### Is an orthogonal diagonalization of a real symmetric matrix unique?

Posted:

**Mon Jul 24, 2017 12:27 am UTC**If we have a real symmetric matrix M which is diagonalized into QLQ^T where Q is orthogonal and L is diagonal (this is possible by spectral theorem), then is Q and L necessarily unique? Is there another orthogonal diagonalization using some Q' != Q or L' != L?

I ask this because it seems many explanations of how to perform PCA with SVD rely on this as an unstated fact. I can't tell if this is because i'm misreading or because I forgot some important theorem from my linear algebra class

Here's my attempt at proof:

suppose it is possible:

M = Q'L'Q'^T

= (QQ^T) Q'L'Q'^T (QQ^T)

= Q (Q^T Q'L'Q'^T Q) Q^T

now it suffices to prove L != (Q^T Q'L'Q'^T Q) . I have the fact that L' and L is diagonal, so if i can show that Q^T Q'L'Q'^T Q is not diagonal, then I would be done, but i'm not sure how to

I ask this because it seems many explanations of how to perform PCA with SVD rely on this as an unstated fact. I can't tell if this is because i'm misreading or because I forgot some important theorem from my linear algebra class

Here's my attempt at proof:

suppose it is possible:

M = Q'L'Q'^T

= (QQ^T) Q'L'Q'^T (QQ^T)

= Q (Q^T Q'L'Q'^T Q) Q^T

now it suffices to prove L != (Q^T Q'L'Q'^T Q) . I have the fact that L' and L is diagonal, so if i can show that Q^T Q'L'Q'^T Q is not diagonal, then I would be done, but i'm not sure how to