That is Finally, remember the properties of the trace and get Application to Error Sum of Squares Recall that where is multivariate standard normal. The function is a quadratic form. In other words we might as well assume that the matrix Q is symmetric. In other words we have to diagonalize Q . To find the eigenvalues of Q we can solve The characteristic polynomial is whose two roots are 2 and 7.
T is an eigenvector as is any non-zero multiple of that vector.
To get a normalized eigenvector you divide through by the length of the vector, that is, by . The second eigenvector may be found similarly. Corresponding to the other two eigenvalues there are actually many possibilities. The third solution would then be perpendicular to this, making the first two entries equal. The key point in the , however, is that the distribution of the quadratic form Z T QZ depends only on the eigenvalues of Q and not on the eigenvectors.