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As far as I understand, if I multiply some normally distributed data by a transformation matrix T , data will be have maximum spread in the direction of the Eigen-Vector of the maximum Eigen-Value of this transformation matrix. Also we can see that the covariance matrix of the data can be seen as C = TT' , where (') is the transpose as seen here : https://robotics.stackexchange.com/questions/2556/how-to-rotate-covariance and here : http://www.lucidarme.me/?p=946 Now the PCA tells us that the direction of the maximum spread of Data will turn out to be the direction of the eigen-vector of the maximum eigenvalue of the covariance matrix itself.

Does this mean that the eigen-vectors of the covariance matrix C are the same as the eigen-vectors of T ? Can we prove this in a general case where T = RS , where S is scale component and R is a rotation component ? Or what am I doing wrong here ? I have read somewhere that matrix A has the same eigen-vectors as A'A iff A'A = AA' , but that is not the case for a general T = RS.

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I have actually found the answer. The answer is that the direction of the maximum elongation/spread of data is NOT ALWAYS the same as the direction of the eigen vector that has the max eigen value. This is true only if the eigen vectors are orthogonal. So now everything is consistent. Try matrix A = [10 1 ; 0 11] in matlab syntax. If you multiply this matrix by a vector b = [0 1]' , you get an elongation greater than the maximum eigen value of this matrix.

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