What does it mean for a matrix to be orthogonally diagonalizable?
I'm a little confused as to when a matrix is orthogonally diagonalizable.
I understand that if symmetric, it's always orthogonally diagonalizable, but in what other cases can you orthogonally diagonalize a matrix?
$\endgroup$ 34 Answers
$\begingroup$I assume that by $A$ being orthogonally diagonalizable, you mean that there's an orthogonal matrix U and a diagonal matrix $D$ such that
$$A = UDU^{-1} = UDU^T.$$
A must then be symmetric, since (note that since $D$ is diagonal, $D^T = D$!)$$A^T = \left(UDU^T\right)^T = \left(DU^T\right)^TU^T = UD^TU^T = UDU^T = A \text{.}$$
$\endgroup$ 2 $\begingroup$A square matrix is said to be orthogonally diagonalizable if there exist an orhtogonal matrix $P$ such that $P^{-1}AP$ is a diagonal matrix.
A square matrix $A$ is orthogonally diagonalizable $\Leftrightarrow$ $A$ is symmetric.
$\endgroup$ $\begingroup$- For a complex inner product space, a matrix $A$ is orthogonally diagonalizable iff $A A^* = A^* A$.
- For a real inner product space, a matrix $A$ is orthogonally diagonalizable iff $A^T = A$.
Notice that the condition in (2) is more strict than (1) in that (2) $\implies$ (1).
Orthogonal diagonalizability of matrix $A \in \mathbb{F}^{n \times n}$ means there exists an orthonormal basis for $\mathbb{F}^n$ consisting of eigenvectors of $A$.
$\endgroup$ $\begingroup$From here:
An $n \times n$ matrix is called orthogonally diagonalizable if there is an orthogonalmatrix $U$ and a diagonal matrix $D$ for which $A = UDU^{-1} = UDU^\top.$Thus, an orthogonally diagonalizable matrix is a special kind of diagonalizable matrix: not only can we factor $A=PDP^{-1},$ but we can find an matrix that $U=P$ that works. In that case case, the columns of $U$ form an basis for $\mathbb R^n.$
Hence, it adds the word "orthogonal" to $P:$ the eigen decomposition of a square matrix $A$ allows $A=PDP^{-1},$ with $P$ containing eigenvectors and $D$ eigenvalues - a square matrix is similar to a diagonal matrix. But what does 'orthogonal' add to $P$?
Having been shown on other answers that $A$ must be symmetric, I want to add the following intuitive way of thinking of it:
A square matrix $A$ is orthogonally diagonalizable if its eigenvectors are orthogonal *which is the case for any symmetrical matrix). The eigenvalues dilate the space subtended along an orthogonal grid.
$\endgroup$
