive semi-definite. 
If the eigenvalues are small, it means the data mostly lies in a smaller dimensional subspace.
While the eigenvalues correspond to the transformed variables' variance, the eigenvectors contain the associated relationships between the original variables.
The transformed variables may represent some physically meaningful relationship, they may account for variations due to other unmeasured characteristics, or account for artifacts such as noise and non-linearities. It is impossible to characterize a variable with non-zero variance without considering the physical source of the data.
In the height and weight example, it is clear that while there may be a strong relationship, many other variables are relevant too. Hence if we compute the orthogonally diagonalized covariance for height and weight measurements, we would expect that one transformed variable has a dominant eigenvalue (variance) and the other much smaller. The dominant variable has no direct physical interpretation, but the corresponding eigenvector represents the relationship between  height and weight. The other variable 'accounts' (in some sense) for the differences from this relationship. If the other eigenvalue is much smaller than the dominant one, then the data lies mostly close to a line, which we might expect.
On the flip side, if one is trying to generate 'random' data with a specified covariance, it is much easier to generate uncorrelated data first and then transform it to obtain the desired correlation.
A: I have a pacific example, which is stronger than what you said.
Let x and y be two random variables varuable, with cov (x,y) =  a, and var(x)=var(y)=1
Now consider cov (x, y-bx)  = cov (x,y) - b var (x) = a - b 
Then if you let a equal to b, then it has y-bx has no correllation with x. The moral of the story 
You could have replaced x and y with two arbitary random variables, with different variances and covariances. You just need to choose the correct b.
I am sure, i can give you linear combination of x,y which is independent of any combination you give me. This shouldnt be too surprising, this is not saying y-bx and x are indepedent (unless x and y are jointly Gaussian). Covariance = 0 does not say as much about random variables as you think.
