What is the relationship between congruence transformation and diagonalization?

[venkat]

i was reading goldstein (oscillations)

in it, it is said that

Va = (lambda)*Ta

where V is the potential energy matrix, T is the kinetic eneregy matrix, lambda is an eigenvalue and a is the corresponding eigenvector(of displacement from equilibrium)

and it is said that the matrix of eigenvectors A diagonalises the matrix V through a congruence transformation A'VA = L
where L is the diagonal matrix with its diagonal elements as eigenvalues. and A' is the transpose of A. this equation is solved by taking
|V-LI| = 0

here I is the identity matrix and 0 is the zero matrix.
but this secular equation is allowed only when the matrix A diagonalises V through a similarity transformation, isn't it?

it is the equation
Va = (lambda)*a

which will yield
|V-LI| = 0 and

not the equation Va = (lambda)*Ta isn't it?

 

[HallsofIvy]

Yes, that's true. IF a matrix has a "complete set of eigenvectors" (there exist a basis for the vetor space consisting of eigenvectors of the matrix), THEN the matrix A, having those eigenvectors as columns, diagonalizes V: A^-1VA= L. Notice that I used A^-1 rather than A'. Of course, if we can construct an orthonomal basis of eigenvectors, then A will be an orthogonal matrix and A^-1= A'.

 

[venkat]

it's true that A diagonalises V .

but my question is:

since a'Ta = 1 , and not a'a = 1 ,( so that a is not orthogonal)
isn't A' not equal to (A inverse)??

...by the way, how did you write (A inverse )...the mathematical way?!

 

[venkat]

can somebody help me? i have no one to discuss things with as i am studying physics on my own...that's why i put up the question here