Every diagonalizable matrix is orthogonally diagonalizable Hi so I came a across a statement in my book saying that a Matrix is symmetric iff it is orthogonally diagonalizable. Would that not mean every diagonalizable matrix is symmetric? Take a matrix that is diagonalizable, use Gram-Schimdt to make them orthogonal, normalize and now we can orthogonally diagaonalize it, hence it is symmetric. I must be wrong somewhere, so i would appreciate if you could point out my mistake in reasoning.
 A: The eigenspaces of a matrix (with distinct eigenvalues, say) need not be mutually-orthogonal, so generally there's no way to choose an orthonormal eigenbasis. (Gram-Schmidt doesn't preserve eigenspaces.)
A: A matrix $A$ is called symmetric if $A=A^T$.
A matrix $A$ is called diagonalizable if there exists an invertible matrix $B$ and a diagonal matrix $D$ such that $BAB^{-1}=D$.
A matrix $A$ is called orthogonally diagonalizable if there exists an orthogonal matrix $U$ (i.e., $UU^T=I$) and a diagonal matrix $D$ such that $UAU^{-1}=D$.
Note that if you "take Gram-Schmidt to make them orthogonal", you replace the original matrix with a different matrix, and the above definitions do not fit any more.
A: 
Would that not mean every diagonalizable matrix is symmetric? 

No, but it does meant that every orthogonally diagonalizable matrix is symmetric.  For example, the matrix
$$
\pmatrix{1&1\\0&0}
$$
is diagonalizable, but not symmetric.  As such, it is not orthogonally diagonalizable.

Take a matrix that is diagonalizable, use Gram-Schimdt to make them orthogonal, normalize and now we can orthogonally diagaonalize it, hence it is symmetric.

I have no idea what you're describing here, so it's hard to find any "mistake in your reasoning".
