Is every symmetric matrix diagonalizable?

I know that

Hermitian matrices are always diagonalizable and real symmetric matrices are real Hermitian matrices and therefore diagonalizable.

But, it is always not the case that a symmetric matrix is a Hermitian matrix.

So my question is I think every real symmetric matrix is diagonalizable, but is it true for every symmetric matrix?

Also,

$1$ and $-1$ are only possible eigenvalues for real orthogonal matrix?

• Interesting relevant fact: as a consequence of Takagi's factorization, every complex symmetric matrix $M$ can be factored in the form $M = AA^T$. – Omnomnomnom Dec 18 '14 at 3:35
• – user1551 Dec 18 '14 at 6:33

The matrix $A = \begin{bmatrix}i&1\\1&-i\end{bmatrix}$ is (complex) symmetric but has Jordan form $A = VJV^{-1}$ where $J = \begin{bmatrix}0&1\\0&0\end{bmatrix}$ and $V = \begin{bmatrix}i&1\\1&0\end{bmatrix}$. So, not every (complex) symmetric matrix is diagonalizable.
The rotation matrix $R = \begin{bmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}$ is real orthogonal and has eigenvalues $\cos\theta \pm i\sin\theta$ which are not $\pm 1$ if $\theta$ isn't a multiple of $\pi$. So, $\pm 1$ are not the only possible eigenvalues for a real orthogonal matrix. However, you can say that the eigenvalues will all lie on the unit circle and other than $\pm 1$, they will come in complex conjugate pairs.