Re: Linear Algebra, Please help.

takaoueda_at_juno.com
Date: 12/15/04 

Date: 15 Dec 2004 08:23:56 -0800

Dave wrote:
> Question:
> Given a symmetric nxn matrix A, show there is
> an orthogonal mx U of determinant 1 with
> U^(-1)AU= D a diagonal matrix.

> This is what I have so far:

> Suppose X1.....Xn is a basis of eignvalues
> of the nxn matrix A of eigen values
> lambda....(lambda)n.

>Set C= (X1.....Xn) (c is invertible)

>Claim that if D is the diagonal (lambda)...(lambda)n
> then CD=AC

The reason is that the jth column of CD is
(lambda)jX_j and the jth column of AC is AX_j.
Therefore,

>D=C^(-1)AC

>if {X_1,...,X_n} is orthonormal, C is orthogonal.

Therefore, if all (lamda)i are different, let
Y_i = X_i/|X_i|. Let U = (Y_1,..., Y_n), and
if determinant(U) = -1, change Y_1 to -Y_1.
Then, by Santos's help, U is a desired matrix.

Here is a solution for the general case. I say
"direct product" of two matrices for the matrix
consisting of these matrices as diagonal blocks
and zeroes for all other elements.

Let X_1 be an eigenvector of unit length for an
eigenvalue lamda_1 of A. Create an orthonormal
basis {X_1, X_2,...., X_n} for R^n using the
Gram-Schmidt process, and let C = (X_1,X_2,....,X_n},
which is an orthogonal matrix. Let U_1 = C. Then
B = (U_1)^TAU_1 is symmetric, B_{11} = lamda_1,
B_{i1} = 0 for i not 1, and B_{1j} = 0 for j not 1.
That is, B is the direct product of the 1\times 1
matrix Lamda_1I and the (n-1)\times(n-1) matrx B'.

Repeat the process for B', then we obtain an
(n-1)\times(n-1) orthogonal matrix C' and the
direct product of the 1\times 1 matrix lamda_2I
and the (n-2)\times(n-2) matrx B". Let U_2 be
the direct product of the 1\times 1 identity
matrx and C'.

In this way we obtain the orthogonal matrix
U_{1},...,U_{n-1}. A desired orthogonal matrix
U is U_1...U_{n-1} or (-U_1)....U_{n-1}. Note
that the eigenvalue lamda_2 of B' is an eigenvalue
of A. Therefore we also proved we also proved
any n\times n symmetric real matrix has linearly
independent n eigenvectors.

Takao