In Linear Algebra Why is following correct:
Given a symmetric matrix $A$ on the field of the real numbers, why is that true that there exists an unitary matrix $P$ such that $PAP^t$ is a diagonal matrix?
I know that from the Spectral Theorem there exists a unitary matrix $P$ such that $PAP^{-1}$ is diagonal. However how can I conclude that $PAP^t$ is diagonal as well?
Thank you
Best Answer
It seems to me that this problem comes down to getting the definitions straight. So here they are:
Symmetric matrix: a matrix $A$ is called symmetric when $A=A^T$. If $A$ is a matrix of real entries, then we can say that $A$ is an instance of a "Hermitian" matrix
Hermitian matirix a matrix $A$ is Hermitian when $A=A^*$, that is, $A$ is equal to its own conjugate transpose.
Depending on the context, real symmetric matrices and complex Hermitian matrices might be referred to as self-adjoint.
Orthogonal matrix: a matrix $U$ is called orthogonal when $UU^T=1$. If $U$ is a matrix of real entries, then we can say that $U$ is an instance of a "unitary" matrix
Unitary matrix a matrix $U$ is called unitary when $UU^*=1$. That is, the inverse of $U$ is its conjugate transpose.
The spectral theorem could then be understood as follows:
Suppose that $A$ is any normal matrix (that is, $AA^*=A^*A$). We can then find a matrix $P$ such that $PAP^*$ is diagonal.
If $A$ happens to be a real and symmetric matrix, then we can find a $P$ that is not only unitary, but also itself real. That is, there is a real matrix $P$ such that $PAP^*$ is diagonal. Since $P$ is real, this is the same as saying $PAP^T$ is diagonal.