All symmetric matrices are diagonalizable.

If a matrix is orthogonally diagonalizable, then ‘A’ must be symmetric.

Definition: A$n\times n$ matrix A is called orthogonally diagonalizable if there is an orthogonal matrix$U$ and a diagonal matrix$D$ for which $A=UD{U}^{-1}$.

Thus, an orthogonally diagonalizable matrix is a special kind of diagonalizable matrix, not only can we factor $A=PD{P}^{-1}$, but we can find a matrix$U=P$ that works.

In this case, the columns of$U$ form a basis for Orthonormal to know which matrices are orthogonally diagonalizable.

Definition: A is called an if symmetric matrix $A=A^{-1}$

Example: If A is any matrix (square or not), then$A{A}^{-1}$ is square.$A{A}^{-1}$ Is also symmetric because ${\left(A^{T}A\right)}^T=A^T{A}^{TT}=A{A}^{-1}$.

Every symmetric matrix has an orthogonal basis consisting of eigenvectors. Therefore, a symmetric matrix is not only diagonalizable but also orthogonally diagonalizable.

Hence, the statement is TRUE.