Prove that that every symmetric real matrix is diagonalizable?

[complexhuman]

How can I prove that that every symmetric real matrix is diagonalizable?

Thanks in advance

 

[HallsofIvy]

That's a MAJOR theorem so at best I can just outline the proof here.

The more general theorem is that every self-adjoint linear transformation is diagonalizable- and symmetric (real) matrices, thought of as linear transformations on Rⁿ, are self- adjoint. A self-adjoint linear transformation is one such that
<Ax,y>= <x,Ay> where <x, y> is the innerproduct on the vector space.

First show that the eigenvalues of a self-adjoint linear transformation must be real:
If λ is an eigenvalue of A, take x to be a unit length eigenvector. Then λ= λ<x, x>= <λx,x>= <x, λx>= complexconjugate(λ)<x,x>= complexconjugate(λ). Since λ equals its complex conjugate, it is real.

Second, show that eigenvectors corresponding to distinct eigenvalues are orthogonal: If Ax= λx and Ay= μ y then λ<x, y>= <λx,y>= <Ax, y>= <x, Ay>= <x,μy>= μ<x,y> (since μ is real). Then (λ- μ)<x,y>= 0. Since λ and μ are distinct, we must have <x,y>= 0.

So: given any self-adoint linear transformation (or symmetric matrix) we can write the vector space as a direct sum of orthogonal subspaces, every vector of which is an eigenvector corresponding to the same eigenvalue. Choose any orthonormal basis for each subspace and show that the matrix for A, restricted to that subspace, is simply λI- a diagonal matrix with only λ on the diagonal.

Finally, the union of the orthonormal bases for each subspace will be an orthonormal basis for the entire space and the matrix corresponding to A in that basis is diagonal.

 

[tacman]

!

To prove that every symmetric real matrix is diagonalizable, we first need to understand the definitions of symmetric matrix and diagonalizable matrix.

A symmetric matrix is a square matrix where the elements are equal to their corresponding elements across the main diagonal. In other words, if A is a symmetric matrix of size n x n, then A[i,j] = A[j,i] for all i,j = 1,2,...,n.

A diagonalizable matrix is a square matrix that can be transformed into a diagonal matrix by a similarity transformation. In other words, if A is a diagonalizable matrix of size n x n, then there exists an invertible matrix P such that P^-1 * A * P = D, where D is a diagonal matrix.

Now, to prove that every symmetric real matrix is diagonalizable, we can use the Spectral Theorem for symmetric matrices. This theorem states that every real symmetric matrix can be diagonalized by an orthogonal matrix.

An orthogonal matrix is a square matrix whose columns and rows are orthogonal unit vectors. This means that the dot product of any two columns (or rows) of an orthogonal matrix is equal to 0, and the length of each column (or row) is equal to 1.

So, if we take a symmetric matrix A, we can find an orthogonal matrix P such that P^-1 * A * P = D, where D is a diagonal matrix. This is because the columns of P will be the eigenvectors of A, and since A is symmetric, its eigenvectors are orthogonal.

Therefore, every symmetric real matrix can be diagonalized by an orthogonal matrix, which means it is diagonalizable. This completes the proof.