91Ó°ÊÓ

Spectral Theorem:

A matrix A is orthogonally diagonalizable if and only if A is symmetric.

Consider a n x n symmetric matrix A.

Therefore, the eigen values of the matrix A are real and the matrix A is diagonalizable.

Thus, by spectral theorem, there exists an orthogonal basis of orthogonal eigenvectors.

Suppose v₁, v₂,....v_nbe the orthogonal vectors of the matrix A, then the matrix S whose columns are the vectors v₁,v₂,.....v_n, orthogonally diagonalize the matrix A.

Thus, S is an orthogonal matrix.

Since S is an orthogonal matrix, therefore S^-1= S^T.

Also, matrix S orthogonally diaonalize the matrix A, therefore

S^-1AS=D, where D is a diagonal matrix.

From S^-1 = S^T ;

S^-1AS = S^TAS

=D

Thus, S^TAS is a diagonal matrix that orthogonally diaonalize the matrix A.

Hence, the given statement is TRUE.